$ \left( \ u_{tt} \ = \ c^2u_{xx}, \quad u(0,x) \ = \ 0, \quad u_t(0,x) \ = \ 0 \,\text{ decay }\right) \quad \Longrightarrow \quad u \equiv 0 $ ? I have to show the following:

Let $u$ be a classical solution of the following system:
$$
u_{tt} \ = \ c^2u_{xx}, \quad u(0,x) \ = \ 0, \quad u_t(0,x) \ = \ 0
$$
Satisfyying the decay requirement:
$$
\exists\alpha>\frac12, \quad
\exists C(t)>0, \quad 
\max\{|u_t(t,x)|,|u_x(t,x)|\} \ \leq \ \frac{C(t)}{|x|^\alpha}
$$
when $|x|$ is big enough.
Show that $u \equiv 0$.

What I did
According to a hint the book gave me, I have to use the fact that the following function of the time $t$ is a constant:
$$
\int_{-\infty}^\infty \frac12 (u_t^2+c^2u_x^2)dx
$$
This fact would be true by the decay requirement. I have shown this by establishing that the integrand can be written as a contrary flux. I know that $u \equiv 0$ solves the requirements, but I dont know why it is the only solution. please give me a hint.
 A: Here's a hint: the energy (your integral) at $t=0$ is equal to $0$.
Proof: $$E(0)=\int_{-\infty}^{+\infty}u_t^2(0,x)+c^2u_x(0,x)\,dx = c^2\int_{-\infty}^{+\infty}u_x^2(0,x)\,dx$$
since the first term is zero by the initial conditions. However $u_x(0,x)$ is also equal to zero since $$u_x(0,x) = \lim_{h\to 0}\frac{u(0,x+h)-u(0,x)}{h} = \lim_{h\to 0}\frac{0-0}{h}=0$$
also by the initial conditions. $\blacksquare$
Therefore, not only is $E(t)$ constant, it is identically zero. What does this mean for $u_t^2,\,u_x^2$, taking into account that they are positive? (And what does this, in turn, mean for $u_t,\,u_x$?)
Answer: (mouse over)

 Since $$\int_{-\infty}^{+\infty} \frac12 (u_t^2+c^2u_x^2)\,dx = 0$$ and the integrand is always positive, said integrand must be zero. Therefore $$u_t^2 + c^2u_x^2\equiv 0$$ and a sum of positive terms equal to zero means that each individual term is zero. Finally, $$\left.\begin{align}u_t^2\equiv 0 &\implies u_t\equiv 0 \\ c^2u_x^2 \equiv 0 &\implies u_x\equiv 0 \end{align}\right\}\implies u(t,x) = k ,\,\text{constant}$$ (since the domain of $u$ is presumably connected). But $k$ must be $0$ since $u(0,0) = 0$, for example.

