Why does differentiation with respect to time commute with the Fourier Transform?

Question

In Stein and Shakarchi's Fourier Analysis, they often invoke the fact that differentiation with respect to time commutes with the Fourier transform with respect to space variables.

For example, on page 147 (Chapter 5, Proof of Theorem 2.1(i)), they show that $u(x, t) = \int_{-\infty}^{\infty} \hat{f}(\xi)e^{-4\pi^2 t\xi^2} e^{2\pi i\xi x}d\xi$ is infinitely differentiable by commuting the derivative with the integral, i.e. $\frac{\partial }{\partial t} = \int_{-\infty}^{\infty} \frac{\partial}{\partial t}\hat{f}(\xi)e^{-4\pi^2 t\xi^2} e^{2\pi i\xi x}d\xi$ and $\frac{\partial }{\partial x} = \int_{-\infty}^{\infty} \frac{\partial}{\partial x}\hat{f}(\xi)e^{-4\pi^2 t\xi^2} e^{2\pi i\xi x}d\xi$.

Another example is on page 185, motivating the solution to the Wave equation in $\mathbb{R}^d$, they say that "differentiation with respect to $t$ commutes with the Fourier transform in the space variables".

My question is, what is the justification for this? I can't seem to find in the book where they prove this claim.

In short, why is $\frac{\partial}{\partial y}FT(f(x, y)) = FT(\frac{\partial}{\partial y}f(x, y))$ where the Fourier transform, $FT$, is taken w.r.t. $x$?

EDIT

Many answers below point to the Liebniz rule, which I am familiar with. However, I think we cannot directly use the Liebniz rule to solve this problem since we are dealing with integration over an unbounded set while the Liebniz rule only applies to integration over a bounded set.

In particular, by the Liebniz rule, for any function $f$ that is continuously differentiable,

$$\begin{align*} \frac{\partial}{\partial t}\int_{-L}^{L}f(x, t)dx &= \int_{-L}^{L} \frac{\partial}{\partial t}f(x, t)dx \\ \lim_{L\rightarrow \infty}\frac{\partial}{\partial t}\int_{-L}^{L}f(x, t)dx &= \lim_{L\rightarrow \infty} \int_{-L}^{L} \frac{\partial}{\partial t}f(x, t)dx \\ \lim_{L\rightarrow \infty}\frac{\partial}{\partial t}\int_{-L}^{L}f(x, t)dx &=\int_{-\infty}^{\infty} \frac{\partial}{\partial t}f(x, t)dx \end{align*} $$

In particular, to show commutativity between integration and derivation, we want $\lim_{L\rightarrow \infty}\frac{\partial}{\partial t}\int_{-L}^{L}f(x, t)dx =\frac{\partial}{\partial t} \lim_{L\rightarrow \infty}\int_{-L}^{L}f(x, t)dx = \frac{\partial}{\partial t}\int_{-\infty}^{\infty} f(x, t)dx$, how do we do this?

Answer 1

There is nothing that special about the Fourier transform, if you derive in the variable that you are not transforming. Differentiation commutes with bounded linear operators in general. Say, if you have a $C^1$ function $\gamma$ from $\mathbb R$ to a Banach space $X$, and a linear, bounded operator $L$ from $X$ to another B.sp. $Y$, then the composition $L\circ\gamma=:\eta$ is a $C^1$ function from $\mathbb R$ to $Y$ and it satisfies $$ d/dt \,\eta(t)= L(d/dt\, \gamma(t)). $$ This is because when you look at the definition of derivative, the linear operator applied to the incremental ratio $(\gamma(t+h)-\gamma(t))/h$ yields $L((\gamma(t+h)-\gamma(t))/h)=(\eta(t+h)-\eta(t))/h$ by linearity, and then the limit as $h\to 0$ commutes with $L$ since $L$ is continuous.

The above is a nice case, where the curve is $C^1$ on a given space where the operator $L$ acts. In your case, the operator $L$ is the Fourier transform, which is continuous from some spaces to some different spaces, and the curve $y\mapsto f(\cdot,y)$ could be a continuous curve with values in some vector space $X$, but the $y$ derivative could belong to a different space $Z$. The above argument has to be modified accordingly, but the principle is always the same: write the incremental ratio, use the linearity of the Fourier transform to let it slide inside the numerator of the fraction, and then try to show that the limit coincides with the ansatz. You will have to use some continuity property of the Fourier Transform at some point. That is to say that in principle, the linearity of the Fourier transform and its continuity in suitable norms are all you need if you differentiate in the variable that you are not transforming.

Edit. This answer is complementary to those that commute integration and derivative. You have to be careful with that, because often the integral defining the Fourier transform is not defined in the usual sense for every $\xi$, for instance when you define the Fourier transform on $L^2$. In that case you have to use some density arguments to make the argument rigorous.

Answer 2

This comes from a general property about integrals, namely that for a "nice/smooth" function $g(x,t)$, $$ \frac{\mathrm d}{\mathrm dt}\int_Sg(x,t)\mathrm dx= \int_S\frac{\mathrm d}{\mathrm dt}g(x,t)\mathrm dx $$ This follows from the properties of sums of derivatives. Consider a simple Riemann sum. $$ \frac{\mathrm d}{\mathrm dt}\sum_{i=-n^2}^{n^2}\frac{1}{n}\cdot g\left(\frac{i}{n},t\right)=\sum_{i=-n^2}^{n^2}\frac{1}{n}\cdot \frac{\mathrm d}{\mathrm dt}g\left(\frac{i}{n},t\right) $$ Taking the limit $n\to\infty$, we (usually) get the integral above, and and this equality still holds. In the context of the Fourier Transform, we have $$ \frac{\mathrm d}{\mathrm dt}FT(f)(\zeta,t)=\frac{\mathrm d}{\mathrm dt}\int_{-\infty}^\infty f(x,t)e^{2\pi i\zeta x}\mathrm dx=\int_{-\infty}^\infty \frac{\mathrm d}{\mathrm dt}f(x,t)e^{2\pi i\zeta x}\mathrm dx =FT(\tfrac{\mathrm d}{\mathrm dt} f)(\zeta,t), $$ using that $e^{2\pi \zeta x}$ is a constant with respect to $t$.

Answer 3

As commented by Ted Shifrin, dominated convergence implies differentiation under the integral, see this question for example. The integrals there are over all of $\mathbb{R}$. For example, for your $t$ derivative you can apply this once you show that the absolute value of the function $$ (\xi,t) \mapsto -4 \pi^2 \xi^2 \cdot \hat{f}(\xi)e^{-4\pi^2 t\xi^2} e^{2\pi i\xi x} $$ is dominated by some integrable function $g(\xi)$ of $\xi$. Now I had to look up in your book that $t>0$ (you should've probably included this in the question), so we have $t>t_0>0$ for some $t_0$. So we can pick $$ g(\xi)= 4 \pi^2 \xi^2 \cdot |\hat{f}(\xi)|e^{-4\pi^2 t_0\xi^2}, $$ which is integrable as I'll leave to you (use theorem 1.12 on page 143 of your book to see that $\hat{f}$ is integrable).