# How to prove that $\int_{0}^{1}\vert\cos(nx)\vert\,dx\not\to 0$ as $n\to\infty$?

Let $$E=\{f\in\mathcal C^1[0,1]\,:\,f(0)=0\}$$ and note that $$E$$ is a linear subspace of $$\mathcal C^1[0,1]$$. Define the following norms on $$E$$ $$\|f\|_\infty=\sup\limits_{x\in[0,1]}\vert f(x)\vert\quad\text{and}\quad\|f\|_1=\int_{0}^{1}\vert f'(x)\vert\,dx$$ Prove that the norms $$f\to\|f\|_\infty\quad\text{and}\quad f\to\|f\|_1$$ are not equivalent on $$E$$.

I thought of the sequence $$(f_n)_n\subset E$$ defined by $$f_n(x)=\frac{\sin(nx)}{n}.$$ We have that $$f_n(x)\overset{u}{\longrightarrow}0$$, and therefore $$f_n\to0$$ in the $$\|\cdot\|_\infty$$ norm.

Now for the $$\|\cdot\|_1$$ norm we have $$\|f_n-0\|_1=\int_{0}^{1}\vert \cos(nx)\vert\,dx.$$ After plugging several values of $$n$$ it appears that the integral does not converge to $$0$$. In fact it seems to stay always above $$1/2$$.

However I am unable to show that the integral does not converge to $$0.$$ Intuitively I think the integral does not converge to $$0$$ because the areas do not cancel, and as $$n$$ get larger, the function $$\cos(nx)$$ has more zeros inside the interval $$[0,1]$$. But I have no idea how to make this rigorous or if this is the correct way to think about the problem.

Therefore, can anyone give me a hint on how to prove that $$\int_{0}^{1}\vert\cos(nx)\vert\,dx\not\rightarrow 0\quad\text{as}\quad n\to\infty.$$ Thank you for your time and appreciate any help and advice.

$$\textbf{Note}:$$ The beginning was added for context.

• Consider the substitution $x=\frac{u}{n}$. Jul 8 '19 at 22:03
• just out of my own curiosity: What’s $\mathcal C^1[0,1]$? Jul 9 '19 at 0:13
• @ChaseRyanTaylor It is the space of continuously differentiable functions defined on the interval $[0,1]$. Jul 9 '19 at 0:14
• math.stackexchange.com/questions/2887032/…
– zwim
Jul 9 '19 at 0:16
• @ChaseRyanTaylor $f''$ may exist indeed, for membership in the space we only require that the function be at least one-time differentiable and that it's first derivative be continuous. Jul 9 '19 at 0:19

## 2 Answers

Since $$|\cos(nx)| \le 1$$ we have $$\int_0^1 |\cos(nx)|\,dx \ge \int_0^1 \cos^2(nx)\,dx = \frac12 + \frac{\cos n\sin n}{2n} \xrightarrow{n\to\infty} \frac12$$

so $$\int_0^1 |\cos(nx)|\,dx \not\to 0$$.

$$\int_0^1|cos(nx)|dx = \frac{1}{n}\int_0^n|cos(t)|dt \geq \frac{1}{n}\sum_{{k=0}}^{[\frac{n}{2\pi}] -1} \int_{2k\pi}^{2(k+1)\pi}|cos(t)|dt =$$

$$= \frac{1}{n}\sum_{k=0}^{[\frac{n}{2\pi}] -1 }4 = \frac{4}{n}[\frac{n}{2\pi}] \geq \frac{4}{n}(\frac{n}{2\pi} -1 ) = \frac{2}{\pi} - \frac{4}{n}$$

Which for large enough $$n \in N$$ is greater than say $$\frac{1}{\pi}$$