Let $f \in C^1(\mathbb{R})$ be such that both $f$ and $f'$ belong to $L^2(\mathbb{R})$. Show that $\hat{f} \in L^1(\mathbb{R})$

Question

The exercise is the following:

Let $f \in C^1(\mathbb{R})$ be such that both $f$ and $f'$ belong to $L^2(\mathbb{R})$. Show that $\hat{f} \in L^1(\mathbb{R})$.

What I want to prove is that $\int_\mathbb{R} |\hat{f}(\xi)|d\xi < +\infty$ but I do not know how to proceed; I have started saying that $\int_\mathbb{R} |\hat{f}(\xi)|d\xi = \int_\mathbb{R} | \int_\mathbb{R} f(x)e^{-2\pi i \xi x} dx|d\xi $ and then the idea was to use the properties of Fourier transform and of its derivatives such as:

• $f'(x) \hat{\to} 2\pi i \xi\hat{f}(\xi)$
• $-2\pi i x f(x) \hat{\to}\hat{f'}(\xi) $

but these properties require $f$ to be in the Schwartz space, that is not our case.

Any suggestion is appreciated. Thanks in advance for the help.

Answer 1

From the exchanges among @reuns, @RyszardSzwarc and myself here is one possible solution:

The assumptions of the problem imply that

1. $f^2$ (and thus $f$) is in $C_0$. See for instance this posting.
2. $2\pi i s\hat{f}(s)=\widehat{f'}(s)\in L_2$. See for example this posting.

The conclusion follows as indicated by reuns and Ryszard:

$$\hat{f}(s)=(1+s^2)^{-1/2}(1+s^2)^{1/2}\hat{f}(s)$$

and so,

$$\int|\hat{f}|\leq\Big(\int(1+s^2)^{-1}\,ds\Big)^{1/2}\Big(\int(1+s^2)|\hat{f}(s)|^2\,ds\Big)^{1/2}<\infty$$

Answer 2

One observation: The assumption $f\in\mathcal{C}^1$ can be weakened to $f$ is differentiable in $\mathbb{R}$. To see this, define $h=f^2$. Then $h\in L_1$ and $h'=2ff'\in L_1$. It follows that for any numbers $a<b$ \begin{align} h(b)-h(a)=\int^b_a h'(t)\,dt \end{align} (See theorem 7.21 in Rudin, W., Real and Complex Analysis, Third edition, page 149, which does not make reference to absolute continuity). Then the limits $A=\lim_{x\rightarrow\infty}h(x)$ and $B=\lim_{x\rightarrow-\infty}h(x)$ exits. The integrbility of $h=f^2$ implies that $A=B=0$. Hence, $f\in\mathcal{C}_0(\mathbb{R})$.

An application of Plancherel's theorem and Lebesgue integration by parts gives $$\widehat{f'}(t)=\lim_{R\rightarrow\infty}\int^R_{-R}e^{-2\pi x t}f'(x)\,dx=\lim_{R\rightarrow\infty}2\pi i t\int^R_{-R}e^{-2\pi i tx}f(x)\,dx=2\pi it\widehat{f}(t)$$ The convergence on the is in $L_2$, and pointwise along a subsequence $R_k\rightarrow\infty$. Therefore $2\pi i t\widehat{f}(t)=\widehat{f'}(t)\in L_2$. The rest is as indicated in the Community wiki answer