Show that for all $f$ and $g$ in $\mathbb{L}^1(E_n)$ that $f(\circ)g(x-\circ) \in \mathbb{L}^1(E_n)$

Question

Show that for all $f$ and $g$ in $\mathbb{L}^1(E_n)$ that $f(\circ)g(x-\circ) \in \mathbb{L}^1(E_n)$ for a.e. $x \in E_n$.

I couldn't find this question elsewhere (it's difficult to search for as well).

---

We must show that $$ \int_{E_n} | f(t) g(x-t) | dt < \infty. $$

\begin{align} \mathrm{LHS} & = \int_{E_n} | f(t) g(x-t) | dt \\ & \le \int_{E_n} | f(t)| | g(x-t)| dt \\ & \le \int_{E_n} | f(t)| \int_{E_n} | g(x-t)| dx dt \quad \text{(not sure about this line)} \\ & = \int_{E_n} | f(t)| dt \int_{E_n} | g(x-t)| dx \quad \text{by Fubini's theorem}\\ & = \Vert f\Vert_1 \Vert g \Vert_1 \\ & < \infty \end{align} as both $\Vert f\Vert_1$ and $\Vert g \Vert_1$ are finite.

---

Could someone confirm that the line I'm unsure of makes sense? If not (or if so) could you please explain why? I think it's true that $|g(x)| \le \Vert g \Vert_1$, $\forall x \in E_n$ since it's like saying that we add all the values $g$ can take and since $g(x)$ is in there, the inequality should hold.

Answer 1

That line is not true.

To show that $h(x) = \int f(y)g(x-y)dy$ is an $L^1$ function, one can use the following trick:

If $\int h(x) < \infty$, then $h(x)$ is finite a.e.

Now, do the following: $$ \int h(x)dx = \int \int f(y)g(x-y) dydx \leq \int \int |f(y)||g(x-y)|dx dy $$ Now, $$ \int \int |f(y)||g(x-y)|dx dy \leq \int |f(y)| \int |g(x-y)|dx dy $$ Finally, $$ \int |f(y)| \int |g(x-y)|dx dy \leq \int |f(y)| dy \int |g(x-y)|dx \leq ||f||_1\cdot||g||_1 $$ Hence, we have that $h(x)$ is bounded a.e., which is the same thing as the convolution exists a.e.