Prove that if $f$ lies in $L^{1}({\bf R}^{n})$ then $\widehat f$ is uniformly continuous on ${\bf R}^{n},$

Question

My post did not duplicate that post because I did not ask what sequence of functions we apply the theorem to or how to solve this problem . The difference is that I solved this question and of course I showed my work . I just need to someone check the validity of my proof with me so I post my work.

Prove that if $f$ lies in $L^{1}({\bf R}^{n})$ then $\widehat f$ is uniformly continuous on ${\bf R}^{n},$ where ${\widehat f(\xi)}~$ is the so-called fourier transform which is given by $\displaystyle \int_{{\bf R}^{n}} f(x)e^{-2\pi i\langle x~,~\xi\rangle}~dx.$

Here is my working :(I edited)

Fixed any $\xi~,\eta\in{{\bf R}^{n}},~$we estimate the following : \begin{align} |{\widehat f(\xi+\eta)-{\widehat f(\xi)}}|&\le\int_{{\bf R}^{n}}|e^{-2\pi i\langle x,~\xi+\eta\rangle}-e^{-2\pi i\langle x,~\xi\rangle}|~|f(x)|~dx\\ &=\int_{{\bf R}^{n}}|e^{-2\pi i\langle x,~\xi\rangle}|~|e^{-2\pi i\langle x,~\eta\rangle}-1|~|f(x)|~dx\\ &=\int_{{\bf R}^{n}}|e^{-2\pi i\langle x,~\eta\rangle}-1|~|f(x)|~dx~~\color{red}{-(1)}\\ &\le\int_{{\bf R}^{n}}\bigg(1+|e^{-2\pi i\langle x,~\eta\rangle}|\bigg)|f(x)|~dx\\ &=\int_{{\bf R}^{n}}2|f(x)|~dx\\ &<+\infty \end{align} where the second and third equalities hold by the fact $|e^{2\pi i\langle x,~t\rangle}|=1$ and the last strict inequality is according to $f\in L^{1}({{\bf R}^{n}}).$

Now , from $\color{red}{(1)}$ we have $$\sup_{\xi\in{\bf R}^{n}}|{\widehat f(\xi+\eta)-{\widehat f(\xi)}}|\le\int_{{\bf R}^{n}}|e^{-2\pi i\langle x,~\eta\rangle}-1|~|f(x)|~dx$$

Moreover,

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\displaystyle\lim_{\eta\rightarrow0}|e^{-2\pi i\langle x,~\eta\rangle}-1|~|f(x)|=0~~$a.e.$~x\in{{\bf R}^{n}},$

keep in mind that $|f|$ is finite a.e. on ${{\bf R}^{n}}$ since $f\in L^{1}({{\bf R}^{n}})~$.

Hence , combine the result as previous and we see on account of the Lebesgue's Dominated Convergence Theorem that

$$\lim_{\eta\rightarrow 0}\sup_{\xi\in{\bf R}^{n}}|{\widehat f(\xi+\eta)-{\widehat f(\xi)}}|=0~.$$

The last line as above explains the uniformness on ${{\bf R}^{n}}$ of the fourier transform of the given $f$ .

Can anyone check my proof for validity ? Any comment or suggestion will be appreciated . Thanks for patient reading .

Answer 1

Strictly speaking, $\lim_{\eta\rightarrow 0}\left|\widehat{f}(\xi+\eta)-\widehat{f}(\xi)\right|=0$ for each $\xi$ proves no uniform continuity. One should do $\lim_{\eta\rightarrow 0}\sup_{\xi}\left|\widehat{f}(\xi+\eta)-\widehat{f}(\xi)\right|=0$. Luckily, part of your reasoning goes through: \begin{align*} \sup_{\xi}\left|\widehat{f}(\xi+\eta)-\widehat{f}(\xi)\right|\leq\int_{{\bf{R}}^{n}}\left|e^{-2\pi i\left<x,\eta\right>}-1\right||f(x)|dx, \end{align*} as $|e^{2\pi i\left<x,\eta\right>}-1|\rightarrow 0$ pointwise as $\eta\rightarrow 0$ and $\left|e^{-2\pi i\left<x,\eta\right>}-1\right||f(x)|\leq\min\{2\pi|\left<x,\eta\right>|,2\}|f(x)|\leq 2|f(x)|$, where $2|f|\in L^{1}({\bf{R}}^{n})$, so Lebesgue Dominated Convergence Theorem implies \begin{align*} \int_{{\bf{R}}^{n}}\left|e^{-2\pi i\left<x,\eta\right>}-1\right||f(x)|dx\rightarrow 0,~~~~\eta\rightarrow 0, \end{align*} so by Squeeze Theorem we have \begin{align*} \lim_{\eta\rightarrow 0}\sup_{\xi}\left|\widehat{f}(\xi+\eta)-\widehat{f}(\xi)\right|=0. \end{align*}

The fact about $|e^{-2\pi i\left<x,\eta\right>}-1|\leq\min\{2\pi|\left<x,\eta\right>|,2\}$: \begin{align*} |e^{iu}-1|&=\left|(\cos u-1)+i\sin u\right|\\ &=\left|-2\sin^{2}(u/2)+2i\sin(u/2)\cos(u/2)\right|\\ &=2|\sin(u/2)|\cdot|-\sin(u/2)+i\cos(u/2)|\\ &=2|\sin(u/2)|\\ &\leq 2\min\{1,|u/2|\}\\ &=\min\{2,|u|\}. \end{align*} Perhaps, \begin{align*} |e^{iu}-1|&=\left|(\cos u-1)+i\sin u\right|\\ &=\left|\int_{0}^{u}-\sin tdt+i\int_{0}^{u}\cos tdt\right|\\ &=\left|\int_{0}^{u}i^{2}\sin tdt+i\int_{0}^{u}\cos tdt\right|\\ &=\left|\int_{0}^{u}(\cos t+i\sin t)dt\right|\\ &\leq\left|\int_{0}^{u}1dt\right|\\ &=|u|, \end{align*} but $|e^{iu}-1|\leq 1+1=2$, so $|e^{iu}-1|\leq\min\{2,|u|\}$.