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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.

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    $\begingroup$ Consider the substitution $x=\frac{u}{n}$. $\endgroup$
    – Mindlack
    Jul 8 '19 at 22:03
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    $\begingroup$ just out of my own curiosity: What’s $\mathcal C^1[0,1]$? $\endgroup$ Jul 9 '19 at 0:13
  • $\begingroup$ @ChaseRyanTaylor It is the space of continuously differentiable functions defined on the interval $[0,1]$. $\endgroup$
    – Stackman
    Jul 9 '19 at 0:14
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    $\begingroup$ math.stackexchange.com/questions/2887032/… $\endgroup$
    – zwim
    Jul 9 '19 at 0:16
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    $\begingroup$ @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. $\endgroup$
    – Stackman
    Jul 9 '19 at 0:19
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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 $\int_0^1 |\cos(nx)|\,dx \not\to 0$.

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

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

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

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