Show that Fourier transformation is differentiable if $\int|xf(x)|\,d\lambda<\infty$

Question

Let $f\in\mathcal{L}^1(\mathbb{R},\mathcal{M},\lambda)$. Then we define the Fourier transform of $f$, denoted $\hat{f}$, by $$\hat{f}(t)=\int_\mathbb{R}e^{-itx}f(x)\,d\lambda(x),\;\;\;\;\;\;\;t\in\mathbb{R}.$$ Prove that if $\int_\mathbb{R}|xf(x)|\,d\lambda(x)<\infty$, then $\hat{f}$ is differentiable on $\mathbb{R}$ and $$\hat{f}'(t)=\int_{\mathbb{R}}(-ix)e^{-itx}f(x)\,d\lambda(x),\;\;\;\;\;\;\;t\in\mathbb{R}.$$

Any hints on this? I can see that the integrand in the differentiated expression is the derivative of the integrand in the original expression, and that $|(-ix)e^{-itx}f(x)|=|xf(x)|$, but I don't know how to attack this.

PS: I saw that a similar question was asked here, but the answer used the Riemann–Lebesgue lemma and Fubini's theorem, neither of which has been covered in my book when this exercise is given.

Solution based on the comments below: $$\hat{f}'(t)=\lim_{h\to0}\frac{1}{h}\left(\int_\mathbb{R}e^{-i(t+h)x}f(x)\,d\lambda(x)-\int_\mathbb{R}e^{-itx}f(x)\,d\lambda(x)\right) \\=\lim_{h\to0}\int_\mathbb{R}\frac{\cos(-(t+h)x)-\cos(-tx)}{h}f(x)\,d\lambda(x) \\+ i \lim_{h\to0}\int_\mathbb{R}\frac{\sin(-(t+h)x)-\sin(-tx)}{h}f(x)\,d\lambda(x)$$

By the mean value theorem, $$\left|\frac{\cos(-(t+h)x)-\cos(-tx)}{h}f(x)\right|\leq|xf(x)|$$ and $$\left|\frac{\sin(-(t+h)x)-\sin(-tx)}{h}f(x)\right|\leq|xf(x)|.$$ Hence, since $|xf(x)|$ is integrable by assumption, we get by the dominated convergence theorem that $$\hat{f}'(t)=\int_\mathbb{R}\lim_{h\to0}\frac{\cos(-(t+h)x)-\cos(-tx)}{h}f(x)\,d\lambda(x) \\+ i\int_\mathbb{R}\lim_{h\to0}\frac{\sin(-(t+h)x)-\sin(-tx)}{h}f(x)\,d\lambda(x)\\=\int_\mathbb{R}x\sin(-tx)f(x)\,d\lambda(x)+i\int_\mathbb{R}-x\cos(-tx)f(x)\,d\lambda(x)\\=\int(-ix)e^{-itx}f(x)\,d\lambda(x).$$

Answer 1

Your proof by separating $e^{-iu}$ into its real and imaginary parts is fine, well done.

But how would you solve this without separating into real and imaginary parts?

Pretty much in the same way, we only need to consider the difference quotient

$$\frac{e^{-i(t+h)x} - e^{-itx}}{h}$$

and need to show that its modulus is bounded by $C\cdot\lvert x\rvert$ for some constant $C\in (0,+\infty)$. Then the same appeal to the dominated convergence theorem finishes the proof. Now,

$$\left\lvert \frac{e^{-i(t+h)x} - e^{-itx}}{h}\right\rvert = \left\lvert e^{-itx}\frac{e^{-ihx}-1}{h}\right\rvert = \left\lvert \frac{e^{-ihx}-1}{h}\right\rvert$$

by the addition theorem for the exponential function and the fact that $\lvert e^{iu}\rvert = 1$ for all $u\in \mathbb{R}$.

The elegant way to obtain the desired bound now multiplies the fraction with $e^{ihx/2}$, which doesn't change the modulus, to obtain

$$\left\lvert\frac{e^{-ihx/2} - e^{ihx/2}}{h}\right\rvert = \left\lvert \frac{-2i\sin \frac{hx}{2}}{h}\right\rvert = \left\lvert \frac{\sin \frac{hx}{2}}{\frac{h}{2}}\right\rvert \leqslant \left\lvert\frac{\frac{hx}{2}}{\frac{h}{2}}\right\rvert = \lvert x\rvert.$$

If we don't think of that way, we can observe that for $x = 0$, the numerator of the difference quotient is always $0$, so the bound in that case holds for all $C$, hence we need only look at $x\neq 0$, and can consider $u = -hx$ as the variable and need to show that

$$\left\lvert \frac{e^{iu}-1}{u}\right\rvert$$

is bounded. For $\lvert u\rvert \geqslant 1$, we have the bound

$$\left\lvert \frac{e^{iu}-1}{u}\right\rvert \leqslant \frac{\lvert e^{iu}\rvert + 1}{\lvert u\rvert} = \frac{2}{\lvert u\rvert} \leqslant 2,$$

and for $\lvert u\rvert < 1$, the Taylor series of the exponential function yields

$$\left\lvert \frac{e^{iu}-1}{u}\right\rvert = \left\lvert \sum_{k=1}^\infty \frac{i^ku^{k-1}}{k!}\right\rvert \leqslant \sum_{k=1}^\infty \frac{\lvert u\rvert^{k-1}}{k!} < e-1 < 2,$$

and thus, re-expanding $u$,

$$\left\lvert \frac{e^{-ihx}-1}{h}\right\rvert = \left\lvert \frac{e^{iu}-1}{u}\cdot (-x)\right\rvert \leqslant 2\lvert x\rvert.$$

(So the elegant way not only is shorter and prettier, it also yields a better bound. But one doesn't always find the neat way, hence it's good to know that there also is another way.)