Functional Analysis on Nam Le

The Invariant Subspace Problem

Thu, 28 May 2026 00:00:00 +0000

Few questions in functional analysis have attracted sustained attention across as many decades as this one. It sits at the confluence of operator theory, spectral theory, and complex analysis, and every partial result has opened new territory rather than narrowing the problem to a routine case.

Problem (Invariant Subspace Problem)

Does every bounded linear operator $T$ on an infinite-dimensional separable complex Hilbert space $\mathcal{H}$ have a non-trivial closed invariant subspace?

That is, does there always exist a closed subspace $\mathcal{M} \subsetneq \mathcal{H}$ with $\mathcal{M} \neq {0}$ such that $T\mathcal{M} \subseteq \mathcal{M}$?

The problem is rated medium importance on the Open Problem Garden. It is old enough to have accumulated a rich history of partial results, yet still open in the Hilbert space setting after more than seventy years.

Trivial Observations and Why They Run Out #

Two subspaces are always invariant: ${0}$ and $\mathcal{H}$ itself. These are the trivial invariant subspaces; the problem asks whether anything else must exist.

On finite-dimensional spaces the answer is immediate: every operator on $\mathbb{C}^n$ has an eigenvector (by the fundamental theorem of algebra applied to the characteristic polynomial), and the span of any eigenvector is a one-dimensional invariant subspace. This argument fails completely in infinite dimensions, where the spectrum can be continuous and eigenvectors need not exist.

On non-separable Hilbert spaces the problem is also trivial but for a different reason: for any non-zero vector $x \in \mathcal{H}$, the closed linear span $\overline{\operatorname{span}{T^n x : n \geq 0}}$ is a closed invariant subspace, and if $\mathcal{H}$ is non-separable it cannot equal all of $\mathcal{H}$. So the problem is genuinely about separable spaces.

Landscape of Known Results #

Positive Results: Classes with Invariant Subspaces #

Theorem (Aronszajn–Smith, 1954)

Every compact operator on a Banach space of dimension greater than one has a non-trivial closed invariant subspace.

The compact case was already known to von Neumann in the 1930s for Hilbert spaces, but was never published; Aronszajn and Smith gave the first published proof, extended to Banach spaces. The key idea is that a compact operator can be approximated by finite-rank operators, each of which has invariant subspaces, and a limiting argument produces an invariant subspace for the compact operator.

Theorem (Lomonosov, 1973)

If a bounded operator $T$ on a Banach space commutes with a non-zero compact operator, then $T$ has a non-trivial hyperinvariant subspace (a subspace invariant under every operator that commutes with $T$).

Lomonosov’s proof is strikingly short, less than a page, and uses the Schauder fixed-point theorem in an unexpected way. It subsumes both the compact case (an operator commutes with itself) and the polynomially compact case (an operator commutes with $p(T)$, which is compact if $p(T)$ is). For several years it seemed that Lomonosov’s theorem might resolve the problem entirely, until Hadwin, Nordgren, Radjavi, and Rosenthal (1980) exhibited an operator that does not commute with any non-zero compact operator yet still has invariant subspaces.

Theorem (Brown, 1987)

Every subnormal operator on a Hilbert space has a non-trivial invariant subspace.

An operator $T$ is subnormal if it is the restriction of a normal operator on a larger Hilbert space. Normal operators are handled by the spectral theorem, which produces a rich lattice of invariant subspaces; subnormal operators inherit invariant subspaces by restriction. Brown’s proof uses techniques from rational approximation theory (the solution of the Halmos problem on subnormal operators).

Beyond these landmark theorems, invariant subspaces are also known for: hyponormal operators with some additional conditions, operators whose spectrum has interior points, operators satisfying growth conditions on the resolvent, and polynomially bounded operators with spectrum containing the unit circle under further constraints (Liu, 2017; Réjasse, 2023).

Beurling’s Theorem: A Complete Classification #

Theorem (Beurling, 1949)

The closed invariant subspaces of the unilateral shift $S : H^2(\mathbb{D}) \to H^2(\mathbb{D})$, $(Sf)(z) = zf(z)$, are exactly the subspaces of the form $\varphi H^2(\mathbb{D})$ where $\varphi$ is an inner function (i.e. $|\varphi(e^{i\theta})| = 1$ a.e.).

Beurling’s theorem is a landmark because it gives not merely existence but a full classification of all invariant subspaces for a single operator. The shift on $H^2$ is in many senses the canonical operator for the Hilbert space invariant subspace problem: finding a counterexample to the full problem is equivalent to finding an operator with no invariant subspaces, and the shift shows how rich such structure can be even for a single operator.

Negative Results: Counterexamples on Banach Spaces #

Theorem (Enflo, 1975/1987; Read, 1984)

There exist separable Banach spaces and bounded linear operators on them with no non-trivial closed invariant subspace. In particular, Read constructed such an operator on $\ell^1$.

