Compact operator?

Question

Let $T:L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R})$ be the operator $Tf := \chi_{[0, 1]}f$. I would like to know if $f$ is compact.

Thus far I have chosen a bounded sequence $(f_k)_{k\in \mathbb{N}} \subseteq L^2(\mathbb{R})$. As $L^2(\mathbb{R})$ is a Hilbert space, (a subsequence of) $f_k$ has some weak limit $f \in L^2(\mathbb{R})$. Thus, $\left(f_k, \chi_{[0, 1]}\right)_{L^2(\mathbb{R})} \rightarrow \left(f, \chi_{[0, 1]}\right)_{L^2(\mathbb{R})}$ but I want $\left\lVert \chi_{[0, 1]} f_k - \chi_{[0, 1]}f \right\rVert_{L^2(\mathbb{R})} \rightarrow 0$ so I doubt that $T$ is compact...

Answer 1

The unit ball of $L^{2}[0,1]$ is not compact. Let $(g_n)$ be sequence in it with no convergent subsequence. Let $f_n(x)=g_n(x)$ for $ x \in [0,1]$ and $0$ for all other $x$. Then $(f_n)$ is a bounded sequence in $L^{2}(\mathbb R)$ and $(Tf_n)$ has no convergencet subsequence. Hence $T$ is not compact.