# Proof of Young's inequality $\Vert u \star v \Vert_p \le \Vert u \Vert_1 \Vert v \Vert_p.$

I am looking at the proof of Young's inequality, however there's a point I can't figure out.

(Young's inequality) Let $u \in L^1(\lambda^n)$ and $v \in L^p(\lambda^n) , p \in [1,\infty)$. The convolution $u \star v$ defines a function in $L^p(\lambda^n)$ and satisfies $$\Vert u \star v \Vert_p \le \Vert u \Vert_1 \Vert v \Vert_p.$$

Proof. WLOG assume $u,v \ge 0$. Using that $\lambda^n$ is invariant under translations, we see with Jensen's inequality and Tonelli's theorem that \begin{align} \Vert u\star v\Vert_p^p & = \int \Big(\int u(y)v(x-y)dy\Big)^p dx \\ & = \Vert u\Vert_1^p \int \Big( \int v(x-y)\frac{u(y)}{\Vert u\Vert_1} dy\Big)^p dx \\ & \le_{Jensen} \Vert u \Vert_1^p \int \int v(x-y)^p \frac{u(y)}{\Vert u\Vert_1} dy dx\\ & = \Vert u\Vert_1^p \int \Big(\int v(x-y)^p dx\Big) \frac{u(y)}{\Vert u\Vert_1} dy \\ & =\Vert u\Vert_1^p \Vert v\Vert_p^p. \end{align}

In the above proof, I don't understand the inequality step that uses Jensen's inequality. By the Jensen's inequality shouldn't we also have the $p$th power on $\frac{u(y)}{\Vert u\Vert_1}$? How are we able to get rid of this?

It is the Jensen's inequality applied to the measure $\dfrac{u(y)}{\|u\|_{1}}dy$, this measure has total mass $1$, and this is required in Jensen's inequality.