How to show distribution $T$ is a constant 
Let $T$ be a distribution on $\mathbb{R}$. $\tau_a\phi(x)=\phi(x-a)$ and $\langle T,\tau_a\phi\rangle=\langle T,\phi\rangle$ for all $a\in \mathbb{R}$ and all test functions $\phi$. Prove that $T$ is a constant.

I have no idea to prove $\langle T,\phi\rangle=0,$ for all $\phi$.
 A: As in my comment, the goal is to show that there is a constant $c$ such that $\langle T, \phi \rangle = c \int_{-\infty}^\infty \phi(x)\,dx$ for all test functions $\phi$.
First of all, let's identify the constant $c$.  Let $\psi$ be your favorite test function having $\int_{-\infty}^\infty \psi(x)\,dx = 1$ and set $c := \langle T, \psi \rangle$.
Next: for any test function $\phi$, show that as $a \to 0$, we have $\frac1a (\phi - \tau_a \phi) \to \phi'$ in the usual topology of $C^\infty_c(\mathbb{R})$.
Now our hypotheses imply that for any $\phi$,
$$\left\langle T , \frac{1}{a} (\phi - \tau_a \phi) \right\rangle = 0.$$
Since $T$ is a continuous linear functional, we can pass to the limit to conclude $\langle T, \phi' \rangle = 0$.  That means that $T' = 0$ which intuitively should mean that $T$ is constant.
To prove this, let $\phi$ be any test function and set $f(x) = \phi(x) -  \psi(x) \int_{-\infty}^\infty \phi(t)\,dt$, so that $\int_{-\infty}^\infty f(x)\,dx = 0$.  Therefore, if we set $g(x) = \int_{-\infty}^x f(t)\,dt$, you can verify that $g$ is smooth and compactly supported, and by the fundamental theorem of calculus, $g' = f$.  As such, 
$$0 = \langle T, g' \rangle = \langle T, f \rangle = \langle T, \phi \rangle - c \int_{-\infty}^\infty \phi(x)\,dx$$
which is the desired conclusion.
(The key in the last paragraph is that a test function $f$ is the derivative of another test function iff we have $\int_{-\infty}^\infty f(x)\,dx = 0$.)
