Functional Analysis

\(L^p\) Spaces

For any measure space \((X, M, \mu)\), one can define \(L^2(\mu)\) with the inner product \({\left\langle {f},~{g} \right\rangle} \coloneqq\int_X f\mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu d\mu\).

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For \(p^{-1}+ q^{-1}= 1\), with \(1<p<\infty\), there is an isomorphism of Banach spaces \begin{align*} \kappa: L^p(\mu) &\to L^q(\mu) \\ f &\mapsto (g \mapsto \int_X f g d\mu ) .\end{align*}

This is surjective by Radon-Nikodym, and an isometry by Holder’s inequality, which is enough to be an isometric isomorphism.

General Facts

Working with inner products: \begin{align*} {\left\langle {tx + sy},~{z} \right\rangle} &= t{\left\langle {x},~{z} \right\rangle} + s{\left\langle {y},~{z} \right\rangle} \\ {\left\langle {x},~{y} \right\rangle} &= \mkern 1.5mu\overline{\mkern-1.5mu{\left\langle {y},~{x} \right\rangle}\mkern-1.5mu}\mkern 1.5mu \\ x\neq 0 \implies {\left\langle {x},~{x} \right\rangle} > 0 .\end{align*} We define \({\left\lVert {x} \right\rVert} \coloneqq\sqrt{{\left\langle {x},~{x} \right\rangle}}\).

For \(x, y\in H\), \begin{align*} {\left\lvert { {\left\langle {x},~{y} \right\rangle}} \right\rvert} \leq {\left\lVert {x} \right\rVert} {\left\lVert {y} \right\rVert} ,\end{align*} with equality iff \(x, y\) are linearly independent.

\begin{align*} {\left\langle {v},~{w} \right\rangle} = 0 \implies {\left\lVert {v+w} \right\rVert}^2 = {\left\lVert {v} \right\rVert}^2 + {\left\lVert {w} \right\rVert}^2 .\end{align*} More generally, if \(x_i \perp x_j\) for all \(i\neq j\), then \begin{align*} {\left\lVert {\sum_k x_k } \right\rVert}_H^2 = \sum_k {\left\lVert {x_k} \right\rVert}_H^2 .\end{align*}

For all \(x, y\in H\), \begin{align*} 4 {\left\langle {x},~{y} \right\rangle} = {\left\lVert {x+y} \right\rVert}^2 - {\left\lVert {x-y} \right\rVert}^2 +i\qty{{\left\lVert {x+iy} \right\rVert}^2 - {\left\lVert {x-iy} \right\rVert}^2} .\end{align*}

For all \(x, y\in H\), \begin{align*} {\left\lVert {x+y} \right\rVert}^2 + {\left\lVert {x-y} \right\rVert}^2 = 2\qty{{\left\lVert {x} \right\rVert}^2 + {\left\lVert {y} \right\rVert}^2 } .\end{align*}

If \(x_k\to x\) and \(y_k\to y\) in \(H\), then \({\left\langle {x_k},~{y_k} \right\rangle} \to {\left\langle {x},~{y} \right\rangle}\). Proof: \begin{align*} {\left\lvert {{\left\langle {x_k},~{y_k} \right\rangle} - {\left\langle {x},~{y} \right\rangle} } \right\rvert} ={\left\lvert {{\left\langle {x_n - x},~{y_n} \right\rangle} + {\left\langle {x},~{y_n-y} \right\rangle} } \right\rvert} \leq {\left\lVert {x_n - x} \right\rVert}{\left\lVert {y_n} \right\rVert} + {\left\lVert {x} \right\rVert} {\left\lVert {y_n - y} \right\rVert} .\end{align*}

Fourier Coefficients

For any orthonormal set \(\left\{{u_{n}}\right\} \subseteq {\mathcal{H}}\) a Hilbert space (not necessarily a basis), \begin{align*} \left\|x-\sum_{n=1}^{N}\left\langle x, u_{n}\right\rangle u_{n}\right\|^{2}=\|x\|^{2}-\sum_{n=1}^{N}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \end{align*} and thus \begin{align*} \sum_{n=1}^{\infty}\left|\left\langle x, u_{n}\right\rangle\right|^{2} \leq\|x\|^{2} .\end{align*}

Note that this generalizes to uncountable bases, and implies that only finitely many terms \({\left\langle {x},~{u_n} \right\rangle}\) can be nonzero. 


