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.


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.

  • Let \(\left\{{f_k}\right\}\) be Cauchy in \(X\).

  • 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 \(f(x) \coloneqq\lim_k f_k(x)\).

  • 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 \(\varepsilon/2\).

  • Show that \(f\in X\):

    The uniform limit of continuous functions is continuous.

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 \(C([0, 1], {\left\lVert {{-}} \right\rVert}_\infty)\).