# Schwartz functions & differentiation under the integral sign.

Let $\mathscr{S}(\mathbb{R}^n)$ denote the space of Schwartz functions. That is $$\mathscr{S}(\mathbb{R}^n) = \left\{ f \in C^\infty(\mathbb{R}^n, \mathbb{R}) ~\colon \sup_{x\in\mathbb{R}^n} \left|x^\alpha \partial_\beta f(x) \right| < \infty \quad \forall \alpha, \beta \right\}$$ where $\alpha$ and $\beta$ are multi-indices. Take $f(t, x) \in \mathscr{S}(\mathbb{R}^{n+1})$ and define $F ~\colon \mathbb{R} \to \mathbb{R}$ by $$F(t) = \int_{\mathbb{R}^n} f(t, x) ~\mathrm{d}t$$ I would like to know when it is valid to compute $F'(t)$ by "differentiation under the integral sign". In other words, when is it justifiable to say $$F'(t) = \partial_t \int_{\mathbb{R}^n} f(t, x) ~ \mathrm{d}t = \int_{\mathbb{R}^n} \partial_t f(t, x) ~ \mathrm{d}t$$ I'm think I am able to show this when I have a few extra conditions, say if I know that $\partial_t f(t, x)$ is a decreasing function of $t$ for all fixed $x$. However, I feel that it should be true in general and perhaps I'm just missing something important. Any help will be appreciated.

## 1 Answer

The result is in fact true in general. For any $h\in\mathbb{R}$, $h\ne0$, $$\frac{F(t+h)-F(t)}{h}=\int_{\mathbb{R}^n}\frac{f(t+h)-f(t)}{h}\,dx.$$ On the one hand $$\lim_{h\to0}\frac{f(t+h)-f(t)}{h}=\frac{\partial f}{\partial t}(x,t)\quad\forall x\in\mathbb{R}^n.$$ On the other $$\Bigl|\frac{f(t+h)-f(t)}{h}\Bigr|=\Bigl|\frac{\partial f}{\partial t}(x,\tau)\Bigr|$$ for some $\tau$ between $t$ and $t+h$. Since $f$ is a Schwartz function, it is bounded (uniformly in $\tau$) by an integrable function. Now use the dominated convergence theorem.

• Hmm, it is precisely the uniform boundedness of a Schwartz function $f$ by an $L^1$ function that I am having trouble proving in general. – providence Feb 6 '15 at 10:47
• If $\phi$ is an Schwartz function in $\mathbb{R}^n$, then $|\phi(x)|\le C_k(1+|x|)^k$ for all $k$. – Julián Aguirre Feb 6 '15 at 11:25