Show that $[f' (t) ]^\vee = - [f^\vee (t)]'$, where the symbol $\vee$ is defined $\phi ^\vee(s) = \phi (-s)$.

Question

How can we show this equality using with distribution? The symbol $\vee$ used for define the reversed function $\phi^\vee(s)$ by $\phi^\vee(s) = \phi(-s)$.

Answer 1

Maybe the notation is obfuscating things here a bit. The proof of this identity is going to be via the chain rule. To apply the chain rule easily, we need to write $f^\vee$ as a composition of two other functions. I'll introduce a function $\text{rev}$ like so: $$\text{rev}(x) = -x$$

The reason for making this introduction is that now, $f^\vee = f \circ \text{rev}$, where $\circ$ is function composition.

Now we're ready to apply the chain rule. The chain rule says that if $f,g$ are functions (with some conditions), then the composition $h = f \circ g$ has a derivative given by: $$h'(x) = f'(g(x)) \cdot g'(x)$$

In our case, we have $h = f^\vee = f \circ \text{rev}$. So the derivative of $h$ is: $$f'(\text{rev}(x)) \cdot \text{rev}'(x)$$

Well, it's easy to see that $\text{rev}'(x) = -1$. And $f'(\text{rev}(x))$ is (by definition) $(f')^{\vee}(x)$. If you understand that, then the rest should be easy.

Answer 2

First show that $(\phi^\vee)' = -(\phi')^\vee$ for an ordinary smooth function $\phi$. Then, for a distribution $f$ and a test function $\phi$ we have $$ \langle (f')^\vee, \phi \rangle = \langle f', \phi^\vee \rangle = - \langle f, (\phi^\vee)' \rangle = - \langle f, -(\phi')^\vee \rangle = \langle f, (\phi')^\vee \rangle = \langle f^\vee, \phi' \rangle = - \langle (f^\vee)', \phi \rangle . $$

Thus, $(f')^\vee = -(f^\vee)'.$