Consider the operator $T:l^2\to L^2(\mathbb{R})$ defined by:

Question

$$T(\{a_k\})=\sum_1^\infty 2a_k\mathbb{1}_{\big[\frac{k}{4},\frac{k+1}{4}\big)}(x)$$ Show that $T$ is a linear isometry. It is a linear ishomorphism?

Don't know how to start.

Answer 1

$$\|T(\left\{a_k\right\})\|_{L^2}^2= \sum_{k=1}^{\infty}\int_{k/4}^{(k+1)/4}4a_k^2 dx= \sum_{k=1}^{\infty}\frac{1}{4}\cdot 4a_k^2=\|\left\{a_k\right\}\|_{\ell^2}^2$$ So it is an isometry. It is non surjective, because $T(\left\{a_k\right\})$ is always a function which is constant on intervals of the form $\left[\frac{k}{4},\frac{k+1}{4}\right]$, and thus a function like $f(x)=e^{-x^2}\in L^2(\mathbb{R})$ cannot be image of any $\left\{a_k\right\}$. Even more simply, if $\ell^2=\ell^2(\mathbb{N})$, we have $T(\left\{a_k\right\})(x)=0$ for all $x<0$, therefore any $L^2$ function which is non-zero on a positive measure subset of $(-\infty,0]$ cannot be in the range of $T$.

Therefore, it is not an isomorphism.