I'm solving the exercise of Conway's functional analysis. I'm trying this question.
Let $X$ be a compact space and let $\mu$ be a positive Borel measure on $X$. Let $T \in \mathcal{B}(L^p(\mu) , C(X))$ where $1<p<\infty$. Show that if $A : L^p(\mu) \to L^p(\mu)$ is defined by $Af = Tf$ then $A$ is compact.
I think that if we use fact that $L^p(\mu)$ is reflexive and if $A$ is completely continuous, then we can solve this problem because this indicate $A$ is compact operator.
So, I assumed that $f_n \to 0$ weakly, i.e. ($\int_X f_n g d\mu \to 0$ for every $g \in L^q(\mu)$) and tried to show that $\int_X (Tf_n)^p d\mu \to 0$. However, I cannot approach to next step.
Thank you.
Since $L^p(\mu)$ is reflexive, its unit ball $B$ is weakly compact. Since $T$ is continuous for the norm topologies, it is also continuous for the weak topologies and hence $T(B)$ is a weakly compact set in $C(X)$. In particular, since the map $f \mapsto f(x)$ is a continuous linear functional on $C(X)$ for each $x \in X$, for any sequence $f_n$ in $T(B)$ there exists a subsequence $f_{n_k}$ that converges pointwise to $f \in C(X)$.
Additionally, since $T$ is a bounded operator $T(B)$ is a bounded set. Therefore by the bounded convergence theorem $f_{n_k} \to f$ in $L^p(X)$ also, proving the desired result.