$g : \mathbb{R} \to \mathbb{R}$ is integrable over $\mathbb{R}$ and has a compact support.
$f(y) = \sin^2(y) \ln|y|$
Show that the convolution $(f ∗ g)(x)$ is well defined for $x \in \mathbb{R}$ and that it is of class $C^1$ over $\mathbb{R}$.
I know that:$$h(x) = (f ∗ g)(x) = \int_{-\infty}^{\infty} \sin^2(\tau) \ln|\tau| \cdot g(x-\tau) \ d\tau$$
I know that being "well defined" means that it doesn't blow up to $-\infty$ nor $\infty$ anywhere and the class of functions $C^1$ contains functions that have partial derivatives (here, just derivative) and those are continuous.
Now, to tell if $h(x)$ is well defined:
Given that $g: \mathbb{R} \to \mathbb{R}$ is integrable over $\mathbb{R}$, it means that $g$ is measurable.
Since $f(y)=\sin^2(y)\ln\vert y\vert = \sin(y) \cdot \sin(y) \cdot \ln∣y∣$ is a product of 3 measurable functions it is measurable as well (I am not sure if I can say that $\ln∣y∣$ is measurable, because of $y = 0$). We know that the convolution of $2$ measurable functions ($f$ and $g$) is well defined.
Next, to show that $(f∗g)(x)$ is of class $C^1$ over $\mathbb{R}$, we need to demonstrate that it is continuously differentiable.
We have:$$h'(x) = \frac{d}{dx} \int_{-\infty}^{\infty} f(y)g(x-y) dy$$
Using the Leibniz rule for differentiating under the integral sign, we have:$$h'(x) = \int_{-\infty}^{\infty} f(y) \frac{d}{dx}g(x-y) dy$$
Since $g$ is integrable and has compact support, $g′$ exists almost everywhere and is bounded. Therefore, $\frac{d}{dx}g(x−y)$ exists almost everywhere as well.
Now, since $f(y)=sin^2(y)ln∣y∣$ is continuous and $g′$ is bounded, the integrand $f(y)\frac{d}{dx}g(x−y)$ is bounded for all $y$ in the support of $g$. By the Dominated Convergence Theorem, we can conclude that $h′(x)$ exists and is continuous.
Hence, $(f∗g)(x)$ is of class $C1$ over $R$.
Is that correct?