Let $T: L^2[0,\infty)\to L^2[0,\infty)$ $T(f)=2f(x+1)$ is it compact?
I think that is not compact. I think that the range of $T$ is isometric to $L^2[1,\infty)$ which is an infinite dimensional banach space. But a compact operator cannot have such range. Thus $T$ is not compact.
Let $T(g)=f \in Im(T)$ then $\int_{1}^{\infty}|f(x)|=2\int_{0}^{\infty}|g(x+1)|<\infty$ so the image is contained in $L^1[1,\infty)$. now take any function $h\in L^1[1,\infty)$ then $1/2r(x)$ maps to $h$ where we define $r(x)=0$ for $x<1$. and $r(x)=h(x)$ for $x>1$. Thus the image is isometric to $L^1[1,\infty)$ and so it is closed and banach. This is impossible unless $T$ is not compact.