Prove an elementary integral inequality ($L^2$ estimate) that arises with the wave equation How can I prove that  $$\Vert u \Vert_{L^2} \le \Vert u_1 \Vert_{L^2} + t \Vert u_2 \Vert_{L^2},$$ where $$u(t,x) = \frac{u_1(x+t) + u_1(x-t)}{2}+ \frac{1}{2}\int_{-t}^t u_2(x+y)\ dy \ ?$$
$u_1,u_2\in L^1 \cap L^2$.

The problem arises in PDE (with the wave equation).
 A: If $y\in\mathbb R$, define the translation operator $\tau_y$ by $\tau_yv(x)=v(x-y)$.  Using the triangle inequality and Minkowski's inequality for integrals on your formula for $u(t,x)$, we get:
$$
\|u(t,.)\|_{L^2}\le\frac{1}{2}(\|\tau_{-t}u_1\|_{L^2}+\|\tau_tu_1\|_{L^2})+\frac{1}{2}\int_{-t}^t\|\tau_{-y}u_2\|_{L^2}dy.
$$
But by the translation invariance of Lebesgue measure, $\|\tau_yv\|_{L^2}=\|v\|_{L^2}$, so the right hand side becomes:
$$
\|u(t,.)\|_{L^2}\le\|u_1\|_{L^2}+\frac{1}{2}\int_{-t}^t\|u_2\|_{L^2}dy=\|u_1\|_{L^2}+t\|u_2\|_{L^2}.
$$
A: Use the triangle inequality:
$$ \lVert u(t,x) \rVert_2 \leq \frac{1}{2} \lVert u_1(x+t) \rVert_2 + \frac{1}{2} \lVert u(x-t) \rVert_2 + \frac{1}{2}\left\lVert \int_{-t}^t u_2(x+y) \, dy \right\rVert_2  $$
The norm is translation-invariant, so the first two terms are both equal to $\lVert u_1 \rVert_2/2$. For the last term, we use the integral triangle inequality $\left\lVert \int_{\Omega} f(x,y) \, d\mu(y) \right\rVert_2 \leq \int_{\Omega} \lVert f(x,y) \rVert_2 \, d\mu(y)  $, and translation-invariance, to find
$$ \frac{1}{2}\left\lVert \int_{-t}^t u_2(x+y) \, dy \right\rVert_2 \leq \frac{1}{2} \int_{-t}^t \lVert u_2(x+y) \rVert_2 \, dy = t \lVert u_2 \rVert_2. $$
