If multiplication operator is compact?

Question

I have a question: we have a multiplication operator $M:C[0,1]\rightarrow L^1[0,1]$ which looks like this: $x(t)\rightarrow tx(t)$ . I need to know whether this operator is compact or not and why? My expirements with Riess criterion didn't give any answers to me. Hope you will help me with it.

Answer 1

The answer is no.

Fix an $\epsilon$ for which $0 < \epsilon < 1$. Consider the subspace $$ S = \{f : f(x) = 0 \text{ for } 0 \leq x \leq \epsilon\} \subset C[0,1]. $$ Note that $S$ is an invariant subspace of $M$, so that $M|_{S} : S \to S$ where $M|_{S}$ is the restriction of $f$ to $S$. Note that $M|_{S}$ is invertible and therefore cannot be compact.

It follows that $M$ cannot be compact.

Answer 2

I'm assuming you meant $C[0,1] \to C[0,1]$. Let $T$ be the operator of multiplication by $g(t) = \min(2t, 1/t)$. Then $MT$ is the identity on the subspace $V$ of $C[0,1]$ consisting of functions that are $0$ on $[0,1/2]$. That being an infinite-dimensional subspace, $MT$ can't be compact, and therefore $T$ can't be compact.

Answer 3

$\{t\exp(2i\pi n t)\}_{n\ge 0}$ doesn't have any convergent subsequence in $L^1[0,1]$.

For a continuous linear map $M:X\to Y$ on Banach spaces, the definition of compact is that $M$ sends the unit ball of $X$ to a subset of $Y$ whose closure is compact (for example this is the case with $f:\ell^1\to \ell^1, f(x)_j = \frac{x_j}{j+1}$)