Let $H$ be a Hilbert space, and let $A : H \to H$ be a bounded linear operator. Then $(\operatorname{Ran}(A))^\perp = \operatorname{Null}(A^\ast)$: the orthogonal complement of the range of $A$ equals the kernel of its adjoint.
A bounded linear operator $T : H \to H$ on a Hilbert space is self-adjoint
if $T = T^\ast$. This is a wrapper around Mathlib's IsSelfAdjoint.
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If $T \in \mathcal{B}(H)$ is a self-adjoint operator (i.e. $T = T^\ast$), then $\langle Tu, u\rangle \in \mathbb{R}$ for every $u \in H$, and the operator norm satisfies $\|T\| = \sup_{\|x\| = 1} |\langle Tx, x\rangle|$, expressed here in terms of the Rayleigh quotient.
Characterization of eigenvalues: for $T \in \mathcal{B}(H)$ and $\mu \in \mathbb{K}$, $\mu$ is an eigenvalue of $T$ if and only if there exists a nonzero vector $u \in H$ (the associated eigenvector) such that $Tu = \mu u$.
HeineโBorel for finite-dimensional normed spaces over $\mathbb{R}$ or $\mathbb{C}$: a subset $s \subseteq E$ is compact if and only if it is closed and bounded.
In a metric space $X$, a subset $K \subseteq X$ is compact if and only if it is sequentially compact: every sequence in $K$ has a subsequence converging to an element of $K$.
Auxiliary lemma: the kernel of $\mu I - T$ coincides with the eigenspace of $T$ associated to the eigenvalue $\mu$.
Fredholm alternative. Let $A = A^\ast \in \mathcal{B}(H)$ be a self-adjoint compact operator, and let $\mu \in \mathbb{K} \setminus \{0\}$. Then $\operatorname{Range}(\mu I - A)$ is closed (equal to $\operatorname{Null}(\mu I - A)^\perp$), and either $\mu I - A$ is bijective, or its kernel โ the eigenspace of $A$ associated to $\mu$ โ is nontrivial and finite-dimensional.
Maximum principle. Let $A = A^\ast \in \mathcal{B}(H)$ be a self-adjoint compact operator on an infinite-dimensional Hilbert space. Then the nonzero eigenvalues of $A$ can be ordered as $|\lambda_1| \ge |\lambda_2| \ge \cdots$ (counted with multiplicity), with pairwise orthonormal eigenvectors $\{u_k\}$ such that $T u_k = \lambda_k u_k$, $|\lambda_j| \to 0$, and for each $j$, $|\lambda_j| = \sup_{\|u\|=1,\, u \in \operatorname{Span}(u_1,\dots,u_{j-1})^\perp} |\langle Au, u\rangle|$.
Spectral theorem for compact self-adjoint operators (range decomposition form). Let $A = A^\ast \in \mathcal{B}(H)$ be a compact self-adjoint operator on a Hilbert space $H$. Then: (1) the closure of the span of all nonzero eigenspaces of $A$ equals the closure of $\operatorname{Range}(A)$; (2) $\overline{\operatorname{Range}(A)}$ and $\operatorname{Null}(A)$ are complementary closed subspaces of $H$; and (3) the family of eigenspaces of $A$ is an orthogonal family.
Auxiliary lemma: for $t \in \mathbb{R}$ and a nonzero vector $x$, the Rayleigh quotient of $tI - T$ at $x$ equals $t$ minus the Rayleigh quotient of $T$ at $x$.
The spectrum of a self-adjoint operator $T \in \mathcal{B}(H)$ is bounded by the Rayleigh quotient: for every $\mu \in \operatorname{Spec}(T)$, $\inf_{x} \langle Tx, x\rangle / \|x\|^2 \le \operatorname{Re}\mu \le \sup_{x} \langle Tx, x\rangle / \|x\|^2$.
Let $A = A^\ast \in \mathcal{B}(H)$ be a self-adjoint operator. Then $A$ is positive โ i.e. $\langle Au, u\rangle \ge 0$ for all $u \in H$ โ if and only if $\operatorname{Spec}(A) \subseteq [0, \infty)$.
