Exercises

Problems to check your understanding of this chapter. The solutions are folded away — try them yourself first.

Medium

1Medium

Show that $\sum_{n=1}^{\infty}\frac{1}{n^2+x^2}$ converges uniformly on all of $\mathbb{R}$ by the M-test, and prove its sum is continuous on $\mathbb{R}$ .

\left|\frac{1}{n^2+x^2}\right| \le \frac{1}{n^2} =: M_n, \qquad \sum_{n=1}^{\infty} M_n < \infty

The Weierstrass M-Test

Show solution

For every real $x$ , $n^2+x^2 \ge n^2$ , so $\left|\frac{1}{n^2+x^2}\right| \le \frac{1}{n^2} =: M_n$ . Since $\sum M_n = \sum \frac{1}{n^2}$ converges (a $p$ -series with $p = 2$ ), the Weierstrass M-test gives uniform convergence of $\sum_{n=1}^{\infty}\frac{1}{n^2+x^2}$ on $\mathbb{R}$ .
Each term $f_n(x) = \frac{1}{n^2+x^2}$ is continuous on $\mathbb{R}$ (the denominator never vanishes). A uniform limit of continuous functions is continuous, so the sum $S(x) = \sum_{n=1}^{\infty}\frac{1}{n^2+x^2}$ is continuous on all of $\mathbb{R}$ .

Hard

2Hard

Let $s_N(x) = \sum_{n=1}^{N}(1-x)x^n$ for $N \ge 1$ . (a) Show $s_N$ converges uniformly on $[0,a]$ ( $0 < a < 1$ ) via Dini's theorem. (b) Prove it does not converge uniformly on $[0,1]$ .

s_N(x) = x(1-x^N) \uparrow x \quad (0 \le x < 1), \qquad s_N(1) = 0

Dini's Theorem

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For $x \ne 1$ , $\sum_{n=1}^{N}x^n = \frac{x(1-x^N)}{1-x}$ , so $s_N(x) = x(1-x^N)$ (the same formula holds at $x = 1$ , where every term is $0$ ). Each $s_N$ is a polynomial, hence continuous, and $s_{N+1}(x)-s_N(x) = (1-x)x^{N+1} \ge 0$ ( $0 \le x \le 1$ ), so $s_N$ is monotone increasing.
(a) For $0 \le x \le a < 1$ , $x^N \to 0$ , so $s_N(x) \to x$ . The limit $s(x) = x$ is continuous on $[0,a]$ ; a monotone increasing sequence of continuous functions converging pointwise to a continuous limit converges uniformly by Dini's theorem.
(b) The pointwise limit on $[0,1]$ is $s(x) = x$ ( $x < 1$ ), $s(1) = 0$ , which is discontinuous at $x = 1$ ( $\lim_{x\to 1^{-}}s(x) = 1 \ne 0 = s(1)$ ). A uniform limit of continuous functions must be continuous, so convergence on $[0,1]$ cannot be uniform.

3Hard

Let $K_n \ge 0$ be continuous $2\pi$ -periodic kernels satisfying: (i) $\dfrac{1}{2\pi}\int_{-\pi}^{\pi} K_n(t)\,dt = 1$ ; (ii) mass concentration: for every $\delta > 0$ , $\dfrac{1}{2\pi}\int_{\delta \le \lvert t \rvert \le \pi} K_n(t)\,dt \to 0$ as $n \to \infty$ . For $f$ continuous and $2\pi$ -periodic, prove the convolution $(f \ast K_n)(x) = \dfrac{1}{2\pi}\int_{-\pi}^{\pi} f(x-t)K_n(t)\,dt$ converges to $f$ uniformly. (The Fejér kernel satisfies these hypotheses — see the Fejér scene — so this generalizes Fejér's theorem.)

\lvert (f \ast K_n)(x) - f(x) \rvert \le \frac{1}{2\pi}\int_{-\pi}^{\pi} K_n(t)\,\lvert f(x-t) - f(x) \rvert\,dt

= \underbrace{\frac{1}{2\pi}\int_{\lvert t \rvert < \delta} \cdots}_{\text{uniform continuity}} + \underbrace{\frac{1}{2\pi}\int_{\delta \le \lvert t \rvert \le \pi} \cdots}_{\text{tail } \to 0}

Show solution

Since the mass is $1$ , write $f(x) = \dfrac{1}{2\pi}\int_{-\pi}^{\pi} f(x)K_n(t)\,dt$ ( $f(x-t)$ is well defined for $t \in [-\pi,\pi]$ because $f$ is $2\pi$ -periodic). Subtracting and taking absolute values gives, for every $x$ , \[\lvert (f \ast K_n)(x) - f(x) \rvert \le \frac{1}{2\pi}\int_{-\pi}^{\pi} K_n(t)\,\lvert f(x-t) - f(x) \rvert\,dt.\] Split the integral into $\lvert t \rvert < \delta$ and $\delta \le \lvert t \rvert \le \pi$ .
Let $\varepsilon > 0$ . As a continuous periodic function, $f$ is uniformly continuous on $\mathbb{R}$ : there is $\delta > 0$ such that $\lvert t \rvert < \delta$ implies $\lvert f(x-t) - f(x) \rvert < \varepsilon/2$ for all $x$ . Since $K_n \ge 0$ with mass $1$ , the first integral is $\le \varepsilon/2$ , independently of $x$ .
The tail integral is bounded via $\lvert f(x-t) - f(x) \rvert \le 2\lVert f \rVert_\infty$ : \[\frac{1}{2\pi}\int_{\delta \le \lvert t \rvert \le \pi} K_n(t)\,\lvert f(x-t) - f(x) \rvert\,dt \le 2\lVert f \rVert_\infty \cdot \frac{1}{2\pi}\int_{\delta \le \lvert t \rvert \le \pi} K_n(t)\,dt \to 0\] as $n \to \infty$ , by hypothesis (ii). So for large $n$ the tail is $<\varepsilon/2$ independently of $x$ . Together, $\lVert f \ast K_n - f \rVert_\infty < \varepsilon$ — uniform convergence. The Fejér kernel $K_N$ satisfies $K_N \ge 0$ , mass $1$ , and the tail bound $K_N(t) \le \dfrac{1}{N\sin^2(\delta/2)}$ on $\delta \le \lvert t \rvert \le \pi$ (the Fejér scene's problems), so Fejér's theorem is a corollary of this general result.