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Let $X$ be the space of all continuously differentiable functions on $[-1,1]$, with the norm $\|f \|_\infty$. On $X$ we also define the norm $\|f \| = \|f \|_\infty + |f'(1)|$ for all $f$ in $X$.

  • If $(f_n)_n$ converges to $0$ with respect to $\|f \|$, does it follow that it converges with respect to $\|f\|_\infty$?
  • If $(f_n)_n$ converges to $0$ with respect to $\|f\|_\infty$, does it follow that it converges w.r.t. $\|f\|$?

For the first question I thought the answer is yes. And my proof is: if $f_n$ converges to $0$ w.r.t. $\|f\|$, then for every $\epsilon$, eventually $\|f_n-0 \| = \|f_n \|_{\infty} + |f'_n(1)|<\epsilon$ and since $\|f'_n(1)| \geq 0$, we have $|f_n-0|_{\infty} \leq |f_n-0| = \|f_n \|_{\infty}+ |f'_n(1)|<\epsilon$. So, $(f_n)_n$ converges wrt $\|f\|_{\infty}$. But I am not sure if it is correct.

I would be glad if you could help me and give a hint for the second one (it seems obvious but probably I am missing something).

Thank you very much. And sorry, it is an easy question but I am new to analysis.

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1 Answer 1

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For the first question, what you did is correct but you can make it a little shorter. If you have two norms $N_1$ and $N_2$ such that $N_1(f)\leq N_2(f)$ for all $f$ and a sequence $\{f_n\}$ converges to $0$ for $N_2$ then this sequence will converge to $0$ for $N_1$.

For the second, you can consider $f_n(x)=\frac{\sin nx}n$. The idea if the following: the $\lVert\cdot\rVert_{\infty}$ goes to $0$ since $\lVert f_n\rVert\leq \frac 1n$, but since $f'_n(x)=\cos (nx)$, we have $f'_n(1)=\cos n$ and the sequence $\{f_n\}$ cannot be convergent for the norm $\lVert\cdot\rVert$. Considering $\frac 1{\sqrt n}x^n$, we can see that we can have the convergence to $0$ for the uniform norm, but $f'_n(1)=\sqrt n$, hence $\lVert f_n\rVert\to +\infty$.

Since a function can oscillate between its bounds, the norm $\lVert \cdot \rVert_{\infty}$ doesn't give any information about the derivative.

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