Posts ¹⁰

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

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 $|f| = 1$.