Show That the Series Converges Uniformly to a Continuous Function

A sequence $\{f_n\}_{n=1}^{\infty}$ of real-valued functions on a set $X$ , $f_n : X\to \mathbb{R}$ , is said to be uniformly convergent on $X$ to a limit function $f:X\to \mathbb{R}$ if for each $\epsilon > 0$ there exists an $N\in \mathbb{N}$ such that

$|f(x) - f_n(x)|< \epsilon$

holds for all $n\geq N$ and $x\in X$ .

The crucial condition which distinguishes uniform convergence from pointwise convergence of a sequence of functions is that the number $N$ in the definition depends only on $\epsilon$ and not on $x$ . It follows that every uniformly convergent sequence of functions is pointwise convergent to the same limit function, thus uniform convergence is stronger than pointwise convergence.

The definition of the uniform convergence is equivalent to the requirement that

$\displaystyle\lim_{n\to \infty} \displaystyle\sup_{x\in X} |f(x) - f_n(x)| = 0 .$

If we denote by $||g||_{\infty}:= \displaystyle\sup_{x\in X}|g(x)|$ the supremum norm of a function $g:X\to \mathbb{R}$ , the last condition becomes

$\displaystyle\lim_{n\to \infty} ||f-f_n||_{\infty} = 0 .$

Find a sequence of functions which converges pointwise but not uniformly.

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Let $f_n(x) = x^n$ be a sequence of functions $[0,1]\to \mathbb{R}$ . Then $\displaystyle\lim_{n\to \infty} f_n(x) = \begin{cases} 0, & x\in [0,1) \\ 1, & x = 1 \\ \end{cases}$
so the sequence $\{f_n\}$ converges pointwise to the function $f(x) = \begin{cases} 0, & x\in [0,1) \\ 1, & x = 1 \\ \end{cases}$

However, given an $\epsilon > 0$ , for $x\in[0,1)$ we have $|f_n(x) - f(x)| = |x^n| < \epsilon \iff 0\leq x < \epsilon^{\frac{1}{n}}< 1$ so there can not exist an $N\in \mathbb{N}$ for which the estimate will hold uniformly (that is, for all $x\in [0,1]$ ) and the convergence is not uniform. The last argument can be replaced by observing that $||f-f_n||_{\infty} = \displaystyle\sup_{x\in [0,1]} |f(x) - f_n(x)| = 1$ for all $n\in \mathbb{N}$ (for $x$ arbitrary close to $1$ we have that $|x^n|$ is arbitrary close to 1). This provides a quicker way to conclude that the convergence is not uniform, since $\displaystyle\lim_{n\to \infty} ||f-f_n||_{\infty} = 1 \neq 0$ .

The following theorem shows that the uniform convergence behaves well under addition and multiplication:

Let $f_n,g_n : X\to \mathbb{R}$ be two sequences of functions $\{f_n\}_{n=1}^{\infty}, \{g_n\}_{n=1}^{\infty}$ , respectively, which converge uniformly to functions $f:X\to \mathbb{R}$ and $g:X\to \mathbb{R}$ on $X$ . Then:

• the sequence $\{f_n + g_n\}_{n=1}^{\infty}$ converges uniformly to $f+g$ on $X$

• if $||f||_{\infty}, ||g||_{\infty}<\infty$ then the sequence $\{f_n \cdot g_n\}_{n=1}^{\infty}$ converges uniformly to $f\cdot g$ on $X$

The notion of uniform convergence extends to series of functions:

A series $\displaystyle\sum_{n=1}^{\infty} f_n(x)$ , with $f_n:X\to \mathbb{R}$ is said to converge uniformly to a function $f:X\to \mathbb{R}$ on $X$ if the sequence $\{s_n\}_{n=1}^{\infty}$ of partial sums, $s_n(x) = f_1(x) + \ldots + f_{n}(x)$ converges uniformly to $f$ .

Fortunately, there exists a criterion for the uniform convergence of series which is easy to apply in practice:

Weierstrass M-test

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Let $\{f_n\}_{n=1}^{\infty}$ be a sequence of real valued functions on a set $X$ such that there is a sequence $M_n$ of positive numbers satisfying $|f_n(x)|\leq M_n \ \forall n\in\mathbb{N}, x\in X$ If $\displaystyle\sum_{n=1}^{\infty} M_n < \infty$ , then the series $\displaystyle\sum_{n=1}^{\infty} f_n(x)$ converges uniformly on $X$ .

