Showing non-compactness of an operator on $L^2(-1,1)$ by finding a bounded sequence with no convergent subsequence

Question

I want to see if the following operator $$\begin{align}T: L^2(-1&,1)\longrightarrow L^2(-1,1), \\&Tf\longrightarrow\chi _{[0,1]}f \end{align}$$ where $\chi _{[0,1]}$ is the characteristic function of the interval $[0,1]$, is a compact operator, with respect to the canonical norm of $L^2$. My guess is that it isn't because if I start with a bounded sequence with no convergent subsequence $f_n$ , then $Tf_n$ surely won't have a convergent subsequence as well. Is that reasoning correct? And can you help me find one such sequence, as well as helping me prove that it has no convergence subsequence? I first thought of $\sin{\pi nx}$, which I think has no convergent subsequence, but don't know how to prove that claim.

Answer 1

Let's recall that a operator $T: L^2((-1,1)) \rightarrow L^2((-1,1))$ is called compact if for every bounded sequence $(\xi_n)_{n\in \mathbb{N}}\subseteq L^2((-1,1))$ the sequence $(T\xi_n)_{n\in \mathbb{N}}\subseteq L^2((-1,1))$ admits a convergent subsequence.

We start by constructing a sequence $(\varphi_n)_{n\in \mathbb{N}}\subseteq L^2([0,1))$ that does not admit any convergent subsequence in $L^2([0,1))$. For this we note that $L^2([0,1))$ is a Hilbert space and hence, Gram-Schmidt let's us construct an orthonormal set $(\varphi_n)_{n\in \mathbb{N}}$. As $\langle \varphi_n, \varphi_m \rangle_{L^2([0,1))}=\delta_{n,m}$ we get for $n\neq m$

\begin{align*} \Vert \varphi_n - \varphi_m \Vert_{L^2([0,1))}^2 &= \Vert \varphi_n \Vert_{L^2([0,1))}^2 - 2 \text{Re}(\langle \varphi_n, \varphi_m \rangle_{L^2([0,1))}) + \Vert \varphi_m \Vert_{L^2([0,1))}^2 \\ &=2. \end{align*} Thus, no subsequence of $(\varphi_n)_{n\in \mathbb{N}}$ is Cauchy and hence no subsequence is convergent.

Now we define $$ \xi_n(x)=\begin{cases} \varphi_n(x),& x\in [0,1),\\ 0,& x\in (-1,0). \end{cases} $$

We have $\Vert \xi_n \Vert_{L^2((-1,1))} = \Vert \varphi_n \Vert_{L^2([0,1))}$, thus $\xi_n\in L^2((-1,1))$. Furthermore, as $T(\xi_n)=\varphi_n$, we get that $(T(\xi_n))_{n\in \mathbb{N}}$ does not admit any convergent subsequence and hence $T$ is not compact.

If one prefers to use linear maps instead of sequences, one could also phrase the proof the following way. The maps $$ i:L^2([0,1)) \rightarrow L^2((-1,1)), \varphi \mapsto \left( x \mapsto \begin{cases} \varphi(x),& x\in [0,1),\\ 0,& x\in (-1,0) \end{cases} \right) $$ and $$ p: L^2((-1,1)) \rightarrow i:L^2([0,1)), \xi \mapsto \xi\vert_{[0,1)}$$ are bounded. Furthermore, we have $id_{L^2([0,1)} = p\circ T \circ \circ i$. However, composition of bounded linear operator with a compact operator yields a compact operator. This would imply that $id_{L^2([0,1))}$ is compact, which is a contradiction (the identity is compact iff the Hilbert space in question is finite dimensional by the argument with the orthonormal family above).