Enflo’s counterexample was the first, constructed in 1975 though not published until 1987 due to its length and complexity. Read’s construction (1984) arrived independently and somewhat earlier in print; a further, more explicit example by Read (1985) lives on the classical space $\ell^1$. These results make clear that the answer to the invariant subspace problem is negative for general Banach spaces. The Hilbert space case remains the central open question precisely because no counterexample on any reflexive Banach space, much less a Hilbert space, has been found.

The Hilbert–Banach Gap #

The separation between Hilbert space and general Banach space behaviour is a recurring theme. Several features of Hilbert spaces that Banach spaces lack suggest why counterexamples might not exist in the Hilbert setting:

The inner product gives every operator an adjoint $T^*$, and the lattice of invariant subspaces of $T$ and of $T^*$ are related by orthogonal complementation.
The spectral theorem for normal operators provides a complete invariant subspace theory for that class, anchoring intuition.
Reflexivity and the existence of unconditional bases in specific Hilbert spaces constrain operator behaviour more than in $\ell^1$.

None of these features has yet been converted into a proof for the general case.

Recent Proof Attempts #

The problem has attracted renewed attention in recent years.

In May 2023, Per Enflo, the same mathematician who produced the first Banach space counterexample, posted a preprint to arXiv (2305.15442) claiming a positive resolution for all separable Hilbert spaces. The original preprint was 13 pages; a substantially expanded version (52 KB) appeared in April 2024. Enflo himself has been cautious about the result, noting that expert review is ongoing. As of this writing the preprint has not received a definitive verdict from the community.

In July 2023 an independent preprint by Neville (arXiv:2307.08176) also claimed a positive solution for separable Hilbert spaces.

In September 2024 a peer-reviewed article in Axioms by Khalil, Yousef, Alshanti, and Abu Hammad announced a proof, but basic errors were identified shortly after publication (Ghatasheh, arXiv:2411.19409, November 2024).

The problem therefore remains officially open. The cluster of recent attempts reflects both its difficulty and its continued centrality in functional analysis.

Research Directions #

1. Cyclic Vectors and the Spectral Radius Formula #

A vector $x \in \mathcal{H}$ is cyclic for $T$ if $\mathcal{H} = \overline{\operatorname{span}{T^n x : n \geq 0}}$. An operator with a non-trivial invariant subspace cannot have every non-zero vector be cyclic. The contrapositive is: if every non-zero vector is cyclic, then $T$ is a counterexample.

Read’s Banach-space constructions proceed by building hypercyclic operators whose orbits are dense. On Hilbert spaces, Hilbert space geometry severely constrains the density of orbits. Making this constraint quantitative, via growth estimates on $|T^n x|$ or on the resolvent $|(T-\lambda)^{-1}|$, might close the gap between known positive results and the general case.

2. Dual Algebra Techniques #

A powerful modern approach studies the dual algebra $\mathcal{A} _T$, the weak-$*$ closure of the polynomials in $T$ as a subalgebra of $\mathcal{B}(\mathcal{H})$. If $\mathcal{A} _T = \mathcal{B}(\mathcal{H})$ (the operator is reflexive in this sense), one can sometimes extract invariant subspaces from the structure of the algebra. Results along these lines have been obtained for $C _{00}$ contractions (Bercovici, Foiaş, Pearcy) and for polynomially bounded operators under spectral conditions (Liu, 2017). The key open question is whether every Hilbert space contraction is reflexive in this sense, or whether the dual algebra approach can be made to work for all contractions via Sz.-Nagy–Foiaş theory.

3. Contractions and the Sz.-Nagy–Foiaş Calculus #

Every contraction ($|T| \leq 1$) on a Hilbert space admits a minimal unitary dilation (Sz.-Nagy’s dilation theorem), and Foiaş developed a functional calculus for contractions based on $H^\infty(\mathbb{D})$. The rich structure of this calculus has produced invariant subspace theorems for $C_{11}$ contractions and for contractions whose spectrum is rich enough. The question is whether the calculus can be pushed to all contractions; the general invariant subspace problem for contractions is equivalent to the full problem (by rescaling), so this is not a simplification but a different vantage point that has been productive.

4. Almost Invariant Half-Spaces #

A weaker notion, studied by Androulakis, Popov, Tcaciuc, and Troitsky, asks for almost invariant half-spaces: closed subspaces $\mathcal{M}$ of infinite dimension and infinite codimension such that $T\mathcal{M} \subseteq \mathcal{M} + \mathcal{F}$ for some finite-dimensional subspace $\mathcal{F}$. These exist for every operator on any infinite-dimensional Banach space. Whether every operator on a Hilbert space has a genuinely invariant (not just almost invariant) infinite-dimensional subspace of infinite codimension remains open and is a concrete intermediate target.