The space of invertible bounded linear operators $GL(H) = \{T \in \mathcal{B}(H) : T \text{ invertible}\}$ is an open subset of $\mathcal{B}(H)$.
The Hilbert space $L^2([0,1])$ of square-integrable real-valued functions on the unit interval, equipped with the restriction of Lebesgue measure to $[0,1]$.
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The $k$-th orthonormal sine basis function on $[0,1]$: $u_k(x) = \sqrt{2} \sin(k\pi x)$. These form an orthonormal eigenbasis of the Green's operator with Dirichlet boundary conditions on the unit interval.
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The $k$-th eigenvalue of the Green's operator on $[0,1]$ with Dirichlet boundary conditions: $\lambda_k = \dfrac{1}{k^2 \pi^2}$.
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The square root of the $k$-th eigenvalue of the Green's operator, $\sqrt{\lambda_k} = \dfrac{1}{k\pi}$. This is the $k$-th eigenvalue of $A^{1/2}$.
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The square root $A^{1/2}$ of the Green's operator, defined via its action on the orthonormal eigenbasis $\{b_k\}$: if $f = \sum_k c_k\, b_k$ where $c_k = \langle b_k, f\rangle$, then $A^{1/2} f = \sum_k \frac{1}{k\pi}\, c_k\, b_k$.
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The Green's kernel for the Dirichlet Laplacian on $[0,1]$: $G(x,t) = \min(x,t)\,(1 - \max(x,t))$. The Green's operator $f \mapsto \int_0^1 G(x,t) f(t)\,dt$ is the inverse of $-\frac{d^2}{dx^2}$ subject to $u(0)=u(1)=0$.
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Key computation showing that the sine basis functions are eigenfunctions of the Green's operator: for $x \in [0,1]$, $\int_0^1 G(x,t)\, u_k(t)\, dt = \lambda_k\, u_k(x)$, where $G$ is the Green's kernel, $u_k(x) = \sqrt{2}\sin(k\pi x)$ and $\lambda_k = 1/(k^2\pi^2)$.
Spectral decomposition of the Green's operator on $L^2([0,1])$. If $A$ is the self-adjoint operator on $L^2([0,1])$ given by the Green's kernel $Af(x) = \int_0^1 G(x,t)\, f(t)\, dt$, and $\{b_k\}_{k \in \mathbb{N}^+}$ is the Hilbert basis of sine functions $b_k(x) = \sqrt{2}\sin(k\pi x)$, then:
- $\mathrm{Null}(A) = \{0\}$;
- each $b_k$ is an eigenvector with eigenvalue $\lambda_k = 1/(k^2\pi^2)$;
- every $f \in L^2([0,1])$ satisfies $Af = \sum_{k=1}^\infty \lambda_k\, \langle b_k, f\rangle\, b_k$.
Existence for the Sturm-Liouville Dirichlet problem on $[0,1]$: given a continuous nonnegative potential $V \in C([0,1])$ with $V \geq 0$ and a continuous right-hand side $f \in C([0,1])$, there exists a $C^2$ function $u$ on $[0,1]$ satisfying $-u''(x) + V(x)\, u(x) = f(x)$ for all $x \in [0,1]$, with the Dirichlet boundary conditions $u(0) = u(1) = 0$.
Uniqueness for the Sturm-Liouville Dirichlet problem on $[0,1]$: any two $C^2$ solutions $u_1, u_2$ of $-u'' + Vu = f$ on $[0,1]$ with the Dirichlet boundary conditions $u(0) = u(1) = 0$ (where $V \geq 0$ is continuous) must agree on $[0,1]$. The proof uses the convexity of $u^2$ when $u'' = V u$ with $V \geq 0$.
Existence and uniqueness for the Sturm-Liouville Dirichlet problem on $[0,1]$: given a continuous nonnegative potential $V \in C([0,1])$ with $V \geq 0$ and a continuous right-hand side $f \in C([0,1])$, there exists a unique $C^2$ function $u$ on $[0,1]$ such that $-u'' + Vu = f$ on $[0,1]$, with the Dirichlet boundary conditions $u(0) = u(1) = 0$.