real analysis - Show $f$ is integrable

Let $f$ be such that $\int_0^{\infty} |f(s)|e^s ds< \infty.$

Now, I want to argue that for $x,y$ sufficiently large and $\lambda < 1$ fixed we have that

$$\int_0^{\infty} \int_x^y e^{\lambda z} e^{s} |f(s+z)|dz ds$$ can be made arbitrarily small for $\lambda <1.$

In other words: $\forall \varepsilon >0 \exists N: \left(x,y >N \Rightarrow \int_0^{\infty} \int_x^y e^{\lambda z} e^{s} |f(s+z)|dz ds< \varepsilon \right)$

But how can I show this rigorously? Does anybody have an idea

Answer

As often, choosing the right order of integration is the key to success: Let $C =\int_0^\infty e^s |f (s)|\,ds $.

We then have for $y\geq x\geq R $ that

\begin{eqnarray*}
\int_0^\infty \int_x^y e^{\lambda z}e^s |f (s+z)|\,dz\,ds &=&\int_x^y e^{\lambda z}\int_0^\infty e^s |f (s+z)|\,ds \,dz \\
(w =s+z)&=& \int_x^y \int_z^\infty e^{w-z}|f (w)|\,dw \\
&\leq& C \cdot \int_x^y e^{(\lambda -1)z} \,dz.
\end{eqnarray*}

Now, I leave it to you to verify

$$
\int_x^y e^{(\lambda -1)z}\,dz \to 0
$$
as $R \to\infty $ where $y\geq x\geq R $.