5. Hyperinvariant Subspaces #

A subspace is hyperinvariant for $T$ if it is invariant under every operator that commutes with $T$. Every hyperinvariant subspace is invariant, so existence of a hyperinvariant subspace implies a positive answer to the invariant subspace problem. Lomonosov’s 1973 theorem gives hyperinvariant subspaces when $T$ commutes with a compact operator. The hyperinvariant subspace problem, does every operator on a Hilbert space (other than scalar multiples of the identity) have a hyperinvariant subspace?, is also open and may be harder than the invariant subspace problem itself.

References #

Aronszajn, N. & Smith, K. T. (1954). Invariant subspaces of completely continuous operators. Annals of Mathematics, 60(2), 345–350.
Beurling, A. (1949). On two problems concerning linear transformations in Hilbert space. Acta Mathematica, 81, 239–255.
Brown, S. (1987). Hyponormal operators with thick spectra have invariant subspaces. Annals of Mathematics, 125(1), 93–103.
Enflo, P. H. (1987). On the invariant subspace problem for Banach spaces. Acta Mathematica, 158, 213–313.
Enflo, P. H. (2023). On the invariant subspace problem in Hilbert spaces. arXiv:2305.15442.
Lomonosov, V. I. (1973). Invariant subspaces of operators commuting with compact operators. Functional Analysis and Its Applications, 7(3), 213–214.
Read, C. J. (1984). A solution to the invariant subspace problem. Bulletin of the London Mathematical Society, 16(4), 337–401.
Read, C. J. (1985). A solution to the invariant subspace problem on the space $\ell^1$. Bulletin of the London Mathematical Society, 17(4), 305–317.
Radjavi, H. & Rosenthal, P. (2003). Invariant Subspaces (2nd ed.). Dover.
Bercovici, H., Foiaş, C., & Pearcy, C. (1985). Dual Algebras with Applications to Invariant Subspaces and Dilation Theory. AMS.

Brezis' first open problem - An elliptic equation involving the critical exponent in 3D

Sat, 18 Apr 2026 00:00:00 +0000

Yamabe problem #

Yamabe problem: Suppose $(\mathcal{M}, g_0)$ is a compact closed Riemannian manifold with dimension $N \geq 3$, does there exist a conformal metric $g = u^{\frac{4}{N-2}}g_0$ which has constant scalar curvature $R_g \equiv C$?

Find $u > 0$ on $\mathcal{M}$ such that $$ -\frac{4(N-1)}{N-2}\Delta_{g_0}u + R_{g_0}u = Cu^{\frac{N+2}{N-2}}\qquad\text{on }\mathcal{M}. $$

Some results:

Trudinger [1968]: if $g$ has non-positive scalar curvature.
Aubin [1976]: $N \geq 6$ and $(\mathcal{M}, g)$ not locally conformally flat.
Schoen [1984]: any dimension, the remaining cases, assuming the Positive Mass Theorem by Schoen-Yau [1979].

A special case #

Consider the special case where $\mathcal{M}$ is a bounded domain $\Omega$ in $\mathbb{R}^{N}$: $$ \begin{cases} -\Delta u = u^{\frac{N+2}{N-2}}\qquad\text{in }\Omega, \\ u > 0\qquad\text{in }\Omega, \\ u = 0\qquad\text{on }\partial\Omega. \end{cases} $$

Pohozaev [1965]: if $\Omega$ is star-shaped, then there is no nontrivial solution.

Brezis-Nirenberg problem #

Consider a lower-order perturbation: $$ \begin{cases} -\Delta u = u^{\frac{N+2}{N-2}} + \lambda u\qquad\text{in }\Omega, \\ u > 0\qquad\text{in }\Omega, \\ u = 0\qquad\text{on }\partial\Omega. \end{cases} $$

Some results:

Pohozaev’s result also yields nonexistence when $\lambda \leq 0$ and $\Omega$ is star-shaped.
If a positive solution exists, then necessarily $\lambda < \lambda_1$, where $\lambda_1$ is the first eigenvalue of $-\Delta$ on $\Omega$ with zero Dirichlet boundary condition.

Hence, for positive solutions on star-shaped domains, $$ 0 < \lambda < \lambda_1. $$

Brezis’ Open Problem 1.1 #

Let $N=3$, and let $\Omega = B_1 \subset \mathbb{R}^3$ be the unit ball. Consider $$ \begin{cases} -\Delta u = u^5 + \lambda u \qquad \text{in } B_1, \\ u = 0 \qquad \text{on } \partial B_1. \end{cases} $$ We ask whether this problem admits a nontrivial positive solution $u \not\equiv 0$.

Here the exponent $5 = \frac{N+2}{N-2}$ is the critical Sobolev exponent when $N=3$, and this is exactly the source of the main compactness difficulty.