Riemann Zeta function

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For $x>1$ let $\zeta(x):= \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^x}$ denote the Riemann Zeta function on $(1,\infty)$ . Fix $a>1$ and consider the Zeta function on $[a,\infty)$ as a series $\displaystyle\sum_{n=1}^{\infty} f_n(x)$ of functions $f_n(x) = \frac{1}{n^x}$ on $[a,\infty)$ . For each $n\in \mathbb{N}$ , $x>a$ implies $|f_n(x)|=\frac{1}{n^x}\leq \frac{1}{n^a}$ and the series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^a}$ converges so the Weierstrass M-test implies that $\zeta(x)$ is uniformly convergent on $[a,\infty)$ .

When $[0,1]$ and $\mathbb{R}$ are equipped with their natural topologies, the sequence of functions $f_n(x): [0,1]\to \mathbb{R}, f_n(x):= x^n$ consists of continuous functions while their pointwise limit $f(x) = \begin{cases} 0, & x\in [0,1) \\ 1, & x = 1 \\ \end{cases}$ is discontinuous. However, if the convergence is uniform such phenomena do not appear:

Uniform limit theorem

Let $X$ be a topological space and $\mathbb{R}$ equipped with the natural topology. If a sequence $\{f_n\}_{n=1}^{\infty}$ of continuous functions $f_n: X\to \mathbb{R}$ converges uniformly to a function $f:X\to \mathbb{R}$ , then $f$ is continuous.

Restricting the sequence of functions $f_n:= x^n$ on the set $(0,1)$ produces the continuous pointwise (but not uniform) limit $f(x)\equiv 0$ on $(0,1)$ , which shows that the continuity of the limit function is not a sufficient condition to imply the uniform convergence of a sequence of continuous functions. There is, however, a partial result in this direction, Dini's theorem which, under additional assumptions that the underlying space $X$ is compact and that the sequence $\{f_n\}_{n=1}^{\infty}$ is monotone, shows that the continuity of the limit function implies uniform convergence.

Riemann Zeta function is continuous on $(1,\infty)$

$\zeta(x)$ is the uniformly convergent limit of its partial sums on each $[a,\infty)$ , $a>1$ , which are continuous functions. Hence, by the Uniform limit theorem $\zeta(x)$ is continuous at each $a>1$ .

Consider the functions $f_n: [0,\infty]\to \mathbb{R}$ defined by $f_n(x):= min\{n,x\}$ . For every $x\in [0, \infty]$ there is a natural number $N\in \mathbb{N}$ such that $f_n(x)=x$ for each $n\geq N$ (take $N>x$ ). It follows that the pointwise limit of $\{f_n\}$ is the function $f:[0,\infty]$ given by $f(x)=x$ . Functions $f_n$ are all bounded functions ( $0\leq f_n(x)\leq n \ \forall x\in[0,\infty]$ ) but the limit function $f$ is unbounded. In the case of uniform convergence of bounded functions, the limit function is again bounded:

Uniform convergence of bounded functions

If a sequence $\{f_n\}$ of bounded functions $f_n: X\to \mathbb{R}$ converges uniformly to a function $f:X\to \mathbb{R}$ , then $f$ is bounded.

Geometric series

For $x\in (-1,1)$ the series $\displaystyle\sum_{n=0}^{\infty} x^n$ converges pointwise to the function $f(x) = \frac{1}{1-x}$ . The partial sums $s_n(x) = 1 + x+ \ldots + x^n$ are bounded functions while the limit function $f(x)$ is unbounded on $(-1,1)$ . Hence the geometric series does not converge uniformly on $(-1,1)$ .

It can however be shown that the geometric series converges uniformly on each interval $[-a,a]$ for $a\in (0,1)$ .

