Is the proof below correct? Any feedback is much appreciated.
Let $f_1,f_2 \in \mathcal{C}^{\infty}_{c}(\mathbb{R}),$ the set of infinitely differentiable functions with compact support in $\mathbb{R}.$ Suppose $f_1=f_2$ as distributions, that is, $$\int_{\mathbb{R}}f_1(x)g(x)\space dx=\int_{\mathbb{R}}f_2(x)g(x)\space dx, \space \text{for all} \space g \in \mathcal{C}^{\infty}_{c}(\mathbb{R}).$$ Show that $f_1=f_2$ as functions, that is, $f_1(x)=f_2(x)$ for all $x.$
I have the following.
Suppose that $f_1 \ne f_2$ as functions. Then there is some $y \in \mathbb{R}$ such that $f_1(y) \ne f_2(y).$ Suppose without loss of generality that $f_1(y) > f_2(y).$ It follows that $f_1(y)-f_2(y)>0.$ Now we observe that $f_1-f_2 \in \mathcal{C}^{\infty}_{c},$ so that in particular $f_1-f_2$ is continuous and has compact support. By hypothesis $$\int_{\mathbb{R}}[f_1(x)-f_2(x)]\cdot g(x)\space dx=0, \space \text{for all} \space g \in \mathcal{C}^{\infty}_{c}.$$ From continuity of $f_1-f_2$ follows the existance of a neighborhood of $y,$ say $\mathcal{B}_{\delta}(y)$, so that $f_1(x)-f_2(x)>0$ for all $x \in \mathcal{B}_{\delta}(y)$. Choose $g \in \mathcal{C}^{\infty}_{c}$ such that $g(x)>0$ for all $x \in \mathcal{B}_{\delta}(y),$ which we can do by translating a bump function. Then $$\int_{\mathbb{R}}[f_1(x)-f_2(x)]\cdot g(x)\space dx>0,$$ and we have a contradiction. Therefore $f_1=f_2$ as functions.
On a side note, how could one extend this result to $\mathcal{C}^{\infty}_{c}(\mathbb{R}^{n})$?
I was thinking that the same idea could work. My issue is with how to deal with the iterated integrals and the vector arguments. Any hints on how to extend this result?