Let $\lambda_1$ be the first Dirichlet eigenvalue of $-\Delta$ on $B_1$. The classical Brezis-Nirenberg theory shows:

If $\lambda \leq 0$, then the only solution is $u \equiv 0$.
If $\frac{1}{4}\lambda_1 < \lambda < \lambda_1$, then there exists a positive radial solution.
If $0 < \lambda \leq \frac{1}{4}\lambda_1$, then any radial solution must be trivial; hence there is no positive radial solution.
If $\lambda > \lambda_1$, there exist sign-changing solutions, but no positive solution.

Therefore the unresolved case is:

Open Problem 1.1. Assume $$ 0 < \lambda \leq \frac{1}{4}\lambda_1. $$ Does there exist a nontrivial solution?
Equivalently, since no positive radial solution can exist in this range, can there exist a non-radial positive solution?

This problem has remained open for decades, even if one restricts further to a smaller interval such as $$ 0 < \lambda < \varepsilon $$ for some sufficiently small $\varepsilon > 0$.

Remarks #

A few points are worth emphasizing:

By the Gidas-Ni-Nirenberg symmetry principle, positive solutions on a ball are often expected to be radial; however, in this regime Brezis observed that any radial solution must vanish, so any eventual positive solution would have to be genuinely non-radial.
This makes dimension $3$ sharply different from higher-dimensional cases, where the Brezis-Nirenberg existence theory is better understood.
The bifurcation picture suggests branches of sign-changing non-radial solutions emerging from higher eigenvalues, but it is not known whether such branches can reach the interval $\left(0,\frac14\lambda_1\right]$.

References #

H. Brezis and L. Nirenberg, Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, Comm. Pure Appl. Math. 36 (1983), 437–477.
H. Brezis, Some of My Favorite Open Problems, Open Problem 1.1.
M. Comte, Solutions of elliptic equations with critical Sobolev exponent in dimension three, Nonlinear Anal. 17 (1991), 445–455.
O. Druet, Elliptic equations with critical Sobolev exponents in dimension 3, Ann. Inst. H. Poincaré Anal. Non Linéaire 19 (2002), 125–142.

Restriction and extension

Wed, 29 Oct 2025 00:00:00 +0000

Considering a smooth compact hyper-surface $\mathcal{S}$ in $\mathbb{R}^d$ with surface measure $d\sigma$. Given $f \in L^1(\mathbb{R}^d)$, the Fourier transform defined as follow: $$ \begin{equation} \hat{f}(x) = \int_{\mathbb{R}^d}e^{-2\pi i x \xi}f(x)dx \end{equation} $$ which by Riemann-Lebesgue is a bounded, continuous function vanishing at infinity.

Since $\hat{f}$ is continuous on $\mathbb{R}^d$, by the Rimann-Lesbegue lemma its restriction to the compact hyper-surface $S \subset \mathbb{R}^d$ is is well-defined pointwise. Specifically, the restriction $\hat{f}\mid_{S}: S \rightarrow \mathbb{C}$ is the continuous function given by $$ \begin{equation} \hat{f}\mid_{S}(\sigma) = \hat{f}(\sigma) = \int_{\mathbb{R}^d}e^{-2\pi i x \xi}f(x)dx \end{equation} $$ for each $\sigma \in S$. This is bounded (as $\hat{f}$ is bounded) and can be integrated against the surface measure $d\sigma$ on $S$.

Thus when we restrict $\hat{f}$ to $S$, we get a meaningful function which has finite $L^q$-norm for every $q$ .

When starting with $f \in L^2(\mathbb{R}^d)$, the Fourier transform $\hat{f}$ is not well-defined point-wise in general, so there is no meaningful way to restrict an arbitrary $L^2$ function to a set of measure zero such as the hyper-surface $S$.

For especially, for any given $f \in L^2(\mathbb{R}^d)$, the Fourier transform is defined in the $L^2$ sense via the Plancherel theorem: $$ \begin{equation} \mathcal{F}: L^2(\mathbb{R}^d) \to L^2(\mathbb{R}^d), \quad | \hat{f} | _{L^2} = | f | _{L^2} \end{equation} $$ It is an isometry. So: $$ \begin{equation} \hat{f} \in L^2(\mathbb{R}^d) \end{equation} $$ Since $\hat{f}$ is only an $L^2$ function — it is not necessarily continuous, and not even bounded, and need not have a pointwise value almost everywhere.

So the expression: $$ \begin{equation} \hat{f}|_S(\sigma) = \hat{f}(\sigma), \quad \sigma \in S \end{equation} $$ does not make sense pointwise for arbitrary $f \in L^2$.

The question arises: what happens for $1 < p < 2$?

Question 1:

For which $p$ and $q$ do we have: $$ \begin{equation} ||\hat{f}|| _{L^q(S, d\sigma)} \lesssim ||f|| _{L^p(\mathbb{R}^d)}, \quad \forall f. \end{equation} $$

This is restriction of Fourier transforms to hyper-surfaces problem in Harmonic analysis.

