Product rule of distribution by multiplicating a $C^\infty$ function

Question

I am reading S. Kesavan's book "Topics In Functional Analysis and Application" and trying to prove the product rule, can you check where I am going wrong?

Let $\psi \in C^\infty (\mathbb{R})$ and $T \in \mathcal{D}'(\mathbb{R})$ , and $\phi \in \mathcal{D}(\mathbb{R})$.

then,

$\begin{align} \dfrac{d}{dx} (\psi T)(\phi) = - \psi T \Big( \dfrac{d\phi}{dx} \Big) = -T \Big( \psi \dfrac{d\phi}{dx} \Big) = -T \Big( \dfrac{d}{dx} (\psi \phi) \Big) + T \left( \dfrac{d\psi}{dx}\phi \right) \end{align}$

by the chain rule of differentiation and linearity of $T$.

Now, can you check the following part where am I going wrong?

$-T \Big( \dfrac{d}{dx} (\psi \phi) \Big) = \dfrac{d}{dx} \Big( T(\psi \phi) \Big) = \dfrac{d}{dx} \Big( \psi T( \phi) \Big) = \psi \dfrac{d}{dx} T(\phi) + \dfrac{d\psi}{dx} T(\phi)$

And, $T \Big( \dfrac{d\psi}{dx}\phi \Big) = \dfrac{d\psi}{dx} T(\phi)$

But the answer in total should be : $ \dfrac{d}{dx} (\psi T)(\phi) = \psi \dfrac{d}{dx} T(\phi) + \dfrac{d\psi}{dx} T(\phi) $

Answer 1

Short answer : $\frac{\text d}{\text d x} \Big( T(\psi\phi) \Big)$ makes no sense.

Long answer : You have : \begin{align} - T\left(\frac{d}{dx} (\psi\phi)\right) &= \left(\frac{dT}{dx}\right)(\psi\phi) \\ &= \left( \psi \frac{dT}{dx}\right)(\phi)\\ \end{align}

and : $$T\left(\frac{d\psi}{dx}\phi\right) = \left( \frac{d\psi}{dx}T\right)(\phi)$$

Which gives the right product rule $$\left( \frac{d}{dx}(\psi T)\right)(\phi) = \left( \psi \frac{dT}{dx}\right)(\phi) + \left( \frac{d\psi}{dx}T\right)(\phi)$$

Notation advice : to avoid such mistakes, I recommend using the following notation : $$\langle T,\phi\rangle \overset{\text{def}}{=} T(\phi)$$

Then, the derivation of the product rule is : \begin{align} \left\langle \frac{d}{dx}(\psi T), \phi \right\rangle &= - \left\langle \psi T , \frac{d}{dx} \phi \right\rangle \\ &= - \left\langle T, \psi \frac{d\phi}{dx} \right\rangle \\ &= \left\langle T, \frac{d\psi}{dx} \phi - \frac{d}{dx}(\phi\psi) \right\rangle \\ &= \left\langle \frac{d\psi}{dx} T ,\phi\right\rangle + \left\langle \frac{dT}{dx}, \phi\psi\right\rangle\\ &= \left\langle \frac{d\psi}{dx} T + \psi \frac{dT}{dx},\phi \right\rangle \end{align}

Answer 2

Note that:

$\begin{align} \dfrac{d}{dx} (\psi T)(\phi) &= -T \Big( \psi \dfrac{d\phi}{dx} \Big) = -T \Big( \dfrac{d}{dx} (\psi \phi) \Big) + T \left( \dfrac{d\psi}{dx}\phi \right) \\ &=\left\langle\frac{dT}{dx}, \psi \phi \right\rangle+\left\langle\frac{d\psi}{dx}T, \phi \right\rangle\\ &=\left\langle\frac{dT}{dx}\psi, \phi \right\rangle+\left\langle\frac{d\psi}{dx}T, \phi \right\rangle. \end{align}$