Exploring Fundamental Theorems and Results in Functional Analysis

Important Theorems and Results in Functional Analysis

Theorem 1.1 (Banach Fixed-Point Theorem). Let $(X,d)$$(X,d)$(X,d)(X, d)$(X,d)$ be a complete metric space, and $T:X\to X$$T:X\to X$T:X rarr XT: X \rightarrow X$T:X\to X$ a contraction mapping with Lipschitz constant $0\le k<1$$0\le k<1$0 <= k < 10 \leq k<1$0\le k<1$. Then $T$$T$TT$T$ has a unique fixed point $x^{\ast }\in X$$x^{\ast }\in X$x^(**)in Xx^{*} \in X$x^{\ast }\in X$ such that $T\left(x^{\ast }\right)=x^{\ast }$$T\left(x^{\ast }\right)=x^{\ast }$T(x^(**))=x^(**)T\left(x^{*}\right)=x^{*}$T\left(x^{\ast }\right)=x^{\ast }$.

Theorem 1.2 (Hahn-Banach Theorem). Let $X$$X$XX$X$ be a real or complex vector space, and let $p:X\to \mathbb{R}$$p:X\to \mathbb{R}$p:X rarrRp: X \rightarrow \mathbb{R}$p:X\to \mathbb{R}$ be a sublinear functional. If $M$$M$MM$M$ is a linear subspace of $X$$X$XX$X$, and $f:M\to \mathbb{R}$$f:M\to \mathbb{R}$f:M rarrRf: M \rightarrow \mathbb{R}$f:M\to \mathbb{R}$ is a linear functional satisfying $f(x)\le p(x)$$f(x)\le p(x)$f(x) <= p(x)f(x) \leq p(x)$f(x)\le p(x)$ for all $x\in M$$x\in M$x in Mx \in M$x\in M$, then there exists a linear extension $F:X\to \mathbb{R}$$F:X\to \mathbb{R}$F:X rarrRF: X \rightarrow \mathbb{R}$F:X\to \mathbb{R}$ of $f$$f$ff$f$ such that $F(x)\le p(x)$$F(x)\le p(x)$F(x) <= p(x)F(x) \leq p(x)$F(x)\le p(x)$ for all $x\in X$$x\in X$x in Xx \in X$x\in X$.

Theorem 1.3 (Uniform Boundedness Principle). Let $X$$X$XX$X$ be a Banach space, and let $\left\{T_{\alpha }\right\}$$\left\{T_{\alpha }\right\}${T_(alpha)}\left\{T_{\alpha}\right\}$\left\{T_{\alpha }\right\}$ be a family of bounded linear operators from $X$$X$XX$X$ to a Banach space $Y$$Y$YY$Y$. If $\underset{\alpha }{sup}\parallel T_{\alpha }x\parallel <\mathrm{\infty }$$\underset{\alpha }{sup} ∥T_{\alpha }x∥<\mathrm{\infty }$s u p_(alpha)||T_(alpha)x|| < oo\sup _{\alpha}\left\|T_{\alpha} x\right\|<\infty$\underset{\alpha }{sup}\parallel T_{\alpha }x\parallel <\mathrm{\infty }$ for all $x\in X$$x\in X$x in Xx \in X$x\in X$, then $\underset{\alpha }{sup}\parallel T_{\alpha }\parallel <\mathrm{\infty }$$\underset{\alpha }{sup} ∥T_{\alpha }∥<\mathrm{\infty }$s u p_(alpha)||T_(alpha)|| < oo\sup _{\alpha}\left\|T_{\alpha}\right\|<\infty$\underset{\alpha }{sup}\parallel T_{\alpha }\parallel <\mathrm{\infty }$.

Theorem 1.4 (Open Mapping Theorem). Let $X$$X$XX$X$ and $Y$$Y$YY$Y$ be Banach spaces, and let $T:X\to Y$$T:X\to Y$T:X rarr YT: X \rightarrow Y$T:X\to Y$ be a bounded linear operator that is onto (surjective). Then $T$$T$TT$T$ maps open sets to open sets, i.e., for any open set $U\subset X$$U\subset X$U sub XU \subset X$U\subset X$, the image $T(U)$$T(U)$T(U)T(U)$T(U)$ is open in $Y$$Y$YY$Y$.

Theorem 1.5 (Closed Graph Theorem). Let $X$$X$XX$X$ and $Y$$Y$YY$Y$ be Banach spaces, and let $T:X\to Y$$T:X\to Y$T:X rarr YT: X \rightarrow Y$T:X\to Y$ be a linear operator. If the graph of $T$$T$TT$T$, denoted by $\mathrm{Graph}(T)$$\mathrm{Graph}(T)$Graph(T)\operatorname{Graph}(T)$\mathrm{Graph}(T)$, is closed in $X\times Y$$X\times Y$X xx YX \times Y$X\times Y$, then $T$$T$TT$T$ is a bounded operator.

Theorem 1.6 (Riesz Representation Theorem). Let H be a Hilbert space. For any bounded linear functional $f$$f$ff$f$ on $H$$H$HH$H$, there exists a unique vector $y\in H$$y\in H$y in Hy \in H$y\in H$ such that $f(x)=⟨x,y⟩$$f(x)=⟨x,y⟩$f(x)=(:x,y:)f(x)=\langle x, y\rangle$f(x)=⟨x,y⟩$ for all $x\in H$$x\in H$x in Hx \in H$x\in H$, where $⟨\cdot ,\cdot ⟩$$⟨\cdot ,\cdot ⟩$(:*,*:)\langle\cdot, \cdot\rangle$⟨\cdot ,\cdot ⟩$ denotes the inner product on $H$$H$HH$H$.

Theorem 1.7 (Spectral Theorem for Self-Adjoint Operators). Let $H$$H$HH$H$ be a complex Hilbert space, and let $T:H\to H$$T:H\to H$T:H rarr HT: H \rightarrow H$T:H\to H$ be a bounded self-adjoint operator. Then there exists an orthonormal basis of $H$$H$HH$H$ consisting of eigenvectors of $T$$T$TT$T$, and the corresponding eigenvalues are real.

Theorem 1.8 (Hilbert-Schmidt Theorem). Let $H$$H$HH$H$ be a separable Hilbert space, and let $T:H\to H$$T:H\to H$T:H rarr HT: H \rightarrow H$T:H\to H$ be a compact self-adjoint operator. Then there exists an orthonormal basis of $H$$H$HH$H$ consisting of eigenvectors of $T$$T$TT$T$, and the corresponding eigenvalues converge to zero.