Is $Tf=\int_0^1\alpha (x)f(x)dx$ a compact operator?

Question

Consider the operator $T:L^1\left([0,1],L^1\left(\mathbb{R^+} \right) \right)\to L^1\left(\mathbb{R^+} \right)$ defined for each $f\in L^1\left([0,1],L^1\left(\mathbb{R^+} \right) \right)$ by $$Tf=\int_0^1\alpha (x)f(x)dx,$$ where $\alpha:[0,1]\to\mathbb{R}$ is a real function.

Is $T$ a compact operator from $L^1\left([0,1],L^1\left(\mathbb{R^+} \right) \right)$ to $L^1\left(\mathbb{R^+} \right)$ when the function $\alpha \neq 0$ ?

Answer 1

No. For simplicity assume that $\alpha$ is bounded. Given $g\in L^1(\mathbb R^+)$ take $f(x)=\alpha(x)g$ so that $Tf = cg$ with $c=\int_0^1 \alpha^2(x)dx \neq 0$. This shows that $T$ is surjective and hence not-compact.