Continuous linear operator that is NOT compact.

Question

Exercise:

Let $U:L^1(0,\infty)\to L^{\infty}(0,\infty)$, defined as \begin{equation}U(f)(x):=\int_0^x f(t)dt\end{equation} Prove that $U$ is linear, continuous but not compact.

My solutions (or what I managed to show):

$U$ is linear is quite trivial. $U$ is continuous: let $(f_n)_{n\geq 1}$ be a sequence in $L^1(0,\infty)$ converging to $f$. To show: $\lim_{n\to \infty}U(f_n)(x)=U(f)(x)$. Since $\lim_{n\to \infty}\int_0^x f_n(t)dt=\int_0^x f(t)dt$, $U$ is continuous.

To show that $U$ is no compact, I have some troubles:

My question:

I tried to start by considering $f_n:=\mathbb{1}_{[n,n+1]}$ so we have $\Vert f_n\Vert_1=1$ and I define $g_n:=U(f_n)$. Then what I want to show is that $g_n$ cannot contain any Cauchy-Subsequence. How can I show this? Thank you in advance

Answer 1

Hint: Fix any $n\in\mathbb{N}$. Then, show that for any $m>n+1$, we have $\|g_n-g_m\|_{\infty} \geq 1$.

Answer 2

Hint: $g_n$ is weak-$*$ convergent.

Answer 3

If $n > m$, then for $x\in [m+1/2,m+1]$, $$ g_n(x) = 0 \text{ but } g_m(x) = x-m \geq 1/2 $$ Hence, $$ \|g_n - g_m\| \geq 1/2 \quad\forall n\neq m $$ so it can't have a Cauchy subsequence.