Proof of Theorem of solution of wave equation in the case $n = 1$

Thu, 31 Jul 2025 00:00:00 +0000

Solution of Brezis Problem 8.24 (1) and (2)

Thu, 31 Jul 2025 00:00:00 +0000

Solution of Evans PDE Problem 13

Thu, 31 Jul 2025 00:00:00 +0000

Collected Lectures on Functional Analysis

Mon, 07 Jul 2025 00:00:00 +0000

📝 An Introduction to Functional Analysis - Laurent W. Marcoux (University of Waterloo)
📝 Functional Analysis: Lecture Notes - Jeff Schenker (Michigan State University)
📝 Functional Analysis Lecture Notes - T.B. Ward (University of East Anglia)
📝 Functional Analysis - Alexander C. R. Belton
📝 Topics in Real and Functional Analysis - Gerald Teschl
📝 Functional Analysis - Christian Remling
📝 Theory of Functions of a Real Variable - Shlomo Sternberg
📝 Functional Analysis - Lawerence Baggett

A lemma of J. L. Lions

Tue, 24 Jun 2025 00:00:00 +0000

This post explores J. L. Lions’ lemma about Banach spaces with compact injection, including applications to functional analysis.

Lemma statement:

Let $X$, $Y$, and $Z$ be three Banach spaces with norms $|| \cdot ||_X$, $|| \cdot ||_Y$, and $|| \cdot ||_Z$. Assume that $X \subset Y$ with compact injection and that $Y \subset Z$ with continuous injection. Prove that

$$ \forall \varepsilon > 0, \exists C_\varepsilon > 0 \text{ satisfying } || u ||_Y \leq \varepsilon || u ||_X + C _{\varepsilon}|| u ||_Z,\quad \forall u \in X $$

Applications:

Prove that for every $\varepsilon > 0$ there exists $C_\varepsilon > 0$ satisfying

$$ \max_{t \in [0,1]} |u(t)| \leq \varepsilon \max_{t \in [0,1]} |u’(t)| + C_\varepsilon ||u ||_{L^1}, \quad \forall u \in C^1([0,1]). $$

Pick $p > 1$. Prove that for every $\varepsilon > 0$ there exists $C = C(\varepsilon, p)$ such that

$$ || u || _{L^\infty(0,1)} \leq \varepsilon || u || _{W^{1,p}(0,1)} + C || u || _{L^1(0,1)}, \quad \forall u \in W^{1,p}(0,1). $$

Proof:

For the initial lemma, just argue by contradiction. Assume the contrary that there exists some $\varepsilon_0 > 0$ and a sequence $(u_n)_{n \in \mathbb{Z}^{+}} \subset X$ such that

$$ || u ||_Y > \varepsilon || u ||_X + C _{\varepsilon}|| u ||_Z $$

Then $u_n \ne 0, \forall n \in \mathbb{Z}^{+}$.

Let $v_n := \dfrac{u_n}{|| u_n||_X}$

Then clearly, $||v_n||_X = 1$ and we have

$$ ||v_n|| _Y > \varepsilon_0 + C _{\varepsilon_0}||v_n||_Z $$

Since $X \subset Y$ with compact injection.

Assume without loss generalization, there is $v \in Y$ such that $|| v_n - v|| _Y \rightarrow 0$ as $n \rightarrow \infty$. In particular, we have $(||v_n||) _{n \in \mathbb{Z}^{+}}$ bounded. It follows that $||v_n|| \rightarrow 0$ as $n \rightarrow \infty$.

And because $Y \subset Z$ with continuous injection, we obtain:

$$ ||v_n - v||_Z \rightarrow 0 \quad \text{as} \quad n \rightarrow \infty $$

Then $v = 0$ and $||v_n||_Y \rightarrow 0$ as $n \rightarrow \infty$

On the other hand, we also have

$$ \lim_{n \rightarrow \infty} > \varepsilon_0 + \varepsilon_0\lim_{n \rightarrow \infty}||v_n||_Z $$

Consequently,

$$ 0 > \varepsilon_0 > 0 $$ which is a contradiction. The two application are more or less immediate after using the given lemma. The proof is completed.

Complex Hahn-Banach Theorem

Tue, 24 Jun 2025 00:00:00 +0000

Let $X$ be a complex vector space, $X_0$ one of its subspaces, $p: X \to \mathbb{R}_+$ such that

$$ p(\lambda x) = |\lambda| p(x), \quad \forall \lambda \in \mathbb{C}, x \in X \text{ and } p(x + y) \leq p(x) + p(y), \quad \forall x, y \in X, $$

satisfying $|f(x)| \leq p(x)$, $\forall x \in X_0$, where $f: X_0 \to \mathbb{C}$ is linear.

