Functional Analysis 14
The Invariant Subspace Problem
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?
Brezis' first open problem - An elliptic equation involving the critical exponent in 3D
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 $\mathbb{R}^{N}$: $$ \begin{cases} -\Delta u = u^{\frac{N+2}{N-2}}\qquad\text{in }\Omega, \\ u > 0\qquad\text{in }\Omega, \\ u = 0\qquad\text{on }\partial\Omega. \end{cases} $$
Restriction and extension
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 $S$.
Proof of Theorem of solution of wave equation in the case $n = 1$
Solution of Brezis Problem 8.24 (1) and (2)
Solution of Evans PDE Problem 13
Collected Lectures on Functional Analysis
A compact study list, grouped by level and type, with freely accessible material for studying Functional Analysis.
A lemma of J. L. Lions
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 $$ \forall \varepsilon > 0, \exists C_\varepsilon > 0 \text{ satisfying } || u ||_Y \leq \varepsilon || u ||_X + C _{\varepsilon}|| u ||_Z,\quad \forall u \in X $$
Complex Hahn-Banach Theorem
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
Real Hahn-Banach Theorem
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 $f(x) \leq u(x)$, $\forall x \in X$.
Riesz Representation Theorem
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: $$ T(f) = \langle f, g \rangle. $$
The application of Hahn-Banach Theorem 01
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 $1 \leq \dfrac{|x_0 - x|}{\delta}$, $\forall x \in X_0$.
The application of Hahn-Banach Theorem 02
$X'$ = $\{ f: X \to \mathbb{K} \}$ 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 $|f| = 1$.