Properties of A Uniformly Convergent Sequence of Functions

Section 2.2 Properties of A Uniformly Convergent Sequence of Functions

Theorem 2.2.1. The Limit of A Uniformly Convergent Sequence of Continuous Functions is Continuous.

Suppose that $\{f_n(\bx)\}$ is a sequence of functions defined on a set $E$ which converges uniformly on $E$ to a limit function $f(\bx)$ on $E\text{,}$ and that $\bx_0\in E$ is such that $\lim_{\bx\in E, \bx \to \bx_0} f_n(\bx) :=L_n$ exists for each $n\text{.}$ Then both $\lim_{\bx\in E, \bx \to \bx_0} f(\bx)$ and $\lim_{n\to \infty}L_n$ exist and equal each other. Namely

\begin{equation*} \lim_{\bx\in E, \bx \to \bx_0} \lim_{n\to \infty} f_n(\bx)=\lim_{n\to \infty} \lim_{\bx\in E, \bx \to \bx_0} f_n(\bx)\text{.} \end{equation*}

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As a consequence, suppose that $\{f_n(\bx)\}$ is a sequence of continuous functions defined on a set $E$ which converges uniformly on $E$ to a limit function $f(\bx)$ on $E\text{,}$ then $f(\bx)$ is continuous on $E\text{.}$

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Proof.

We will first show that $\{L_n\}$ is Cauchy as $n\to\infty$ and that $\{f(\bx)\}$ is Cauchy as $\bx \to \bx_0\text{.}$ The latter means that $\epsilon > 0\text{,}$ there exists a neighborhood of $B(\bx_0)$ of $\bx_0$ such that $\vert f(\bx)-f(\by)\vert \lt \epsilon$ for all $\bx, \by \in B(\bx_0)\text{.}$

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For any $\epsilon \gt 0\text{,}$ by the uniform convergence assumption, there exists some $N\in \bbN$ such that

\begin{equation*} \vert f_n(\bx)-f(\bx)\vert \lt \epsilon \text{ and $ \vert f_n(\bx)-f_N(\bx)\vert \lt \epsilon$ for all $x\in E, n\ge N$}\text{.} \end{equation*}

Using $\lim_{\bx\in E, \bx \to \bx_0} f_N(\bx) :=L_N\text{,}$ there exists some neighborhood $B(\bx_0)$ of $\bx_0$ such that

\begin{equation*} \vert f_N(\bx)-L_N \vert\lt \epsilon \text{ for all $x\in B(\bx_0)$.} \end{equation*}

Then for all $x\in B(\bx_0), n\ge N\text{,}$ we have

\begin{equation*} \vert f_n(\bx)-L_N \vert\le \vert f_n(\bx)-f_N(\bx) \vert+ \vert f_N(\bx)-L_N \vert \lt 2\epsilon. \end{equation*}

Using $\lim_{\bx\in E, \bx \to \bx_0} f_n(\bx) :=L_n$ we obtain

\begin{equation*} \vert L_n-L_N \vert\le 2\epsilon \text{ for all $n\ge N$.} \end{equation*}

This shows that $\{L_n\}$ is Cauchy as $n\to\infty\text{,}$ so has a limit. Call it $L\text{.}$

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In the same setting, for $\bx, \by \in B(\bx_0)\text{,}$

\begin{align*} \vert f(\bx)-f(\by)\vert \amp \le \vert f(\bx)-f_N(\bx)\vert + \vert f_N(\bx)-f_N(\by)\vert + \vert f_N(\by)-f(\by)\vert\\ \amp \le 2\epsilon + \vert f_N(\bx)-f_N(\by)\vert. \end{align*}

But $\vert f_N(\bx)-f_N(\by)\vert \le \vert f_N(\bx)-L_N\vert +\vert L_N-f_N(\by)\vert\le 2\epsilon\text{,}$ so we have $\vert f(\bx)-f(\by)\vert \le 4\epsilon\text{,}$ for $\bx, \by \in B(\bx_0)\text{,}$ which shows that $\{f(\bx)\}$ is Cauchy as $\bx \to \bx_0$ and $\lim_{\bx \in E, \bx\to \bx_0} f(\bx)$ exists.

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Now for any $n\ge N\text{,}$ sending $\bx\to \bx_0$ in $\vert f_n(\bx)-f(\bx)\vert \lt \epsilon\text{,}$ we get

\begin{equation*} \vert L_n-\lim_{\bx \in E, \bx\to \bx_0}f(\bx)\vert \le \epsilon. \end{equation*}

Then sending $n\to \infty$ in the above we get

\begin{equation*} \vert \lim_{n\to \infty} L_n-\lim_{\bx \in E, \bx\to \bx_0}f(\bx)\vert \le \epsilon. \end{equation*}

Since this holds for any $\epsilon \gt 0\text{,}$ we conclude that $\lim_{n\to \infty} L_n=\lim_{\bx \in E, \bx\to \bx_0}f(\bx)\text{.}$

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In some contexts (often when dealing with series of functions), the limit function $f(x)$ in the definition of uniform convergence may not be known; then we have the following

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Theorem 2.2.2. Cauchy Criterion for Uniform Convergence.

