Exercise 8.15

We want to show that

I’ll show this by induction. It is true for the case $N=0$ since both sides equal 1, so assume that

Then we have

Dividing both sides by $(1-\cos <!--\; FUNCTION\; APPLICATION\; -->x)$ gives the desired result.

(a) Since $\cos <!--\; FUNCTION\; APPLICATION\; -->x\le 1$ for all $x$, we have $1-\cos <!--\; FUNCTION\; APPLICATION\; -->(N+1)x\ge 0$ and $1-\cos <!--\; FUNCTION\; APPLICATION\; -->x\ge 0$, hence $K_N(x)\ge 0$.

(b) For $n\ne 0$,

since $e^{\mathit{in\pi }}=e^{-\mathit{in\pi }}$ for all all integers $n$. Hence

and so

(c) Since $1-\cos <!--\; FUNCTION\; APPLICATION\; -->x$ monotonically decreases from 2 to 0 on $[-\pi ,0]$ and monotonically increases from 0 to 2 on $[0,\pi ]$, we have $1-\cos <!--\; FUNCTION\; APPLICATION\; -->x\ge 1-\cos <!--\; FUNCTION\; APPLICATION\; -->\delta$ on $[-\pi ,-\delta ]\cup [\delta ,\pi ]$. Also, the maximum value of $1-\cos <!--\; FUNCTION\; APPLICATION\; -->(N+1)x$ is 2, so

for $0<\delta \le |x|\le \pi$.

By (78) in the text, we have

Let $𝜀>0$. Since $f(x)$ is uniformly continuous on $[-\pi ,\pi ]$, there is a $\delta >0$ such that $|f(x-t)-f(t)|<𝜀∕2$ if $|t|\le \delta$. Also, $f(x)$ has a maximum value $M$ on $[-\pi ,\pi ]$. Using (a), (b), and (c), we have

for large enough $N$. That is, ${\sigma }_N(f;x)\to f(x)$ uniformly on $[-\pi ,\pi ]$.