A problem from distribution theory. Let $f$, $g\in C(\Omega)$, and suppose that $f \neq g$ in $C(\Omega)$. How can we prove that $f \neq g$ as distributions?
Here's the idea of my proof.
$f$ and $g$ are continuous functions, so they will be locally integrable.
Now, take any $\phi \neq 0 \in D(\Omega)$. Let us suppose that $\langle T_f,\phi\rangle =\langle T_g,\phi \rangle $ and $f\neq g$
$\langle T_f,\phi \rangle = \langle T_g,\phi \rangle$
$\implies \int_\Omega f(x) \phi(x) dx = \int_\Omega g(x) \phi(x) dx$
$\implies \int_\Omega \phi(x)[f(x)-g(x)] dx =0$
i.e., the area under above function is zero.
We know that $\phi$ is non zero, we just need to prove that this integral will be zero only when $f(x)-g(x)=0$ and it will make contradiction to our supposition that $f$ and $g$ are not equal.
[Please help me prove the last point]

Let me add a little more detail to my comment.
You are correct up to the $\int_{\Omega} \phi(x)[f(x)-g(x)]dx = 0$.
Now, for $\varepsilon > 0$ and $x_0 \in \Omega$, let $B_{\varepsilon}(x_0)$ denote the ball of radius $\varepsilon$ centered at $x_0$. Then, you can find a test function $\psi$ with the following properties:
(i) $\psi(x) > 0 \, \forall x \in B_{\varepsilon}(x_0)$ 
(ii) $\psi(x) = 0 \, \forall x \notin B_{\varepsilon}(x_0)$
(iii) $\int_{\Omega} \psi = \int_{B_{\varepsilon}(x_0)}\psi = 1.$
Now, choose $\phi = \psi$ and for any $x_0 \in \Omega$
$$
0 = \int_{\Omega} \psi(x)[f(x)-g(x)]dx = \int_{B_{\varepsilon}(x_0)} \psi(x)[f(x)-g(x)]dx \approx f(x_0)-g(x_0)
$$
As $\varepsilon \rightarrow 0$, the equation is exact, i.e. $f(x_0) - g(x_0) = 0$. Since $x_0$ was arbitrary, $f = g$ in $\Omega$, a contradiction.