Suppose that $\{f_n(\bx)\}$ is a sequence of functions defined on a set $E\text{.}$ Suppose that for any $\epsilon \gt 0\text{,}$ there exists $N\in \bbN$ such that for any $n, m\ge N$

\begin{equation*} \vert f_n(\bx)-f_m(\bx)\vert \lt \epsilon \text{ for all $x\in E$,} \end{equation*}

then $\{f_n(\bx)\}$ converges uniformly on $E\text{.}$

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Suppose that $\sum_{n=1}^{\infty}a_n(x)$ is a series of functions defined on a set $E$ and that $\vert a_n(x)\vert \le c_n(x)$ for all $x\in E$ and all $n\text{.}$ Suppose further that the series $\sum_{n=1}^{\infty}c_n(x)$ converges uniformly on $E\text{.}$ Then the series $\sum_{n=1}^{\infty}a_n(x)$ converges uniformly on $E\text{.}$

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Remark 2.2.3.

Theorem 2.2.1 implies that the answer to question (a) above would be affirmative, if we assume that $f_{n}(x)\to f(x)$ uniformly over $E\text{.}$

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One often cannot apply Theorem 2.2.1 directly on the entire interval on which the question is being asked. For example, we know that the sequence of functions $s_n(x)=\sum_{m=1}^n \frac{x^m}{m}$ converges for every $x\in (-1,1)\text{,}$ but does not converge uniformly on $(-1,1)$ (can you provide a proof?). Using the Cauchy Criterion for Uniform Convergence we can also see that, for any $0< r <1\text{,}$ the sequence of $s_n(x)=\sum_{m=1}^n \frac{x^m}{m}$ converges uniformly on $[-r,r]\text{,}$ therefore, making the limit $\sum_{m=1}^{\infty} \frac{x^m}{m}$ a continuous function on $(-r,r)\text{.}$ Since $r$ is arbitrary subject to $0< r <1\text{,}$ and any $x_0\in (-1,1)$ can be included in some $(-r,r)$ with $0< r <1\text{,}$ this shows that the limit $\sum_{m=1}^{\infty} \frac{x^m}{m}$ a continuous function on $(-1, 1)\text{.}$ (The limit $\sum_{m=1}^{\infty} \frac{x^m}{m}$ fails to be a uniformly continuous function on $(-1,1)\text{,}$ even though each $s_n(x)$ is uniformly continuous on $(-1,1)\text{.}$)

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Exercise 2.2.4.

Does the series $\sum_{m=1}^{\infty} m x^{m}$ converge uniformly over $(-1, 1)\text{?}$ Does it define a continuous function on $(-1, 1)\text{?}$

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Exercise 2.2.5.

Is it true that if a sequence of continuous functions $f_{n}$ on $E$ converges uniformly to $f$ on $E\text{,}$ then $f$ is bounded on $E\text{?}$

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Remark 2.2.6.

Although the $M$-test is an often used tool to check for uniform convergence, there are cases where the $M$-test doe not apply directly, as in the case of

\begin{equation*} \sum_{k=1}^{\infty} \frac{ \sin (kx)}{k}\text{,} \end{equation*}

as this series only converges conditionally. Here one needs to check directly whether the Cauchy Criterion for Uniform Convergence holds for the partial sums of this series, namely, we need to examine whether, for any $\epsilon>0\text{,}$ we can find $N$ such that for all $n>m\ge N\text{,}$ we have

\begin{equation*} |\frac{ \sin (m+1)x }{m+1} + \cdots + \frac{\sin nx}{n}| < \epsilon \quad \text{ for all } x \text{ in a specified set}? \end{equation*}

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This series can be analyzed by applying the summation by parts via Abel’s Lemma. Set $s_m(x)= \sum_{k=1}^{m} \sin (kx)\text{.}$ Then, using that each $\sin (kx)$ is the imaginary part of $e^{ikx}$ and a simple formula for the sum of the latter, we have

\begin{equation*} s_m(x)=\frac{\cos \frac x2 -\cos \left(m+\frac 12\right) x}{2\sin \frac x2} = \frac{\sin \frac{mx}{2} \sin \frac{(m+1)x}{2}}{\sin \frac x2}\text{.} \end{equation*}

