Let $X=Y=C([0,1])$ be the space of continuous functions $f:[0,1]\to \mathbb{R}$. The normed space $(X,\Vert \cdot \Vert_X) $ with $\Vert f\Vert_X:=\sup_{0\leq t\leq 1}\vert f(t)\vert$ is complete and the normed space $(Y,\Vert \cdot \Vert_Y)$ with $\Vert f\Vert_Y:=\sqrt{\int_0^1\vert f(t)\vert^2dt}$ is not complete.
Let A be the bijective bounded linear operator given by the identity map\begin{equation} A:id:X\to Y\end{equation} Then I want to show that its inverse is unbounded.
The goal hier is to show that the hypothesis that the target space in the Inverse Operator Theorem must be complete. I have the operator norm \begin{equation} \Vert A^{-1}\Vert=\sup_{f\in C([0,1])}\frac{\sup_{0\leq t\leq t}\vert f(t)\vert}{\sqrt{\int_0^1\vert f(t)\vert^2dt}} \end{equation} How can I show that this is not bounded? Thank you for your help.
Take the sequence $f_n=t^n$. You will find that the ratio $\|f_n\|_X/\|f_n\|_Y\to\infty$ as $n\to\infty$.