- Suppose that f is in \(L ^ { 1 } ( \mathbb { R } )\) . Prove directly (i.e., without citing properties of the Fourier transform) that the function
\[ \widehat { f } ( t ) = \int _ { \mathbb { R } } e ^ { - i x t } f ( x ) d x \]
is uniformly continuous and \({ \widehat { f } } ( t ) \to 0 { \mathrm { ~ a s ~ } } t \to \infty\)
Solution
Write \(\norm f_1=\int_\RR\abs f\). The integral defining \(\widehat f(t)\) converges absolutely, since its integrand has modulus \(\abs{f(x)}\).
1. \(\widehat f\) is uniformly continuous.
Let \(\eps>0\); we may assume \(\norm f_1>0\), since otherwise \(\widehat f=0\). Choose \(R\ge1\) with \(\int_{\abs x>R}\abs f<\eps/4\). Since \(\abs{e^{iu}-1}\le\abs u\) for real \(u\), and \(\abs{e^{-ixt}-e^{-ixs}}\le2\), \[\abs{\widehat f(t)-\widehat f(s)}\le\int_{\abs x\le R}\abs x\abs{t-s}\abs{f(x)}\,dx+2\int_{\abs x>R}\abs f\le R\abs{t-s}\norm f_1+\frac\eps2.\] So \(\abs{t-s}<\eps/(2R\norm f_1)\) implies \(\abs{\widehat f(t)-\widehat f(s)}<\eps\), with a bound independent of \(t\) and \(s\).
2. \(\norm{f(\cdot+h)-f}_1\to0\) as \(h\to0\).
Given \(\eta>0\), choose \(\varphi\in C_c(\RR)\) with \(\norm{f-\varphi}_1<\eta\) [@Fol13]. Translation invariance of the integral gives \(\norm{f(\cdot+h)-f}_1\le2\eta+\norm{\varphi(\cdot+h)-\varphi}_1\). For \(\abs h\le1\) both functions on the right vanish outside one bounded interval, and uniform continuity of \(\varphi\) makes their difference tend uniformly to \(0\), so its integral tends to \(0\). Letting \(h\to0\) and then \(\eta\to0\) proves the claim.
3. \(\widehat f(t)\to0\) as \(\abs t\to\infty\).
For \(t\ne0\), the substitution \(x=u+\pi/t\) and \(e^{-i\pi}=-1\) give \(\widehat f(t)=-\int e^{-iut}f(u+\pi/t)\,du\). Averaging with the definition, \[\abs{\widehat f(t)}=\frac12\abs{\int e^{-ixt}\bigl(f(x)-f(x+\pi/t)\bigr)\,dx}\le\frac12\norm{f(\cdot+\pi/t)-f}_1,\] which tends to \(0\) by step 2.
4. Q.E.D.
Steps 1 and 3 are the two claims.