    
    
      
        
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      Let \(S_{N} = \sum_{n=1}^N {\left\langle {x},~{u_{n}} \right\rangle} u_{n}\) 

\begin{align*}
{\left\lVert {x - S_{N}} \right\rVert}^2 
&= {\left\langle {x - S_{n}},~{x - S_{N}} \right\rangle} \\
&= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_{N}} \right\rVert}^2 - 2\Re{\left\langle {x},~{S_{N}} \right\rangle} \\
&= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_{N}} \right\rVert}^2 - 2\Re {\left\langle {x},~{\sum_{n=1}^N {\left\langle {x},~{u_{n}} \right\rangle}u_{n}} \right\rangle} \\
&= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_{N}} \right\rVert}^2 - 2\Re \sum_{n=1}^N {\left\langle {x},~{ {\left\langle {x},~{u_{n}} \right\rangle}u_{n}} \right\rangle} \\
&= {\left\lVert {x} \right\rVert}^2 + {\left\lVert {S_{N}} \right\rVert}^2 - 2\Re \sum_{n=1}^N \overline{{\left\langle {x},~{u_{n}} \right\rangle}}{\left\langle {x},~{u_{n}} \right\rangle} \\
&= {\left\lVert {x} \right\rVert}^2 + \left\|\sum_{n=1}^N {\left\langle {x},~{u_{n}} \right\rangle} u_{n}\right\|^2 - 2 \sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 \\
&= {\left\lVert {x} \right\rVert}^2 + \sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 - 2 \sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 \\
&= {\left\lVert {x} \right\rVert}^2 - \sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2
.\end{align*}

    
    
  
        
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      By continuity of the norm and inner product, we have 

\begin{align*}
\lim_{N\to\infty} {\left\lVert {x - S_{N}} \right\rVert}^2 
&= \lim_{N\to\infty} {\left\lVert {x} \right\rVert}^2 - \sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 \\
\implies {\left\lVert {x - \lim_{N\to\infty} S_{N}} \right\rVert}^2 &= {\left\lVert {x} \right\rVert}^2 - \lim_{N\to\infty}\sum_{n=1}^N {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2\\
\implies 
{\left\lVert {x - \sum_{n=1}^\infty {\left\langle {x},~{u_{n}} \right\rangle} u_{n}} \right\rVert}^2 &= {\left\lVert {x} \right\rVert}^2 - 
\sum_{n=1}^\infty {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2
.\end{align*}

    
    
  
        
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      Then noting that \(0 \leq {\left\lVert {x - S_{N}} \right\rVert}^2\), 

\begin{align*}
0 &\leq 
{\left\lVert {x} \right\rVert}^2 - 
\sum_{n=1}^\infty {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 \\
\implies 
\sum_{n=1}^\infty {\left\lvert {{\left\langle {x},~{u_{n}} \right\rangle}} \right\rvert}^2 &\leq 
{\left\lVert {x} \right\rVert}^2 \hfill\blacksquare
.\end{align*}

    
    
  
        
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      Let \(\left\{{u_n}\right\}_{n\in A}\) be an orthonormal set in a Hilbert space \({\mathcal{H}}\). TFAE:
    

  
    
    
      
        
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      Completeness: \(\left\{{u_n}\right\}\) is a complete basis, i.e. \({\left\langle {x},~{u_n} \right\rangle}=0\) for all \(n\) implies \(x=0\)
    
    
  
        
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      Parseval’s identity: 

\begin{align*}
\sum_{n\in A} {\left\lvert { {\left\langle {x},~{u_n} \right\rangle} } \right\rvert}^2 = {\left\lVert {x} \right\rVert}^2_{{\mathcal{H}}}
.\end{align*}

    
    
  
        
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      Every \(x\in {\mathcal{H}}\) can be expressed uniquely as 

\begin{align*}
x = \sum_{n\in A} {\left\langle {x},~{u_n} \right\rangle}u_n
,\end{align*}
 where the sum has only countably many nonzero terms.
    
    
  
        
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      If \(\Lambda\) is a continuous linear functional on a Hilbert space \(H\), then there exists a unique \(y \in H\) such that 

\begin{align*}
\forall x\in H,\quad \Lambda(x) = {\left\langle {x},~{y} \right\rangle}
.\end{align*}

    

  

    
    
      
        
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          Define \(M \coloneqq\ker \Lambda\).
        
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          Then \(M\) is a closed subspace and so \(H = M \oplus M^\perp\)
        
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          There is some \(z\in M^\perp\) such that \({\left\lVert {z} \right\rVert} = 1\).
        
