Prove that if $f,g \in L^1(-\infty,+\infty)$ and $$h(x)=\int^{+\infty}_{-\infty}f(x-e^y)g(y)dy$$ then $h \in L^1(-\infty,+\infty)$ and $$||h||_1 \leq ||f||_1 ||g||_1$$ I would be thankful if I see a detailed proof of these kind of questions that you have that you have to somehow start from a double integral and separate them as multiplication of two integrals, especially in cases like this where $f$ and $g$ does not necessarily belong to $L^{\infty}(\mathbb{R})$.
Thank you in advance
The function $k(x,y) = f(x-e^y)g(y)$ is Lebesgue-measurable. By Tonelli's theorem,
$$\begin{align} \int_{-\infty}^\infty \int_{-\infty}^\infty\bigl\lvert f(x-e^y)g(y)\bigr\rvert\,dy\,dx &= \int_{-\infty}^\infty \left(\int_{-\infty}^\infty\bigl\lvert f(x-e^y)\bigr\rvert\,dx\right)\lvert g(y)\rvert\,dy\\ &= \left(\int_{-\infty}^\infty \lvert f(x)\rvert\,dx\right)\left(\int_{-\infty}^\infty \lvert g(y)\rvert\,dy\right)\\ &= \lVert f\rVert_1 \lVert g\rVert_1\\ &< \infty. \end{align}$$
Hence $k \in L^1(\mathbb{R}^2)$.
Hence by Fubini's theorem, for almost all $x$, the function $y\mapsto f(x-e^y)g(y)$ is integrable, and the almost everywhere defined function
$$h(x) = \int_{-\infty}^\infty f(x-e^y)g(y)\,dy$$
is also integrable ($h \in L^1(\mathbb{R})$), and we have
$$\int_{-\infty}^\infty h(x)\,dx = \int_{-\infty}^\infty f(x)\,dx\cdot \int_{-\infty}^\infty g(y)\,dy$$
as well as
$$\lVert h\rVert_1 = \int_{-\infty}^\infty \lvert h(x)\rvert\,dx \leqslant \int_{-\infty}^\infty \int_{-\infty}^\infty \bigl\lvert f(x-e^y)g(y)\bigr\rvert\,dy\,dx = \lVert f\rVert_1\lVert g\rVert_1.$$