Let $G$ be a locally compact, Hausdorff, non-abelian group and $K$ a compact subgroup of $G$. Let $f \in C_c(G)$ then prove that $\tilde f \in C_c(G)$ where $ \tilde f(x):=\int_K \int_Kf(k_1xk_2) dk_1dk_2$
My attempt: Let's first get our notations straight: $L_x(f)(y):=f(x^{-1}y)$ and $R_xf(y):=f(yx)$.
Note that for $x,y$ in some neighborhood of identity $e$, we have, $|\tilde f(x) - \tilde f(y)|=|L_{x^{-1}} \circ \tilde f(e)- L_{x^{-1}} \circ \tilde f(xy)|$, so now using uniform continuity of the left translation $x \mapsto L_x$ we get that it is enough to prove that $\tilde f$ is continuous at $e$.
So for $x$ in some neighborhood of identity ( say $U$ ), we have, $$|\tilde f(x)-\tilde f(e)|=\Big|\int_K \int_K f(k_1xk_2) dk_1 dk_2-\int_K \int_K f(k_1k_2) dk_1 dk_2\Big|$$ $$\le \int_K \int_K \Big|f(k_1xk_2)-f(k_1k_2)\Big|dk_1dk_2$$ $$= \int_K \int_K \Big|R_{k_2} \circ \Big(f(k_1x)- f(k_1)\Big)\Big|dk_1dk_2$$ $$= \int_K \int_K \Big|f(k_1x)- f(k_1)\Big|dk_1dk_2$$ $$= \int_K \Big|f(k_1x)- f(k_1)\Big|dk_1$$ $$<\varepsilon \int_K dk_1 \text{( Using Uniform continuity of } f \in C_c(G) )$$ $$=\varepsilon$$
For the compact support, note that $x \in Supp \tilde f \implies \tilde f(x) \ne 0 \implies \int_K \int_Kf(k_1xk_2)dk_1dk_2 \ne 0$ $\implies \exists k_1,k_2 \in K \text{ s.t. } f(k_1xk_2) \ne 0 \implies k_1xk_2 \in Supp f \implies x \in {k_1}^{-1} Supp f {k_2}^{-1} \subset K^{-1} Supp f K^{-1}$
So we get that $Supp \tilde f \subset K^{-1} Supp f K^{-1}$, hence $\tilde f \in C_c(G)$
Q.E.D
Is my solution correct? Please point out mistakes if any.