A property of the convolution of operator For any $f\in C_c^\infty(\mathbb{R})$, let $T_f$ denote the bounded operator on $L^2(\mathbb{R})$ given by convolution by $f$: that is,
$$(T_f g)(x):=\int f(y)g(x-y)\,dy,$$
for $g\in L^2(\mathbb{R})$. Let $\chi\in C_c^\infty(\mathbb{R})$ act on $L^2(\mathbb{R})$ by multiplication. Then is it true that the operators
$$T_f\circ\chi,\qquad\chi\circ T_f$$
are compact operators on $L^2(\mathbb{R})$?
Thoughts: I thought about using the Fourier transform to turn the convolution operator $T_f$ into a multiplication operator, but then $\chi$ becomes a convolution operator, which leads back to the original problem.
 A: I will tackle the case $\chi \circ T_f$. Since $L^2(\mathbb{R})$ is a Hilbert space, it suffices to check that $\chi \circ T_f$ is completely continuous, that is, for any sequence $\left\{g_n\right\}\subset L^2(\mathbb{R})$,
$$g_n\rightharpoonup 0\text{ weakly} \Rightarrow (\chi\circ T_f)g_n\to 0\text{ strongly} $$
Since $\chi\in C_c^{\infty}(\mathbb{R})$, let $X$ be the support of $\chi$, which is a compact set and hence has finite Lebesgue measure. We have
\begin{align*}\left\|(\chi \circ T_f)g_n\right\|_2^2&=\int_{X}|\chi(x)|^2|(f*g_n)(x)|^2dx\leq \|\chi\|_{\infty}^2\int_X\left|(f*g_n)(x)\right|^2dx=(1)\end{align*}
Since $g_n\rightharpoonup 0$ weakly in $L^2(\mathbb{R})$, we have 
$$\lim_{n\to +\infty}(f*g_n)(x)=\lim_{n\to +\infty}\int_{\mathbb{R}}f(x-y)g_n(y)dy= 0,\qquad \forall f\in L^2(\mathbb{R}),\;\text{a.e. }x\in \mathbb{R}$$
So the integrand $|(f*g_n)(x)|^2$ in $(1)$ converges to $0$ for a.e. $x\in \mathbb{R}$. It remains to show that there is a dominant in $L^1(X)$ to conclude by the dominated convergence theorem. Again, since $g_n\rightharpoonup 0$ weakly in $L^2(\mathbb{R})$, the sequence $\left\{g_n\right\}$ is bounded in $L^2(\mathbb{R})$ and so for some $C>0$ we have by Young's theorem (or just Cauchy-Schwarz)
\begin{align*}|(f*g_n)(x)|^2\leq \|f\|_{2}^2\|g_n\|_2^2\leq C\|f\|_{2}^2,\qquad \text{a.e. }x\in \mathbb{R}\end{align*}
which surely belongs to $L^1(X)$ since $X$ is compact.
Remark. There was no need to assume $f\in C_c^{\infty}(\mathbb{R})$. The above argument works for all $f\in L^2(\mathbb{R})$. What matters is that $\chi\in C_c^{\infty}(\mathbb{R})$ so that when we calculate the $L^2$ norm we can reduce the integration domain to a compact subset $X$ which allows us to find a dominant function in $L^1(X)$. I suppose that the requirements swap when proving that $T_f\circ \chi$ is compact - you will then only need $f\in C_c^{\infty}(\mathbb{R})$ and $\chi \in L^2(\mathbb{R})$.
