$L:L^2([0,1])\rightarrow L^2([0,1])$, $Lf(x)=\frac1{x+1}f(x)$ not compact

Question

Let $L:L^2([0,1])\rightarrow L^2([0,1])$ given by $Lf(x)=\frac1{x+1}f(x)$. I want to show that $L$ is continuous but not compact.

The boundness follows by $\|Lf\|_2\leq \|f\|_2 $ since $\frac1{x+1}\leq 1$. So we get $\|L\|\leq 1$.

But why is $L$ not compact? I know an operator is called compact if every bounded sequence $x_n$ has a subsequence $x_{n_k}$ such that $Lx_{n_k}$ converges. Which sequence does fail here?

Answer 1

Note that $L$ is invertible with $L^{-1} (f) = (1+x) f$. Thus the composition $L\circ L^{-1}$ is the identity. If $L$ is compact, then so is $I$, which is not true.