If two distributions vanish on the same set of test functions, then one is a constant multiple of the other

From Chapter 1, Exercise 14, Strichartz's book: Distribution theory & fourier transforms

1. Suppose $f$ & $g$ are distributions such that $\langle f,\phi\rangle=0 \Leftrightarrow \langle g,\phi \rangle = 0$. Show that $\langle f,\phi \rangle =c\langle g,\phi\rangle$ for some constant c.

Does the question mean that if the condition is true for some test functions $\phi$ the result $\langle f,\phi \rangle=c \langle g,\phi\rangle$ can be deduced for all such test functions $\phi$? I think that one has to exploit the continuity of distributions like one has for functions wherein continuous functions are completely defined if they are defined on a dense subset. But apart from that I don't know how to proceed.

The assumption: $\forall \phi$ (in the test class) $\langle f,\phi\rangle=0 \Leftrightarrow \langle g,\phi \rangle = 0$.

The desired conclusion: $\exists c$ such that $\forall \phi$ the equality $\langle f,\phi \rangle =c\langle g,\phi\rangle$ holds.

Plan of attack:

1. Dispose of the trivial case in which $\forall \phi$ $\langle g,\phi\rangle=0$.
2. Pick $\phi_0$ such that $\langle g,\phi_0\rangle\ne 0$. Let $c=\langle f,\phi_0 \rangle /\langle g,\phi_0\rangle$; this is the only choice that could work, right?
3. If $\phi$ is any test function, let $b=\langle g,\phi \rangle /\langle g,\phi_0\rangle$ and observe that $\langle g,\phi-b\phi_0\rangle=0$. The rest should be easy.
• I couldn't understand why we are interested in case 3? Isn't it enough to look for case 1&2? Nov 9, 2015 at 19:14
• These are not cases, they are steps of a proof. I like to split proofs into small enumerated chunks. Step 1: dispose of trivial case. Step 2: pick a candidate for constant $c$. Step 3: prove that the candidate works.
– user147263
Nov 9, 2015 at 19:16
• Thanks, but I couldn't understand how we observed $\langle g,\phi-b\phi_0\rangle=0$ Nov 11, 2015 at 14:16