Ive been looking for bounds on Fourier coefficients on bounded measurable functions but can only find for bounded variation. Is there nothing to say about the coefficients of the former? and why if that is the case?
It certainly looks like they are bounded on finite measure spaces
$\int_{[0,1]}fe^{-xn}d\lambda<\mid\mid f\mid \mid$
What about the converse, given bounded sequence is there an $L^{\infty}$ function?
Best Answer
In order to get bound on Fourier coefficients on any $L^p$ function, we first must see how is Fourier transform making sense on that $L^p $ space and what is the image of that $L^p $ space under the Fourier transform.
So the entire story of Fourier transform starts from $L^1$ and we have that image of $L^1$ under Fourier transform is contained in $L^\infty$ ( by Riemann-Lebesgue lemma it's actually contained inside $C_0$, in fact image of $L^1$ forms a dense sub-algebra of $C_0$) and moreover, $$||\hat{f}||_{L^\infty} \le ||f||_{L^1}$$ i.e. Fourier transform is a bounded Linear transformation from $L^1 \to L^\infty$ with operator norm $\le 1$. In case of $L^2$, we have Plancherel hence, Fourier transform is a Unitary operator on $L^2$. Then you note by Hausdorff-Young inequality, that Fourier transform is actually a bounded linear transformation from $L^p \to L^{q}$ where ( $1 \le p \le 2$ and $\frac{1}{p}+\frac{1}{q}=1$ ) with operator norm $\le 1$ ( by Riesz-Thorin interpolation ) .
But what happens when $p > 2$ ? If we have something like a Hausdorff-Young inequality in this case, then we should move forward, right? As that should give us some bound etc. But Alas! no such inequality exists! In fact , the first question that should bother us is where does the Fourier transform takes this $L^p $ spaces?
Recall that for $L^p,p>2$, the Fourier transform is defined basically in terms of tempered distributions. So let's compute the Fourier transform of an $L^\infty$ function by hand, say $f \equiv 1$. As $f \in L^1_{loc}(\Bbb R)$ we have the tempered distribution $T_f$ defined by $T_f(\phi)=\int_{\Bbb R} f \phi ,\forall \phi \in \mathscr{S}(\Bbb R)$ where $\mathscr{S}(\Bbb R)$ is the space of all Schwarz class functions on $\Bbb R$. As $f \equiv 1$, we get that the Fourier transform $$\hat{T_f}(\phi)=T_f(\hat{\phi})=\int_{\Bbb R} \hat{\phi}(x)dx=\phi(0)=\delta_0(\phi), \forall \phi \in \mathscr{S}(\Bbb R)$$ Then as you note, the fourier transform of the $L^\infty$ function $f \equiv 1$ is the dirac measure at 0, i.e. the Fourier transform is not even a function! So you shouldn't be expecting any bound for Fourier coefficients on $L^\infty$ functions, isn't it?
Then there is definitely a bound, and it's trivially true, since $L^p \subset L^1, \forall p$, so basically you're working with the initial bound on $L^1$