Giving a bound of the norm of a convolution Let $f:(-1,1)\to \mathbb{R}$ be a smooth function with compact support. Suppose that $f(x)\geq 0$ for all $x$ and $\int f =1$. Extend $f$ to $\mathbb{R}$ by $f(x) = 0$ if $x\not \in (-1,1)$. Show that $\|f\ast g\|_{p}=\|g\|_{p}$ for all $g\in L^p(\mathbb{R})$, $1< p < \infty$.
My best attempt showed that $\|f\ast g\| \leq \|g\|\cdot \int |f(x)|^p$ but I don't know how to reach the result.
The problem gives you as a hint to use Jensen's theorem.

Well first there's a typo in the question; you mean to ask how to prove $||f*g||_p
\le||g||_p$, right?
And there's some funny business in the question itself; smoothness and compact support are irrelevant. (And they don't make it any easier as far as I can see.) All that matters is $f\ge 0$ and $\int f=1$.
That means if you define a measure $\mu$ by $\mu(E) = \int_E f$ then $\mu$ is a probability measure, and in general $\int h d\mu=\int hf$.
Now since clearly $|g*f|\le|g|*f$ you can assume that $g\ge0$. Define $\phi:(0,\infty)\to(0,\infty$ by $\phi(t)=t^p$. Jensen says exactly that
$$(g*f(x))^p=\phi\left(\int g(x-t)d\mu(t)\right)\le\int\phi(g(x-t))d\mu=\int g(x-t)^pf(t)dt.$$
Now Fubinificate.
