Riemann-Lebesgue lemma

If $f\in L^1({\mathbb{R}}^n)$, then its Fourier transform is continuous and tends to zero as $|\xi |\to \infty$. Approximate $f$ in $L^1$ by a compactly supported smooth function; integration by parts makes the approximant's transform decay, while $\Vert \hat{f}-\hat{g}{\Vert }_{\infty }\le \Vert f-g{\Vert }_1$ transfers the conclusion.

For a continuous real function $f$ on $[a,b]$, uniform step approximation on a compact interval gives a finite step function $g={\sum }_j{c}_j{1}_{I_j}$ with small uniform error. Each cosine integral over an interval has absolute value at most $2/|\omega |$. ThereforeFirst choose $g$ to make the first term small, then let $|\omega |$ grow with $g$ fixed. This proves the cosine version of the Riemann-Lebesgue lemma without differentiating $f$; the sine and complex-exponential versions follow in the same way.

The functions $f_n(t)=\sin (nt)$ converge weakly to zero in $L^1([0,2\pi ])$. Indeed, every $g\in L^{\infty }$ also belongs to $L^1$ on this finite interval, and the Riemann-Lebesgue lemma gives ${\int }_0^{2\pi }g(t)\sin (nt)dt\to 0$. However,so $L^1([0,2\pi ])$ does not have the Schur property.

If a radial function on ${\mathbb{R}}^n$ behaves like $r^a$ near zero and like $r^{-b}$ at infinity, then its $q$th power is locally integrable at zero exactly when $aq>-n$ and integrable at infinity exactly when $bq>n$. This follows from the radial measure factor $r^{n-1}dr$.

For $2\le p<\infty$,Thus every function in $L^2\cap L^{\infty }$ belongs to every $L^p$ between them.