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I will try to write my proof of the Riemann lebesgues Lemma, correct me if something is wrong.

If $f\in L_{1}(R)$ then its fourier transform $f^{*}$ maps to $C_{0}(R)$, a space of continuous functions that converge to zero in infinity, that is $\lim_{|\zeta|\to\infty} |f^{*}(\zeta)|=0.$

Proof: Since compactly supported functions are dense in $L_{1}$, there exists a function g such that $||f-g||<\varepsilon$, therefore $$\lim_{|\zeta|\to \infty}|f^{*}|=\lim_{|\zeta|\to \infty}|\int_{-\infty}^{+\infty}e^{-2\pi i x \zeta}(f(x)-g(x)+g(x))dx|\leq$$ $$\lim_{|\zeta|\to \infty}\int_{-\infty}^{+\infty}|f(x)-g(x)||e^{-2\pi i x \zeta}|dx+\lim_{|\zeta|\to \infty}|\int_{-\infty}^{+\infty}g(x)e^{-2\pi i x \zeta}dx|$$ Now the first part is less than $\varepsilon$ and since $g$ has compact support, $-\infty +\infty$ becomes some $a,b$ and since $g$ is contious... Can someone help me to proceed with the second part? I assume since g is continuous it can be approximated and because of $\cos (2\pi x \zeta)$ and $\sin (2\pi x \zeta)$ it will go to zero, please help.

Thank you so much!

Vuk Stojiljkovic
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  • I fixed it! Thanks for the comment. Hopefully someone reaches out to help me. I believe this is the way but I cant seem to deal with the second part to go to zero... – Vuk Stojiljkovic Mar 09 '24 at 15:23
  • Proving it for $C_c$ functions directly still seems hard without more ‘tricks’. I suggest you approximate better using $C^1_c$ functions atleast so that you can integrate-by-parts. Alternatively, you can even use Stone-Weierstrass to uniformly approximate by trigonometric polynomials integrated over a large compact interval $[-R,R]$ (and the result in this case is an obvious direct calculation). See here and the link within. – peek-a-boo Mar 10 '24 at 02:05
  • @peek-a-boo Using $C_{c}^{1}$ functions it should be easy, since by partial integration one part will vanish due to $g$ being 0 on the boundaries, then going inside with abs value will leave M as the Max since g' exists on (a,b) and |e^{...}| which is 1 by module, we will have $\frac{1}{\zeta}$ in the front and therefore it will go to zero. Another Idea I came up with is that by using step functions, since g is integrable there exists step function s.t |g-l|<\varepsilon , and then we will have | \int l \cosx| , and here since l is a step function, it is 1 on some set A s.t $A\subset (a,b)$ – Vuk Stojiljkovic Mar 10 '24 at 08:49
  • and zero outherwise, therefore it is $\sum_{i=1}^{i=n}|\int_{x_{i-1}}^{x^{i}}c_{n}\cos(2\pi x \zeta)dx|$ which when solved we get $\leq\sum_{i=1}^{n}\frac{2c_{n}}{\zeta}$ which goes to zero since $|\zeta|\to \infty$. I believe this would work too, any thoughts? – Vuk Stojiljkovic Mar 10 '24 at 08:51
  • Yup sounds like what I said in the link. – peek-a-boo Mar 10 '24 at 15:21

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