Trying to prove that the first derivative of dirac's delta function is odd.

Question

I'm trying to prove the oddness of $\delta'(x)$, I have managed to show that $\int \delta'(x)f(x) dx$ vanishes if f(x) is an even function, which shows $\delta'(x)$ is an odd distribution. However, if I integrate directly, I get $\int \delta'(x)f(x)dx = -\int \delta(x)f'(x)dx$ And $\int\delta'(-x)f(x)dx = f(x)\int\delta'(-x)dx - \int f'(x)\int\delta'(-x)dx dx$

Now the problem is in this part

$\int\delta'(-x)dx = \int-\frac{d}{du}\delta(u)du = -\delta(u)$ I'm unsure whether this last part is true, it seems kinda fishy to turn the derivative $\frac{d}{dx}$ to $\frac{d}{du}$ when I apply the change of variable $u = -x$ Is this correct? Is there a better way to approach this? Thanks for reading. Edit: Assume the integrals are from $-\infty$ to $\infty$.

Answer 1

Let's denote by $g$ the reflection operator, i.e., $g(x)= -x$. $\delta^{'}$ is said to be odd if $\delta^{'} \circ g= - \delta ^{'}$ in the sense of distribution, i.e., if $\langle \delta^{'} \circ g, f \rangle = - \langle \delta ^{'}, f \rangle$ for any test function $f$ (see page 70 in here) . This is satisfied by $\delta^{'}$. Allowing an abuse of notation:

$$ \begin{aligned} \int (\delta ^{'} \circ g) \; f \; dx &= \int \delta ^{'}(-x) f(x) \; dx \\ & = \int \delta ^{'} (f \circ g) dx \\ & = - \int \delta (f \circ g)^{'} dx \\ & = + \int \delta(x) f {'} (-x) dx \\ & = f^{'}(0) \\ \end{aligned} $$ where we used the chain rule in in third equality. The right-hand-side reads: $$ \begin{aligned} -\langle \delta^{'} , f \rangle &= -\int \delta^{'} (x) f (x) dx \\ &= \int \delta (x) f^{'} (x) dx \\ & = f^{'}(0) \end{aligned} $$ Hence the derivative of the dirac delta is odd.

Answer 2

Define $P:C_c^\infty \to C_c^\infty$ by $(P\varphi)(x)=\varphi(-x).$ Note that, by the chain rule, $(P\varphi)'(x) = -\varphi'(x).$

For $\varphi,\psi\in C_c^\infty,$ $$ \langle P\varphi, \psi \rangle = \int_{-\infty}^{\infty} P\varphi(x) \, \psi(x) \, dx = \int_{-\infty}^{\infty} \varphi(-x) \, \psi(x) \, dx = \{ x := -y \} = \int_{+\infty}^{-\infty} \varphi(y) \, \psi(-y) \, (-dy) \\ = \int_{-\infty}^{\infty} \varphi(y) \, \psi(-y) \, dy = \int_{-\infty}^{\infty} \varphi(y) \, P\psi(y) \, dy = \langle \varphi, P\psi \rangle. $$ Therefore we extend $P$ to distributions by for $u\in\mathcal{D}'$ setting $\langle Pu, \varphi \rangle = \langle u, P\varphi \rangle.$ We call a distribution odd iff $Pu=-u.$

Now, $$ \langle P(\delta'), \varphi \rangle = \langle \delta', P\varphi \rangle = -\langle \delta, (P\varphi)' \rangle = -(P\varphi)'(0) = -(-\varphi'(0)) = -\langle \delta, -\varphi' \rangle = -\langle \delta', \varphi \rangle, $$ so $P(\delta') = -\delta',$ i.e. $\delta'$ is odd.