$\langle u,\phi\rangle=0$ for all $\phi$ implies $u=0$

Question

Proving the injectivity of a function, I arrived to

$\forall \phi\in H^s \int_{\mathbb{R}^d} u(x)\phi(x)dx=0$, where $u\in H^{-s}$.

I know that if $\phi\in S$ and $u\in S'$, by the definition of the $0$ tempered distribution, this implies that $u=0$. But we don't have that. I also know that $S\subset H^s$ for all real s continuously, but I think that this has no relationship. Why can we assure that the above formula implies $u=0$? being $u\in H^s$.

Thank you in advance!

Answer 1

You have

$$\int_{\Bbb R^d} u(x)\phi(x)dx = \int_{\Bbb R^d} \hat u(\xi)\bar{\hat \phi}(\xi)d\xi =\int_{\Bbb R^d} \hat u(\xi) (1+|\xi|^2)^{-s/2} \bar{\hat \phi}(\xi)(1+|\xi|^2)^{s/2}d\xi. $$

Now you have $\hat u(\xi) (1+|\xi|^2)^{-s/2}$ is a $L^2$ function. Take, for example, arbitrary $\phi\in S$, therefore, functions of the form $\bar{\hat \phi}(\xi)(1+|\xi|^2)^{s/2}$ cover all $S$. Thus, $\hat u(\xi) (1+|\xi|^2)^{-s/2}$ is orthogonal in $L^2$ to $S$, therefore, $$\hat u(\xi) (1+|\xi|^2)^{-s/2}=0 \quad \mbox{in the sense of }\quad L^2(\Bbb R^d).$$

This implies that $\hat u=0$ and, therefore, $u=0$.

Answer 2

$u\in H^{-s}$, therefore by definition of $H^{-s}$, $u\in S'$, the space of tempered distributions (I did not realize this fact when asking the question).

Therefore $(1+|\xi|^2)^s\bar{\hat{u}}=0$ in the sense of tempered distributions. If we multiply by $(1+|\xi|^2)^-s$, and since multiplying by this function with this crescence rate let S' stable, we find that $\bar{\hat{u}}=0$ as tempered distribution, and then $u\equiv 0$.