Lebesgue conduct integral

Question

I. Suppose $f\in \mathcal{L^1}(R^n),g\in \mathcal{L^1}(R^n)$, then conduct integral $f*g$ is defined as $f*g(x)= \int_{R^n}f(x-y)g(y)dy$ for all $x$.

My task is to prove following statements.

(1) $f*g \in \mathcal{L^1}( R^n)$;

(2) When $g \in \mathcal{C}_0^1(R^n)$, then $f*g \in \mathcal{C}^1(R^n)$.

II. $\mathcal{L}^1(R)$ stands for the family of all Lebesgue integrable functions on $R$.

Let $ \hat{f}(\xi)=\int_R e^{-ix\xi}f(x)dx $ for $f\in\mathcal{L}^1(R)$. Answer the following questions:

(1) Show that $\hat{f}(\xi)$ is a bounded continuous function on $R$.

(2) Let $(f*g)(x)=\int_R f(x-y)g(y)dy $ for $f,g\in \mathcal{L}^1(R)$. Show that $f*g\in\mathcal{L}^1(R)$.

(3) Show that $\hat{f*g}=\hat{f}\hat{g}$ for $f,g\in \mathcal{L}^1(R)$.

Answer 1

Hint: start considering the integral $\int_{[-R,R]}|f*g(x)|dx$, and notice that this can be bounded by $\int_{[-R,R]}\int_{\Bbb R}|f(x-y)g(y)|dydx$. Now use Fubini-Tonelli's theorem for non-negative functions in order to switch the integrals.

For the other part, you can use dominated convergence theorem.

Answer 2

For II(2), it is same as I(1). And for $(1)$, $e^{-ix\xi}\le 2$, then I can prove $\hat{f}\le \int_R 2f(x)dx<\infty$, then it is bounded. As to continuous?