Under these conditions, there exists a linear functional $F: X \to \mathbb{C}$ such that $F|_{X_0} = f$ and

$$ |F(x)| \leq p(x), \quad \forall x \in X. $$

Proof: Since $f$ is linear, it follows that $\text{Re } f: X_0 \to \mathbb{R}$ is linear and $$ \text{Re } f(x) \leq |f(x)| \leq p(x), \quad \forall x \in X_0. $$

By the Real Hahn-Banach Theorem there exists $g: X \to \mathbb{R}$ a linear functional such that $g$ is an extension for $\text{Re } f$ and $g(x) \leq p(x)$, $\forall x \in X$. We also have $g(x) = -g(-x) \geq -p(x)$ so $|g(x)| \leq p(x)$, $\forall x \in X$.

Define now $F(x) = g(x) - i g(ix)$, $\forall x \in X$. This is obviously linear and if $x \in X_0$ we have $$ F(x) = g(x) - i g(ix) = \text{Re } f(x) - i \text{Re } i f(x) = \text{Re } f(x) + i \text{Im } f(x) = f(x), \quad \forall x \in X_0. $$

For the last part we have $|F(x)| = e^{i\theta} F(x) = F(e^{i\theta} x) = g(e^{i\theta} x)$, because this is a real number. Furthermore, we have $g(e^{i\theta} x) \leq p(e^{i\theta} x) = p(x)$. Combining the two above, we get $$ |F(x)| \leq p(x), \quad \forall x \in X, $$ which solves the theorem.

Real Hahn-Banach Theorem

Tue, 24 Jun 2025 00:00:00 +0000

Suppose $X$ is a vector space over $\mathbb{R}$, $p: X \to \mathbb{R}$ has the following properties:

$p(X) = \lambda p(x)$, $\forall x \in X$, $\lambda \in \mathbb{R}_+$ and $p(x + y) \leq p(x) + p(y)$, $\forall x, y \in X$.
Let $X_0$ be a subspace of $X$ and $u: X_0 \to \mathbb{R}$ a linear functional such that $u(x) \leq p(x)$, $\forall x \in X_0$.

Then we can find $f: X \to \mathbb{R}$ a linear functional such that $f|_{X_0} = u$ and $f(x) \leq u(x)$, $\forall x \in X$.

Proof: Let $Y$ is a subspace of $X$, $g: Y \to \mathbb{R}$ is a linear functional which extends $u$ and $g \leq p$ on $Y$

Consider the set $M = { (Y, g) }$. Define an order relation on $M$ like this $(Y_1, g_1) \leq (Y_2, g_2)$ if $Y_1 \subset Y_2$ and $g_2$ is an extension for $g_1$.

We show that in $M$ every chain has an upper bound. Suppose $M_0$ is a totally ordered subset of $M$. Then define $Y_0 = \bigcup_{(Y,g) \in M_0} Y$ and $g: Y_0 \to \mathbb{R}$, $g(y) = g_0(y)$ if $y \in Y_0$ and $(Y_0, g) \in M_0$. This function is well defined, and $Y_0$ is a subspace of $X$ because the set $M_0$ is totally ordered.

Furthermore, from the definition for $g_0$, we have that $g_0 \leq p$. Therefore $(Y_0, g_0) \in M$, and is obviously an upper bound for $M_0$. By Zorn’s Lemma, we find that $M$ has at least one maximal element $(Z, h)$.

Suppose $X \neq Z$. Then we can find $x_0 \in X \setminus Z$. Define $W = \text{Span}{Z, x_0} = \mathbb{R} \cdot x_0 \oplus Z$. Therefore, $W$ is a linear subspace in $X$. Let $y, z \in Z$. Then $$ h(y) + h(z) = h(y + z) \leq p(y + z) = p(y - x_0 + x_0 + z) \leq p(y - x_0) + p(x_0 + z) $$ Therefore, we have $$ h(z) - p(-x _0 + z) + h(y) - p(y - x _0) \leq - h(y) + p(x _0 + y), \quad\forall y, z \in Z $$

Therefore, we can say $$ a = \sup_{z \in Z} (h(z) - p(-x_0 + z)) \leq - \inf_{y \in Z} (-h(y) + p(x_0 + y)) $$ Pick one $c \in [a, b]$ and define $h_1(z) = \lambda c + h(y)$, where $z = \lambda x_0 + y$ (unique representation), $h_1$ is linear, and extends $h_1$ on $W$, which means that it extends $u$ on $X_0$.

We can check that $(W, h_1) \in M$ and the maximal element $h_1$ is the requested functional element, which is a contradiction.

Therefore $Z = X$, and the maximal element $h_1$ is the requested functional.

