Show that the normed space $\left(C^1(\mathbb T),\lVert \text{.}\rVert\right)$ where $C^1$ is the space of continuously differentiable functions, and $ \lVert f\rVert:=\lvert f(0)\rvert+\lVert f'\rVert_{L^2(\mathbb T)}$ is not a Banach space
- We define $\mathbb T:=\mathbb R/2π\mathbb Z$, the topological space of $2π$-periodic functions, in other words $\mathbb T$ can be seen as the interval $[0,2π]$ but the points $0$ and $2π$ are the same if we move leftwords to $2π.$
Edit: I've solved it in the answers.
(New approach):
To show that $C^1(\mathbb T)$ is not complete under this norm we must find a $Cauchy$ sequence $f_n$ which does not converge in this norm. We define $f_n$ by setting $f_n(0):=0$ and specifying $f_n'$ ,a continuous function, taking care that $\int_0^{2π}f_n'=0$ (otherwise $f_n$ will not be periodic). Define $f'$ on $[-π,π)$ to be the following function (where the height is $\pm1$):
Then $f'_n$ converges (in $L^2$) to the function that is $-1$ on $(-π,0)$ and $+1$ on $(0,π)$ The idea is that this is not a continuous function. Suppose then that
$f_n\rightarrow f \in C^1(\mathbb T) $ in the $\lVert $.$\rVert $ norm This implies that $f'_n \rightarrow f'$ in the $L^2$ norm. But $L^2$ limits are unique, hence $f'$ , which is a continuous function, must be equal a.e. to the function $g(x)=-\mathbb 1_{(-π,0)}(x) + \mathbb 1_{(0,π)}(x).$ But no continuous function $f'$ can do this: suppose $f'=g$ a.e. To be precise let us say that $f=g'$ except on a set $E⊆\mathbb T$ of measure $0$. Then $(f')^{-1}\big((-1,1)\big)$ is an open set and, by the indermediate value theorem, it is not empty, hence it is of positive measure. Therefore $(f')^{-1}\big((-1,1)\big)\setminus E$ is also of positive measure, hence nonempty, and $f'=g$ on this set. This is impossible since $g$ only takes the values $\pm1$.