Sequences and Series

Sequences of functions

definition (limsup of functions):

For f:A→R, lim sup

example (?):

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

slogan:

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

proposition (The Cauchy condensation test):

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

proof (showing a useful trick):

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

proposition (Comparison Test):

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.
proposition (Sufficient condition for Taylor convergence):

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).

proposition (p-tests):

Let n be a fixed dimension and set B = \left\{{x\in {\mathbb{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*}

proposition (Small Tails for Series of Functions):

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

corollary (Term by Term Continuity Theorem):

\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*}

proposition (Cauchy criterion for sums):

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*}

theorem (Lagrange and Cauchy Remainders):

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

proposition (Testing Uniform Convergence: The Sup Norm Test):

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.

remark (Negating the Sup Norm test):

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).

proposition (C(I) is complete):

The space X = C([0, 1]), continuous functions f: [0, 1] \to {\mathbb{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.

proof:

    
    
      
        
  • 
          
    
    
      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 {\mathbb{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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remark:
    

      In other cases, you may need to show the limit is bounded, or has bounded derivative, or whatever other conditions define X.
    

  
theorem (Weierstrass Approximation):
    

      If [a, b] \subset {\mathbb{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).