Estimation of Energy for the Wave Equation using Gronwall's Inequality 
Given the initial value problem :
$$\matrix{u_{tt}-u_{xx} &=& f(x,t)&\;\;\forall x\in \mathbb R\,,\;t\in
 [0,T]\\u(x,0) &=& g(x)&\\u_t(x,0) &=& h(x)&}$$
where the initial data and $u(\;,t)$ are supported for every $\;t\in
 [0,T]$.
We also define the energy as: $$E(t)=\frac{1}{2} \int_{\mathbb R}
 (u_t)^2\;+\;(u_x)^2\;\;dx$$
Prove that: $$\max_{\;τ\;\in [0,T]} \sqrt{E(t)} \lt \sqrt{E(0)}\;+\;\sqrt
 2 \int_{0}^t {\Vert f(\;.τ) \Vert}_{L^2} \;dτ\;$$

My attempt:
$\frac{dE(t)}{dt}\;=\;\frac{1}{2} \int_{\mathbb R} 2u_xu_{xt}\;+\;2u_tu_{tt}\; dx\;=\;\int_{\mathbb R} u_tu_{tt}\;-\;u_tu_{xx} \;\;dx\;+[u_xu_t]_{-\infty}^{+\infty}\;=\;\int_{\mathbb R} u_tf\;dx\;\le \frac{1}{2} \int_{\mathbb R} (u_t)^2\;+f^2 \;dx\;\le \frac{1}{2} \int_{\mathbb R} (u_t)^2\;+\;(u_x)^2\;dx\;+\;\frac{1}{2} \int_{\mathbb R} f^2 \;dx\;=\;E(t)+\frac{1}{2} \int_{\mathbb R} f^2 \;dx\;$ 
At this point I assume I should use some kind of Gronwall's inequality in order to prove the above, but I have no idea how. I would appreciate if someone could help me through this!
Hints or other solutions than this are welcome. Thanks in advance
 A: The inequality you use $\int_{\mathbb{R}} fu_t\,{\rm d}x \leq \frac{1}{2}\int_{\mathbb{R}} f^2 + u_t^2\,{\rm d}x$ is not strong enough to yield the desired result. If we instead use Cauchy-Schwarz
$$\int_{\mathbb{R}} fu_t\,{\rm d}x \leq \sqrt{\int_{\mathbb{R}} f^2\,{\rm d}x\int_{\mathbb{R}} u_t^2\,{\rm d}x}$$
in the result you have derived, $\frac{dE(t)}{dt} \leq \int_{\mathbb{R}}fu_t\,{\rm d}x$, we obtain
$$\frac{dE(t)}{dt} \leq \sqrt{\int_{\mathbb{R}} f^2\,{\rm d}x\int_{\mathbb{R}} u_t^2\,{\rm d}x} \leq \sqrt{\int_{\mathbb{R}} f^2\,{\rm d}x\int_{\mathbb{R}} u_t^2 + u_x^2\,{\rm d}x} \equiv \|f(\cdot,t)\|_{2} \sqrt{2E(t)}$$
and since $\frac{dE}{dt} = 2\sqrt{E}\frac{d\sqrt{E}}{dt}$ this gives us
$$\frac{d\sqrt{E(t)}}{dt} \leq \frac{1}{\sqrt{2}}\|f(\cdot,t)\|_2 \implies \sqrt{E(t)} \leq \sqrt{E(0)} + \frac{\sqrt{2}}{2}\int_0^t\|f(\cdot,\tau)\|_2\,{\rm d}\tau$$
Note that this is a stronger result than the one you are to show by a factor $\frac{\sqrt{2}}{2}$ instead of $\sqrt{2}$ in the last term.
