Sequences and Series

Sequences of functions

For \(f:A\to {\mathbf{R}}\), \begin{align*} \limsup_{x\to y} f(x) \coloneqq\lim_{{\varepsilon}\to 0} \sup f\qty{A \cap B_{\varepsilon}(y) \setminus\left\{{ y }\right\}} .\end{align*}

A series of continuous functions that does not converge uniformly but is still continuous: \begin{align*} g(x) \coloneqq\sum {1 \over 1 + n^2 x} .\end{align*}

Take \(x = 1/n^2\).

Sequences of number

\(\limsup\) is largest limit of a convergent subsequence, \(\liminf\) is the smallest.

For \(\left\{{a_k}\right\}\) is a non-increasing sequence in \({\mathbf{R}}\) then \begin{align*} \sum_{k\geq 1} a_k < \infty \iff \sum_{k\geq 1} 2^k a_{2^k}<\infty .\end{align*}

Show that \begin{align*} \sum a_k \leq \sum 2^k a_{2^k} \leq 2 \sum a_k \end{align*} using \begin{align*} \sum a_k = a_0 + a_1 + a_2 + a_3 + \cdots \leq \qty{a_1} + \qty{a_2 + a_2} + \qty {a_3 + a_3 + a_3 + a_3} + \cdots \\ \end{align*} where each group with \(a_k\) has \(2^k\) terms.

Series

If \(0\leq a_n \leq b_n\), then

  • \(\sum b_n < \infty \implies \sum a_n < \infty\), and
  • \(\sum a_n = \infty \implies \sum b_n = \infty\).

Given a point \(c\) and some \(\varepsilon>0\), if \(f \in C^\infty(I)\) and there exists an \(M\) such that \begin{align*} x \in N_\varepsilon(c) \implies {\left\lvert {f^{(n)}(x)} \right\rvert} \leq M^n \end{align*} then the Taylor expansion about \(c\) converges on \(N_\varepsilon(c)\).

Let \(n\) be a fixed dimension and set \(B = \left\{{x\in {\mathbf{R}}^n {~\mathrel{\Big\vert}~}{\left\lVert {x} \right\rVert} \leq 1}\right\}\). \begin{align*} \sum \frac 1 {n^p} < \infty &\iff p>1 \\ \int_\varepsilon^\infty \frac 1 {x^p} < \infty &\iff p>1 \\ \int_0^1 \frac 1 {x^p} < \infty &\iff p<1 \\ \int_B \frac{1}{{\left\lvert {x} \right\rvert}^p} < \infty &\iff p < n \\ \int_{B^c} \frac{1}{{\left\lvert {x} \right\rvert}^p} < \infty &\iff p > n \\ .\end{align*}

\begin{align*} \sum f_n < \infty \implies {\left\lVert {f_n} \right\rVert}_\infty \overset{n\to\infty}\longrightarrow 0 .\end{align*}

\begin{align*} f_n \text{ cts},\, {\left\lVert {\sum_{k\leq N} f_n \to F} \right\rVert}_\infty \overset{N\to\infty}\longrightarrow 0 \implies F \text{ cts} .\end{align*}

Uniformly Cauchy iff uniformly convergent, i.e. \begin{align*} {\left\lVert {f_n - f_m} \right\rVert} \overset{m, n\to \infty}\longrightarrow 0 \iff \exists f,\, {\left\lVert {f_n - f} \right\rVert} \overset{n\to\infty}\longrightarrow 0 .\end{align*}

If \(f\) is \(n\) times differentiable on a neighborhood of a point \(p\), say \(N_\delta(p)\), then for all points \(x\) in the deleted neighborhood \(N_\delta(p) - \left\{{p}\right\}\) , there exists a point \(\xi\) strictly between \(x\) and \(p\) such that \begin{align*} x \in N_\delta(p)-\left\{{p}\right\} \implies f(x) &= \sum_{k=0}^{n-1} \frac{f^{(k)}(p)}{k!}(x-p)^k + \frac{f^{(n)}(\xi)}{n!}(x-p)^n \\ \\ &= \sum_{k=0}^{n-1} \frac{f^{(k)}(p)}{k!}(x-p)^k + \int_c^x \frac{1}{n!} {\frac{\partial ^n f}{\partial x^n}\,}(t) (x-t)^n ~dt \end{align*}

Uniform Convergence

\(f_n \to f\) uniformly iff there exists an \(M_n\) such that \({\left\lVert {f_n - f} \right\rVert}_\infty \leq M_n \to 0\).

Negating: find an \(x\) which depends on \(n\) for which \({\left\lVert {f_n} \right\rVert}_\infty > {\varepsilon}\) (negating small tails) or \({\left\lVert {f_n - f_m} \right\rVert} > {\varepsilon}\) (negating the Cauchy criterion).

The space \(X = C([0, 1])\), continuous functions \(f: [0, 1] \to {\mathbf{R}}\), equipped with the norm \begin{align*} {\left\lVert {f} \right\rVert}_\infty \coloneqq\sup_{x\in [0, 1]} {\left\lvert {f(x)} \right\rvert} \end{align*} is a complete metric space. 


    
    
      
        
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      Let \(\left\{{f_k}\right\}\) be Cauchy in \(X\).
    
    
  
        
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      Define a candidate limit using pointwise convergence:
    
    
  
    
    
      Fix an \(x\); since 

\begin{align*}
{\left\lvert {f_k(x) - f_j(x)} \right\rvert}  \leq {\left\lVert {f_k - f_k} \right\rVert} \to 0
\end{align*}
 the sequence \(\left\{{f_k(x)}\right\}\) is Cauchy in \({\mathbf{R}}\). So define \(f(x) \coloneqq\lim_k f_k(x)\).
    
    
  
        
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      Show that \({\left\lVert {f_k - f} \right\rVert} \to 0\): 

\begin{align*}
{\left\lvert {f_k(x) - f_j(x)} \right\rvert} < \varepsilon ~\forall x \implies \lim_{j} {\left\lvert {f_k(x) - f_j(x)} \right\rvert} <\varepsilon ~\forall x
\end{align*}
 Alternatively, \({\left\lVert {f_k-f} \right\rVert} \leq {\left\lVert {f_k - f_N} \right\rVert} + {\left\lVert {f_N - f_j} \right\rVert}\), where \(N, j\) can be chosen large enough to bound each term by \(\varepsilon/2\).
    
    
  
        
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      Show that \(f\in X\):
    
    
  
    
    
      The uniform limit of continuous functions is continuous.
    
    
  
        
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      In other cases, you may need to show the limit is bounded, or has bounded derivative, or whatever other conditions define \(X\).
    

  

    

      If \([a, b] \subset {\mathbf{R}}\) is a closed interval and \(f\) is continuous, then for every \({\varepsilon}> 0\) there exists a polynomial \(p_{\varepsilon}\) such that \({\left\lVert {f- p_{\varepsilon}} \right\rVert}_{L^\infty([a, b])} \overset{{\varepsilon}\to 0}\to 0\).
    

  
    

      Equivalently, polynomials are dense in the Banach space \(C([0, 1], {\left\lVert {{-}} \right\rVert}_\infty)\).