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          Set \(u \coloneqq\Lambda(x) z - \Lambda(z) x\)
        
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  • 
          Check
        
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\begin{align*}\Lambda(u) = \Lambda(\Lambda(x) z - \Lambda(z) x) = \Lambda(x) \Lambda(z) - \Lambda(z) \Lambda(x) = 0 \implies u\in M\end{align*}

    

  
    
    
      
        
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          Compute
        
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\begin{align*}
0 &= {\left\langle {u},~{z} \right\rangle} \\ 
&= {\left\langle {\Lambda(x) z - \Lambda(z) x},~{z} \right\rangle} \\
&= {\left\langle {\Lambda(x) z},~{z} \right\rangle} - {\left\langle {\Lambda(z) x},~{z} \right\rangle} \\
&= \Lambda(x) {\left\langle {z},~{z} \right\rangle} - \Lambda(z) {\left\langle {x},~{z} \right\rangle} \\
&= \Lambda(x) {\left\lVert {z} \right\rVert}^2 - \Lambda(z) {\left\langle {x},~{z} \right\rangle} \\
&= \Lambda(x) - \Lambda(z) {\left\langle {x},~{z} \right\rangle} \\
&= \Lambda(x) -  {\left\langle {x},~{\overline{\Lambda(z)} z} \right\rangle}
,\end{align*}

    

  
    
    
      
        
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          Choose \(y \coloneqq\overline{\Lambda(z)} z\).
        
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          Check uniqueness:
        
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\begin{align*}
{\left\langle {x},~{y} \right\rangle} &= {\left\langle {x},~{y'} \right\rangle} \quad\forall x \\
\implies  {\left\langle {x},~{y-y'} \right\rangle} &= 0 \quad\forall x \\
\implies {\left\langle {y-y'},~{y-y'} \right\rangle} &= 0 \\
\implies {\left\lVert {y-y'} \right\rVert} &= 0 \\
\implies y-y' &= \mathbf{0} \implies y = y'
.\end{align*}

    

  

    

      Let \(U = \left\{{u_{n}}\right\}_{n=1}^\infty\) be an orthonormal set (not necessarily a basis), then
    

  
    
    
      
        
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          There is an isometric surjection
        
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\begin{align*}
\mathcal{H} &\to \ell^2({\mathbb{N}}) \\
\mathbf{x} &\mapsto \left\{{{\left\langle {\mathbf{x}},~{\mathbf{u}_{n}} \right\rangle}}\right\}_{n=1}^\infty
\end{align*}

    

  
    

      i.e. if \(\left\{{a_{n}}\right\} \in \ell^2({\mathbb{N}})\), so \(\sum {\left\lvert {a_{n}} \right\rvert}^2 < \infty\), then there exists a \(\mathbf{x} \in \mathcal{H}\) such that 

\begin{align*}
a_{n} = {\left\langle {\mathbf{x}},~{\mathbf{u}_{n}} \right\rangle} \quad \forall n.
\end{align*}

    

  
    
    
      
        
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          \(\mathbf{x}\) can be chosen such that 

\begin{align*}
{\left\lVert {\mathbf{x}} \right\rVert}^2 = \sum {\left\lvert {a_{n}} \right\rvert}^2
\end{align*}

        
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      Note: the choice of \(\mathbf{x}\) is unique \(\iff\) \(\left\{{u_{n}}\right\}\) is complete, i.e. \({\left\langle {\mathbf{x}},~{\mathbf{u}_{n}} \right\rangle} = 0\) for all \(n\) implies \(\mathbf{x} = \mathbf{0}\).
    

  
    

  

    
    
      
        
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          Given \(\left\{{a_{n}}\right\}\), define \(S_{N} = \sum^N a_{n} \mathbf{u}_{n}\).
        
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          \(S_{N}\) is Cauchy in \(\mathcal{H}\) and so \(S_{N} \to \mathbf{x}\) for some \(\mathbf{x} \in \mathcal{H}\).
        
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          \({\left\langle {x},~{u_{n}} \right\rangle} = {\left\langle {x - S_{N}},~{u_{n}} \right\rangle} + {\left\langle {S_{N}},~{u_{n}} \right\rangle} \to a_{n}\)
        
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          By construction, \({\left\lVert {x-S_{N}} \right\rVert}^2 = {\left\lVert {x} \right\rVert}^2 - \sum^N {\left\lvert {a_{n}} \right\rvert}^2 \to 0\), so \({\left\lVert {x} \right\rVert}^2 = \sum^\infty {\left\lvert {a_{n}} \right\rvert}^2\).
        