Riesz Representation Theorem

Tue, 24 Jun 2025 00:00:00 +0000

1. Riesz Representation Theorem #

Let $H$ be a Hilbert space over $\mathbb{R}$ or $\mathbb{C}$, and $T$ be a bounded linear functional on $H$ (a bounded operator from $H$ to the field $\mathbb{R}$ or $\mathbb{C}$, where $H$ is defined over that field). The following is known as the Riesz Representation Theorem:

Theorem 1:

If $T$ is a bounded linear functional on the Hilbert space $H$, then there exists $g \in H$ such that for every $f \in H$, we have: $$ T(f) = \langle f, g \rangle. $$

Moreover, $|T| = |g|$ (here $|T|$ denotes the operator norm of $T$, while $|g|$ is the Hilbert space norm of $g$).

Now, let’s prove this theorem.

Proof:

Assume that $H$ is separable for now. The proof for any Hilbert space is not much more difficult, but the separable case nicely uses ideas we have developed related to Fourier analysis. Additionally, we will work over $\mathbb{R}$.

Since $H$ is separable, we can choose an orthonormal basis $\phi_j$, $j \geq 1$, for $H$. Let $T$ be a bounded linear functional and set $a_j = T(\phi_j)$. For $f \in H$, set $c_j = \langle f, \phi_j \rangle$, and define $$ f_n = \sum_{j=1}^{n} c_j \phi_j. $$

Since the $\phi_j$ form a basis, we know that $|f - f_n| \to 0$ as $n \to \infty$.

Since $T$ is linear, we have: $$ T(f_n) = \sum_{j=1}^{n} a_j c_j. \tag{1} $$

Since $T$ is bounded, assume with norm $|T| < \infty$, we have: $$ |T(f) - T(f_n)| \leq |T| |f - f_n|. \tag{2} $$

Because $|f - f_n| \to 0$ as $n \to \infty$, we conclude from equations (1) and (2) that: $$ T(f) = \lim_{n\to\infty} T(f_n) = \sum_{j=1}^{\infty} a_j c_j. \tag{3} $$

In fact, the sequence $a_j$ must be square-summable. To see this, first note that since $|T(f)| \leq |T| |f|$, we have: $$ \left|\sum_{j=1}^{\infty} c_j a_j\right| \leq |T| \left(\sum_{j=1}^{\infty} c_j^2\right)^{1/2}. \tag{4} $$

Equation (4) must hold for every square-summable sequence $c_j$ (since any such $c_j$ corresponds to some element in $H$). Fix a positive integer $N$ and define the sequence $c_j = a_j$ for $j \leq N$, $c_j = 0$ for $j > N$. Clearly, such a sequence is square-summable, and equation (4) gives us: $$ \left(\sum_{j=1}^{N} a_j^2\right)^{1/2} \leq |T|. \tag{5} $$

Thus, $a_j$ is square-summable, as the sequence of partial sums is bounded above.

Since $a_j$ is square-summable, the function $g = \sum_{j} a_j \phi_j$ is well-defined as an element of $H$, and $T(f) = \sum_{j} a_j c_j = \langle f, g \rangle$. Finally, equation (5) shows that $|g| \leq |T|$. But from the Cauchy-Schwarz inequality, we also have $|T(f)| = |\langle f, g \rangle| \leq |f| |g|$ or $\frac{|T(f)|}{|f|} \leq |g|$, implying $|T| \leq |g|$, hence $|T| = |g|$. The proof is complete.

2. Application to PDE #

This example illustrates how functional analysis methods are used in PDEs (although the example is for an ODE). Consider the ODE: $$ -f’’(x) + b(x)f(x) = q(x) \tag{6} $$

on the interval $0 < x < 1$, with $b(x) \geq \delta > 0$ for some $\delta$; assume the functions $b$ and $q$ are continuous on $[0, 1]$. We want to find a solution to equation (6) with $f’(0) = f’(1) = 0$ (other boundary conditions could also be applied). If we multiply (6) by a $C^1$ function $\phi$ and integrate the first term, $-f’’\phi$, by parts from $x = 0$ to $x = 1$, we obtain: $$ \int_0^1 (f’(x)\phi’(x) + b(x)f(x)\phi(x)),dx = \int_0^1 q(x)\phi(x),dx. \tag{7} $$

Equation (7) must hold for every $\phi \in C^1([0, 1])$, if $f$ is a $C^2(0, 1)$ solution of equation (6) that is continuous on $[0, 1]$. Conversely, if for a $C^2$ function $f$, we find that (7) holds for every $\phi$, then $f$ must be a solution of equation (6), because if we “undo” the integration by parts in (7), we get: $$ \phi(1)f’(1) - \phi(0)f’(0) + \phi(x)(-f’’(x) + b(x)f(x)) = \phi(x)q(x) $$ for every $\phi$.

A familiar PDE argument then shows that $f’(0) = f’(1) = 0$ and equation (6) must hold.