Note that $s_m(\frac 1m)/m \to 2\sin^2\frac 12\text{,}$ so $s_m(x)$ does not remain bounded uniformly in $m$ near $x=0\text{.}$ However, for any $\pi>\delta>0\text{,}$ when $x$ is restricted to $\left\{x\in [-\pi, \pi]: 0 < \delta < |x|\right\}\text{,}$ there exists some $C>0$ depending on $\delta$ such that $\vert s_m(x)\vert \le C$ uniformly in $m\text{.}$ Then for such $x$ the Abel summation-by-parts formula gives

\begin{align*} \amp |\frac{ \sin (m+1)x }{m+1} + \cdots + \frac{\sin nx}{n}| \\ =\amp \left| \frac{ s_{m+1}(x)-s_m(x)}{m+1}+ \cdots + \frac{ s_n(x)-s_{n-1}(x)}{n}\right| \\ =\amp \left| s_{m+1}(x)\left(\frac{1}{m+1}-\frac{1}{m+2}\right) +\cdots+ s_{n-1}(x)\left(\frac{1}{n-1}-\frac{1}{n}\right)\right| \\ \amp + \left|\frac{s_n(x)}{n}-\frac{s_m(x)}{m+1}\right|\\ \le \amp C\left( \frac{1}{m+1}-\frac{1}{m+2} + \frac{1}{m+2}-\frac{1}{m+3} + \cdots + \frac{1}{n-1}-\frac{1}{n} + \frac 1n +\frac {1}{m+1}\right)\\ \le \amp \epsilon\text{,} \end{align*}

provided $n > m\ge N$ for some sufficiently large $N$ depending on $\epsilon$ and $C$ (therefore depending on $\delta$). Here, a crucial property used is the monotonicity of $\frac 1k\text{,}$ which is used in

\begin{equation*} \left| \frac{1}{k+1}-\frac{1}{k+2}\right|=\frac{1}{k+1}-\frac{1}{k+2} \end{equation*}

which then allows for telescoping. This verifies the Cauchy Criterion for Uniform Convergence on $\left\{x\in [-\pi, \pi]: 0 < \delta < |x|\right\}\text{.}$

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Since this series is $2\pi$ periodic, so the result on $[-\pi, \pi]$ can be extended to $\bbR$ by using the periodic property.

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Theorem 2.2.7. Interchange of Limits and Integration.

Suppose that $\{f_n(x)\}$ is a sequence of Riemann-Stieltjes integrable functions with respect to $\alpha$ on $(a, b)$ with $\alpha(b)-\alpha(a)$ finite, and that $f_n(x)\to f(x)$ uniformly over $(a, b)\text{.}$ Then $f(x)$ is Riemann-Stieltjes integrable functions with respect to $\alpha$ on $(a, b)$ and

\begin{equation} \int_a^b f(x)\, d\alpha(x) =\lim_{n\to \infty} \int_a^b f_n(x)\, d\alpha(x).\tag{2.2.1} \end{equation}

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Suppose that $\{a_n(x)\}$ is a sequence of Riemann-Stieltjes integrable functions with respect to $\alpha$ on $(a, b)$ with $\alpha(b)-\alpha(a)$ finite, and that the series $\sum_{n=1}^{\infty}a_n(x)$ converges uniformly over $(a, b)\text{.}$ Then $\sum_{n=1}^{\infty}a_n(x)$ is Riemann-Stieltjes integrable functions with respect to $\alpha$ on $(a, b)$ and

\begin{equation*} \int_a^b \left( \sum_{n=1}^{\infty}a_n(x) \right)\, d\alpha(x)= \sum_{n=1}^{\infty} \int_a^b a_n(x)\, d\alpha(x). \end{equation*}

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Proof.

For simplicity of presentation we will assume that $\alpha(x)$ is monotone increasing, and set $L=\alpha(b)-\alpha(a)\text{.}$ For any $\epsilon \gt 0\text{,}$ we first use the uniform convergence of $\{f_n(x)\}$ to $f(x)$ on $(a, b)$ to find some $N$ such that for all $n\ge N$ and all $x\in (a, b)\text{,}$ $\vert f_n(x)-f(x)\vert \lt \epsilon/L\text{.}$ So for any partition $\cP$ of $[a, b]$ and any interval $I\in \cP\text{,}$ we have

\begin{equation*} f_n(x) \le \vert f_n(x)-f(x)\vert +f(x) \lt \epsilon/L +M_f(I) \end{equation*}

for all $x\in I\text{,}$ from which it follows that

\begin{equation*} U(f_n, \cP, \alpha)\le U(f, \cP, \alpha)+\epsilon. \end{equation*}

Similarly, we have $U(f, \cP, \alpha)\le U(f_n, \cP, \alpha)+\epsilon\text{,}$ so

\begin{equation*} \vert U(f, \cP, \alpha)- U(f_n, \cP, \alpha)\vert \le \epsilon \quad \forall n\ge N \end{equation*}

and

\begin{equation*} \vert L(f, \cP, \alpha)- L(f_n, \cP, \alpha)\vert \le \epsilon \quad \forall n\ge N. \end{equation*}

