Question about Compact operator and Compactness

Question

Suppose we have a finite measure space $(X,\Sigma,\mu)$ and an integral operator $T: L^p(X)\to L^q(X)$. We also assume that this operator is compact that is for every bounded sequence $\{f_n\}$ in $L^p(X)$ the sequence $\{Tf_n\}$ has a convergent subsequence in $L^q(X).$ Suppose I am saying that for a bounded set $B$ then image set $T(B)$ is compact. So now I am confused about how these two are related. I mean does the compactness of the operator $T$ implies that the compactness of the set $T(B)$ or vice versa? Like suppose my operator T is compact and using the compactness of the operator T if I prove that $T(B)$ is compact then are these two different results or the same results? Please if someone clears this doubt.

Answer 1

Below is a counterexample to the claim that there exists a nonzero operator which could satisfy the property that the image of every bounded set is compact and not just relatively compact:

Let $T : X \to Y$ be a nonzero operator between two normed spaces $X$ and $Y$. Assume by contradiction that every bounded set has closed image.
Because $T$ is nonzero, there exists $x \in X \setminus \{0\}$ such that $T(x) \neq 0$.
Now simply let $x_n := \frac{1}{n} x$. By assumption, the bounded set $B := \{x_n \mid n \in \mathbb{N^*}\}$ ($\mathbb{N}^*$ means the positive integers without zero) has compact image through $T$, hence the set $C := \{T(x_n) \mid n \in \mathbb{N^*}\} = \left\{\frac{1}{n} T(x) \mid n \in \mathbb{N^*}\right\}$ is closed.
However, $\frac{1}{n}T(x)$ converges to $0$ when $n$ tends to $\infty$, meaning that $0$ would belong to $C$, but none of the vectors in $C$ are the vector $0$ since $T(x) \neq 0$, absurd.

Hence every nonzero operator admits at least one (but it's actually an infinity of course) bounded set of non-closed image. Since compact sets are in particular complete and thus closed, your operator cannot have send all bounded sets on compact sets.

What I mentioned in my comments still holds though, that is an equivalent definition of compact operator is that every bounded set is sent on a relatively compact set, aka a st of compact closure. See here for example: A linear operator is compact if and only if the image of any bounded set is relatively compact.