As a countable set, $\mathbb{Q}\cap [0,1]$ can be enumerated by $\{q_i\}_{i=1}^{\infty}$ . Let $f_n:[0,1]\to \mathbb{R}$ be the indicator function of the set $\{q_1,q_2,\ldots, q_n\}$ , that is $f_n(x) = \begin{cases} 1, & x = q_1,q_2,\ldots, q_n \\ 0, & otherwise \\ \end{cases}$ Each $f_n$ is Riemann integrable as a bounded function with finitely many discontinuities on $[0,1]$ . However, the pointwise limit of $\{f_n\}_{n=1}^{\infty}$ is the indicator function of $\mathbb{Q}\cap [0,1]$ , which is known to be non Riemann integrable.

In the example above the pointwise limit of Riemann integrable functions turned out to be non Riemann integrable. With the assumption of uniform convergence the following theorem holds:

Uniform convergence and Riemann integrability

Let $f_n:[a,b]\to \mathbb{R}$ be a sequence of Riemann integrable functions on $[a,b]$ converging uniformly to $f:[a,b]\to \mathbb{R}$ . Then $f$ is Riemann integrable and $\displaystyle\lim_{n\to \infty} \displaystyle\int_{a}^{b} f_n(x) dx = \displaystyle\int_{a}^{b} f(x) dx.$

In particular, if the series $\displaystyle\sum_{n=1}^{\infty} f_n(x)$ converges uniformly, then

$\displaystyle\int_{a}^{b} \displaystyle\sum_{n=1}^{\infty} f_n(x) dx = \displaystyle\sum_{n=1}^{\infty} \displaystyle\int_{a}^{b} f_n(x) dx.$

An expansion for the logarithmic function

As stated above, the geometric series is uniformly convergent in each interval $[-a,a]$ , $a\in (0,1)$ . Applying this to the Riemann integral we can interchange the summation and the integral in the following calculation:

$ln(\frac{1}{1-a}) = \displaystyle\int_{0}^{a} \frac{1}{1-x} dx = \displaystyle\int_{0}^{a} \displaystyle\sum_{n=0}^{\infty} x^n dx = \displaystyle\sum_{n=0}^{\infty} \displaystyle\int_{0}^{a} x^n dx = \displaystyle\sum_{n=0}^{\infty} \frac{a^{n+1}}{n+1}$

Weierstrass Function is continuous

In 1872, Karl Weierstrass published a paper on a function that is continuous and nowhere differentiable, thus giving a new counter-intuitive insight into the relation of continuity and differentiability. He analyzed the function

$W(x) = \displaystyle\sum_{n=0}^{\infty} a^n cos(b^n \pi x)$

and proved, under the assumptions that $a\in (0,1)$ , $b>1$ is an odd integer and $ab>1 + \frac{3}{2}\pi$ , that the function $W(x)$ is continuous and nowhere differentiable. While the proof of non differentiability is beyond the scope of this text, continuity follows directly from the Uniform limit theorem:

For all $n\in \mathbb{N}$ we have $|a^n cos(b^n \pi x)| \leq a^n$ and $\displaystyle\sum_{n=0}^{\infty} a^n < \infty$ for $a\in (0,1)$ so, by the Weierstrass M-Test the defining series for $W(x)$ is uniformly convergent. Functions $a^n cos(b^n \pi x), n\in \mathbb{N}$ are continuous and so are the partial sums of the series for $W(x)$ . By the Uniform limit theorem, $W(x)$ is continuous.

Note: this example also shows that the uniform limit of differentiable functions need not be differentiable.

For $n\in \mathbb{N}$ , let $f_n:[0, \pi]\to \mathbb{R}$ be defined by

$f_n(x):= \sin^n (x)$

Is the sequence $\{f_n\}_{n=1}^{\infty}$ uniformly convergent?

Hint: is the limit function continuous?

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This problem is often posed in calculus textbooks.

(B) only (B) and (C) only (A) only (C) only

Which of the following sequences of functions $\{f_n\}_{n=1}^{\infty}$ is/are uniformly convergent on its/their domain?

(A): $f_n(x) = \cos^{n}(x)$ on $\left[-\dfrac \pi2, \dfrac \pi2\right]$ .

(B): $f_n(x) = \dfrac{x}{n}$ on $\mathbb{R}$ .

(C): $f_n(x) = \dfrac{\sin(nx)}{n}$ on $\mathbb{R}$ .

Notation: $\mathbb R$ denotes the set of real numbers.

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Source: https://brilliant.org/wiki/uniform-convergence/

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