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Next, for any $n\ge N\text{,}$ we find a partition $\cP_n$ of $[a, b]$ such that

\begin{equation*} \int_a^b f_n(x)\, d\alpha(x)-\epsilon \le L(f_n, \cP_n, \alpha) \le U(f_n, \cP_n, \alpha)\le \int_a^b f_n(x)\, d\alpha(x) +\epsilon. \end{equation*}

Now we find

\begin{equation*} L(f_n, \cP_n, \alpha)-\epsilon \le L(f, \cP_n, \alpha)\le U(f, \cP_n, \alpha)\le U(f_n, \cP_n, \alpha)+\epsilon, \end{equation*}

which, when combined with the above estimates, gives

\begin{equation*} \int_a^b f_n(x)\, d\alpha(x)-2\epsilon \le L(f, \cP_n, \alpha)\le U(f, \cP_n, \alpha)\le \int_a^b f_n(x)\, d\alpha(x) +2\epsilon. \end{equation*}

This shows that

\begin{equation*} U(f, \cP_n, \alpha)-L(f, \cP_n, \alpha) \lt 4\epsilon \text{ so $\int_a^b f(x)\, d\alpha(x)$ exists;} \end{equation*}

and since $L(f, \cP_n, \alpha) \le \int_a^b f(x)\, d\alpha(x) \le U(f, \cP_n, \alpha)\text{,}$ we have

\begin{equation*} \vert \int_a^b f(x)\, d\alpha(x) - \int_a^b f_n(x)\, d\alpha(x)\vert \lt 2\epsilon \quad \forall n\ge N. \end{equation*}

This shows that (2.2.1).

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The case for the series follows from the above by applying it to the sequence of partial sums of the series.

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Remark 2.2.8.

The formulation here allows either $a$ or $b$ to be infinite, as long as there is a finite variation $\alpha(b)-\alpha(a)$ over $(a, b)\text{.}$ For instance, for $a\gt 0\text{,}$ the integral $\int_a^{\infty}f_n(x)\, dx$ could be formulated as $\int_a^{\infty}f_n(x)x^p\, d\alpha(x)\text{,}$ with $\alpha(x)=(1-p)^{-1}x^{1-p}\text{.}$ If for some $p\gt 1\text{,}$

\begin{equation} f_n(x)x^p\to f(x)x^p \text{ uniformly over $(a, \infty)$}\text{,}\tag{2.2.2} \end{equation}

we can conclude that

\begin{equation*} \int_a^{\infty}f_n(x)\, dx= \int_a^{\infty}f_n(x)x^p\, d\alpha(x)\to \int_a^{\infty}f(x)x^p\, d\alpha(x)=\int_a^{\infty}f(x)\, dx. \end{equation*}

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Often one needs some modification to (2.2.2). Suppose that for any finite $b \gt a\text{,}$ $f_n(x)\to f(x)$ uniformly over $[a, b]\text{,}$ and that for some $C\gt 0, q\gt 1\text{,}$ we have

\begin{equation*} \vert f_n(x)\vert \le C x^{-q} \text{ for all $x\in (a, \infty)$ and all $n$.} \end{equation*}

Then, it follows that $\vert f(x)\vert \le C x^{-q}$ for all $x\in (a, \infty)\text{,}$ and that for some fixed $p, q\gt p \gt 1\text{,}$

\begin{equation*} \vert f_n(x)-f(x)\vert x^p\le 2C x^{p-q} \end{equation*}

So for any $\epsilon \gt 0\text{,}$ we can find some $b'\gt a$ depending on $\epsilon$ such that $\vert f_n(x)-f(x)\vert x^p \le \epsilon$ for all $x\ge b'$ and all $n\text{.}$ But this is not quite the same as (2.2.2). We next discussion an extension to address such scenarios.

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Exercise 2.2.9.

Identify the pointwise limit $f(x)$ of the sequence $\{ \frac{nx}{1+n^{2}x^{2}}\} $ for $x\in [0, 1]\text{.}$ Does it converge uniformly over $x\in [0, 1]\text{?}$ Does it holds that $\int_{0}^{1} \frac{nx}{1+n^{2}x^{2}} \, dx \to \int_{0}^{1} f(x)\, dx\text{?}$

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Exercise 2.2.10.

Identify the pointwise limit $f(x)$ of the sequence $\{ \frac{n}{n^{2}+x^{2}}\}$ for $x\in [0, \infty)\text{.}$ Does it converge uniformly over $x\in [0, \infty)\text{?}$ Does it holds that $\int_{0}^{\infty }\frac{n}{n^{2}+x^{2}}\, dx \to \int_{0}^{\infty } f(x)\, dx\text{?}$

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