Question about $\|\hat{f} \|_{\infty} \leq \|f\|_{1}$ Let $f \in L^{1}(\mathbb{R})$ and denote $\hat{f}$ its Fourier transform.
It follows by definition of the Fourier transform that $\|\hat{f} \|_{\infty} \leq \|f\|_{1}$. I was trying to sense of when this inequality is strong or useful. Can someone give me an example of $f$ for which the given inequality is 'strong' or an explanation when this inequality is useful? Any comment appreciated.
 A: The inequality is used to show that the Fourier transform:
$$\mathcal{F}: L^1(\mathbb{R},\mathbb{C}) \to L^{\infty}(\mathbb{R})\ , \ f \mapsto \hat{f}$$
is a bounded linear map. Actually, the above isn't quite right. What you need to do is to let the co-domain of $\mathcal{F}$ be $\mathbb{C}^{\mathbb{R}}$ and then use your inequality to show that it does go into the space of essentially bounded maps. But that's not really something to get so upset about, is it?
Indeed, we can rephrase your inequality as:
$$\left|\left|\mathcal{F}f \right|\right|_{\infty} \leq \left|\left|f\right|\right|_1$$
The boundedness of this linear map is actually rather important because we would like to extend this map to other spaces if possible. Typically, you'll work with the Fourier transform on Schwartz space first or some nicely behaved dense subset of $L^2(\mathbb{R})$. You will show that, if $\mathcal{S}(\mathbb{R})$ is Schwartz space, then $\mathcal{F}$ is a bijective endomorphism of Schwartz space. This is one of the few results preceding the proof that $\mathcal{F}$ is, in fact, a unitary isomorphism of $L^2(\mathbb{R})$. The boundedness of the map allows you to apply the appropriate extension theorem.
A: What can one say about the image of $L^p(\Bbb R)$ under the Fourier tranform for $1 < p < 2$. Note that here again as in the case of $L^2(\Bbb R)$, one does not have any concrete integral expression of the Fourier transform, nor does one have any Plancherel theorem (unlike the case of $L^2(\Bbb R)$). In fact combining the inequality you have quoted above, the Plancherel theorem and the Riesz-Thorin interpolation, one not only concludes that the image of $L^p(\Bbb R)$ under the Fourier tranform for $1 < p < 2$ is contained in $L^{p*}(\Bbb R)$ (here $p*$ is the Holder conjugate of $p$) in fact one has the inequality,
$$||\hat{f}||_{L^{p*}(\Bbb R)} \le ||f||_{L^p(\Bbb R)} \:.$$
This is a very important application of the inequality cited in the question.