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Operator Norms

    

      Let \(L:X \to {\mathbb{C}}\) be a linear functional, then the following are equivalent:
    

  
    
    
      
        
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          \(L\) is continuous
        
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          \(L\) is continuous at zero
        
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          \(L\) is bounded, i.e. \(\exists c\geq 0\) such that \({\left\lvert {L(x)} \right\rvert} \leq c {\left\lVert {x} \right\rVert}\) for all \(x\in H\)
        
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      \(2 \implies 3\): Choose \(\delta < 1\) such that 

\begin{align*}
{\left\lVert {x} \right\rVert} \leq \delta \implies {\left\lvert {L(x)} \right\rvert} < 1.
\end{align*}
 Then 

\begin{align*}
{\left\lvert {L(x)} \right\rvert} 
&= {\left\lvert {L\left( \frac{{\left\lVert {x} \right\rVert}}{\delta} \frac{\delta }{{\left\lVert {x} \right\rVert}} x \right)} \right\rvert} \\
&= \frac{{\left\lVert {x} \right\rVert}}{\delta} ~{\left\lvert {L\left( \delta \frac{x }{{\left\lVert {x} \right\rVert}} \right)} \right\rvert} \\
&\leq \frac{{\left\lVert {x} \right\rVert}}{\delta} 1
,\end{align*}
 so we can take \(c = \frac 1 \delta\). \(\hfill\blacksquare\)
    

  
    

      \(3 \implies 1\):
    

  
    

      We have \({\left\lvert {L(x-y)} \right\rvert} \leq c{\left\lVert {x-y} \right\rVert}\), so given \(\varepsilon \geq 0\) simply choose \(\delta = \frac \varepsilon c\).
    

  

    

      If \(H\) is a Hilbert space, then \((H {}^{ \vee }, {\left\lVert {{-}} \right\rVert}_{\text{op}})\) is a normed space.
    

  

    

      The only nontrivial property is the triangle inequality, but 

\begin{align*}
{\left\lVert {L_{1} + L_{2}} \right\rVert}_{^{\operatorname{op}}} = \sup {\left\lvert {L_{1}(x) + L_{2}(x)} \right\rvert} \leq \sup {\left\lvert {L_{1}(x)} \right\rvert} + {\left\lvert {\sup L_{2}(x)} \right\rvert} = {\left\lVert {L_{1}} \right\rVert}_{^{\operatorname{op}}}+ {\left\lVert {L_{2}} \right\rVert}_{^{\operatorname{op}}}
.\end{align*}

    

  

    

      If \(X\) is a normed vector space, then \((X {}^{ \vee }, {\left\lVert {{-}} \right\rVert}_{\text{op}})\) is a Banach space.
    

  

    
    
      
        
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      Let \(\left\{{L_{n}}\right\}\) be Cauchy in \(X {}^{ \vee }\).
    
    
  
        
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      Then for all \(x\in C\), \(\left\{{L_{n}(x)}\right\} \subset {\mathbb{C}}\) is Cauchy and converges to something denoted \(L(x)\).
    
    
  
        
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      Need to show \(L\) is continuous and \({\left\lVert {L_{n} - L} \right\rVert} \to 0\).
    
    
  
        
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      Since \(\left\{{L_{n}}\right\}\) is Cauchy in \(X {}^{ \vee }\), choose \(N\) large enough so that 

\begin{align*}
n, m \geq N \implies {\left\lVert {L_{n} - L_{m}} \right\rVert} < \varepsilon \implies {\left\lvert {L_{m}(x) - L_{n}(x)} \right\rvert} < \varepsilon \quad \forall x {~\mathrel{\Big\vert}~}{\left\lVert {x} \right\rVert} = 1
.\end{align*}

    
    
  
        
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      Take \(n\to \infty\) to obtain 

\begin{align*}m \geq N 
&\implies {\left\lvert {L_{m}(x) - L(x)} \right\rvert} < \varepsilon \quad \forall x {~\mathrel{\Big\vert}~}{\left\lVert {x} \right\rVert} = 1\\
&\implies {\left\lVert {L_{m} - L} \right\rVert} < \varepsilon \to 0
.\end{align*}

    
    
  
        
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      Continuity: 

\begin{align*}
{\left\lvert {L(x)} \right\rvert} &= {\left\lvert {L(x) - L_{n}(x) + L_{n}(x)} \right\rvert} \\
&\leq  {\left\lvert {L(x) - L_{n}(x)} \right\rvert} + {\left\lvert {L_{n}(x)} \right\rvert} \\
&\leq \varepsilon {\left\lVert {x} \right\rVert} + c{\left\lVert {x} \right\rVert} \\
&= (\varepsilon + c){\left\lVert {x} \right\rVert} \hfill\blacksquare
.\end{align*}

    
    
  
        
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