Using Integration by Parts to Justify the Derivative of a Tempered Distribution

Question

From a textbook on Harmonic Analysis:

Question: Why does integration by parts yield

$$ T_{f'}(φ) = ∫_ℝ f'(x)φ(x)dx = - ∫_{ℝ}f(x)φ'(x)dx = -T_f(φ')? $$

Attempt. I know that Integration by Parts does yield that

$$ ∫_{ℝ}f(x)φ(x)dx = ∫_ℝ f'(x)φ(x)dx + ∫_{ℝ}f(x)φ'(x)dx $$

so that the equality would hold if $∫_{ℝ}f(x)φ(x)dx = 0$.

Answer 1

For a Schwartz function we have:

$\displaystyle \lim_{|x|\to \infty} f(x)φ(x)=0$

As a result, in other words, we have:

$\displaystyle f(x)φ(x)|_{-\infty}^{\infty}=0$

In integration by parts we have:

$\displaystyle ∫_ℝ f'(x)φ(x)dx + ∫_{ℝ}f(x)φ'(x)dx=f(x)φ(x)|_{-\infty}^{\infty} =0$

$\displaystyle \to ∫_ℝ f'(x)φ(x)dx=-∫_{ℝ}f(x)φ'(x)dx$

Answer 2

\begin{align*} T_{f'}(\varphi)&=\int_{\bf{R}}f'(x)\varphi(x)dx\\ &=\lim_{N\rightarrow\infty}\int_{-N}^{N}f'(x)\varphi(x)dx\\ &=\lim_{N\rightarrow\infty}\left(f(N)\varphi(N)-f(-N)\varphi(-N)-\int_{-N}^{N}f(x)\varphi'(x)dx\right)\\ &=\lim_{N\rightarrow\infty}-\int_{-N}^{N}f(x)\varphi'(x)dx\\ &=-\int_{\bf{R}}f(x)\varphi'(x)dx\\ &=-T_{f}(\varphi'). \end{align*}