Let $C[0,1]$ denote the vector space of all continuous functions on the interval $[0,1]$, endowed with the supremum norm, and let $D$ denote the subspace
$D:=\{x\in C[0,1]: x'' \text{ is continuous and } x(0)=x'(0)=0 \}\subset C[0,1]$.
Is it true that the linear mapping $T: D\longrightarrow C[0,1]$, given by $Tx(t):=x''(t)+x(t)$, is continuous? And does there exist the inverse $T^{-1}$? If so, what is $T^{-1}$?
I think/guess that the mapping is not continuous. Any hint/comment will be appreciated.
Best Answer
For the first question, let $x_n(t)=t^n$. Then $x_n\in D$, and $Tx=n(n-1)t^{n-2}+t^n$. So $\|x_n\|=1$ for all $n$, while $\|Tx_n\|=n(n-1)+1$ (since $Tx_n$ is increasing). So $$ \frac{\|Tx_n\|}{\|x_n\|}=n(n-1)+1, $$ and so $T$ is not bounded.
An inverse (as an unbounded operator) does exist: the initial value problem $$\tag1x''+x=f,\ x(0)=x'(0)=0$$ always has a unique solution if $f$ is continuous. So $T$ is onto. It is also one-to-one, because the homogeneous problem only has the solution $x=0$.
The linear map $T^{-1}$ maps $f$ to the solution $x$ of the IVP $(1)$. Since a fundamental set of solutions for the homogeneous problem in $(1)$ is given by $\cos t$, $\sin t$, one gets that $$ (T^{-1}f)(x)=\int_0^x(\cos t\sin x-\cos x\sin t)\,f(t)\,dt. $$ The second part of your question was asked earlier today.