We will show that there is a unique solution to equation (7). Such a “solution” does not necessarily need to be twice differentiable as required by equation (6), but it will satisfy equation (7). Equation (7) is often called the “weak” form of the problem.

Define an inner product: $$ \langle g, h \rangle = \int_0^1 (g’(x)h’(x) + b(x)g(x)h(x)),dx $$

on the space $C^1([0, 1])$, and let $H$ denote the completion of this space. This is essentially the procedure used on the third problem of the first exam; the presence of $b(x)$ makes no difference. (Note that we must use $b \geq \delta > 0$ to ensure that $\langle \cdot, \cdot \rangle$ is indeed an inner product, so that $|g| = \sqrt{\langle g, g \rangle} = 0$ if and only if $g \equiv 0$.) The space $H$ is a Hilbert space and can be understood (if needed) as a subspace of $C([0, 1])$.

Define a functional $T : H \to \mathbb{R}$ by: $$ T(\phi) = \int_0^1 q(x)\phi(x),dx $$

You can easily check that $T$ is bounded on $H$ (using Cauchy-Schwarz). From the Riesz Representation Theorem, it follows that there must exist a function $f \in H$ such that: $$ T(\phi) = \langle f, \phi \rangle $$

for every $\phi \in H$. This is exactly equation (7), the weak form of the ODE!

The function $f$ satisfying equation (7) lies in $H$. Under the conditions on $b$ (specifically, $b \geq \delta > 0$ and $|b|_\infty < \infty$ since $b \in C([0, 1])$), the function $f$ lies in the same space defined in the third problem of the first exam. Specifically, $f$ is a continuous function. Proving that $f$ is actually twice differentiable requires more work, along with additional assumptions about the function $q$.

References #

[1] (Original) The Riesz Representation Theorem, MA 466, Kurt Bryan

The application of Hahn-Banach Theorem 01

Tue, 24 Jun 2025 00:00:00 +0000

Suppose $X$ is a normed space and $X_0$ is a closed subspace of $X$ and $x_0 \in X \setminus X_0$. Then we can find $f \in X’$ such that $f(x_0) = 1$ and $f(x) = 0$, $\forall x \in X_0$.

Proof: Since $x_0 \notin X_0$, we can find $\delta > 0$ such that $|x_0 - x| \geq \delta$, $\forall x \in X_0$, which is equivalent to $1 \leq \dfrac{|x_0 - x|}{\delta}$, $\forall x \in X_0$.

Define $Y = \text{Span}{x_0, X_0} = X_0 \oplus \mathbb{K} \cdot x_0$. Then for each $y \in Y$ we can find a unique $\lambda \in \mathbb{K}$ such that $u = \lambda x_0 + x$, $x \in X_0$. Define $u: Y \to \mathbb{K}$ by $u(y) = u(\lambda x_0 + x) = \lambda$. It is well defined and linear.

Furthermore, we have: $$|u(y)| = |\lambda| \leq |\lambda| \frac{|x _0 + x|}{\delta} = \frac{1}{\delta} |y| \quad \text{for} \lambda \neq 0$$ If $\lambda = 0$, then $y \in X_0$ and $u(y) = 0 \leq \frac{1}{\delta} |y|$.

Therefore, we obtain
$$ u(y) \leq \frac{1}{\delta} |y| \quad\forall y \in Y $$ By Hahn-Banach’s Theorem, we can extend $u$ to $f: X \to \mathbb{K}$ such that $f|_Y = u$ and $|f(x)| \leq \dfrac{1}{\delta} |x|$, $\forall x \in X$. Therefore $f(x_0) = u(x_0) = 1$ and $x \in X_0 \Rightarrow f(x) = 0$.

The application of Hahn-Banach Theorem 02

Tue, 24 Jun 2025 00:00:00 +0000

$X'$ = $\{ f: X \to \mathbb{K} \}$ where $f$ is is linear and continuous and $X$ is a Banach space over $\mathbb{K}$. Prove that $X' \neq {0}$, in fact, for every $x \neq 0 \in X$, we can find $f \in X’$ such that $f(x) = |x|$ and $|f| = 1$.

Proof: Pick $x_0 \in X$. Define $X_0 = x_0 \cdot \mathbb{K}$, a subspace of $X$, and $g: X_0 \to \mathbb{K}$, $g(x) = x$, which is linear. Since $g$ and $|\cdot|$ satisfy the conditions of the Hahn-Banach theorem, we can find $f: X \to \mathbb{K}$ such that $f|_{X_0} = g$, $f$ is linear and $f(x) \leq |x|$, $\forall x \in X$. Therefore $f(x_0) = g(x_0) = |x_0|$ and $|f| \leq 1$. The equality $f(x_0) = |x_0|$ guarantees that $|f| = 1$.

Mathematics - Functional Analysis

Thu, 27 Jun 2024 23:14:15 +0800