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Show that $C^1[0,1]$ is not a banach space using the closed graph theory with the maximum norm. First, look at the derivative operator: $D:C^1[0,1]\to C[0,1]$, $D(f)=f'$.

We can check that $D$ is linear and not bounded (by taking an example such as a polynomial $x^{n+1}$). Thus $D$ is not continuos. I'm not sure, if it is possible to show that $D$ has a closed graph (a linear map $T:X\to Y$ has a closed graph if $x_n\subset X$ such that $x_n\to x$ and $T_{x_n}\to y$ then $Tx=y$).So if by contradiction, we assume that $C^1[0,1]$ is banach with the sup norm, then get by the closed graph theory that $D$ is continuos, which is not true according to what we've said.

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$D$ has a closed graph: Let $u_n \to u$ and $D(u_n) = u_n' \to v$ uniformly. We want to show that $v=D(u)=u'$. To this end, notice that for all $t \in [0,1], \ \int_0^t u_n'(s) ds \to \int_0^t v(s) ds$ and thus $ u_n(t) - u_n(0) \to \int_0^t v(s) ds$. In conclusion, $u(t) =u(0) + \int_0^t v(s) ds$ and so $ u \in C^1[0,1]$ and $u'=v$.

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    If $u_n' $ converges uniformly to $v$, then so do the integrals – Evangelopoulos Foivos Dec 20 '20 at 18:34
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    See my edit, is it clear now? – Evangelopoulos Foivos Dec 20 '20 at 19:19
  • Yes,thanks! but if I want to use the sup norm, how then it can be accomplished? –  Dec 21 '20 at 08:19
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    I used the supremum norm, don't get confused by the integrals. We start with $u_n \to u$ and $u_n' \to v$ with the $\textbf{supremum} $ norm, aka uniformly. Now, it is known that if a sequence of functions $f_n$ converges uniformly to a function $f$, then we can "pass the limit under the integral". So, we infer that $\int_0^t u_n'(s) ds \to \int_0^t v(s) ds$ – Evangelopoulos Foivos Dec 21 '20 at 09:39
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    No, the supremum norm is not the same as the integral. I used the fact that if $ || f_n -f||\infty \to 0 \textbf{ then } \int_0^t f_n(s) ds \to \int_0^t f(s) ds$. (To see that this is true, notice that $ | \int_0^t (f_n(s) -f(s) )ds | \leq ||f_n-f||\infty \int_0^t 1 ds \to 0$.) – Evangelopoulos Foivos Dec 21 '20 at 11:41