# Inequality about the Fourier transform: $\Vert u \Vert_{L^k} \le \Vert \mathcal{F}(u) \Vert_{L^m}$ (where $1 \le m \le 2$ and $m,k$ Holder conjugates)

How can I prove the following inequality about the Fourier transform?

$$\Vert u \Vert_{L^k} \le \Vert \mathcal{F}(u) \Vert_{L^m}$$ for $1 \le m \le 2$ and $m,k$ Holder conjugates (that is $1/k + 1/m = 1)$.

I think it follows from the classical Hausdorff-Young inequality, but I don't know how.

Hausdorff-Young $$\Vert \mathcal{F}(u) \Vert_{L^p} \le \Vert u \Vert_{L^q}$$ for $1 \le q \le 2$ and $m,k$ Holder conjugates (that is $1/k + 1/m = 1)$.

• I think your inequality is backwards. The correct inequality is known as the Hausdorff-Young inequality... it can easily be proved using the Riesz-Thorin interpolation theorem. Commented Jul 6, 2017 at 23:50
• @YousufSoliman I know Hausdorff-Young, I think this must be a corollary, but I don't know how to get it. I've edited the question to include it.
– user428573
Commented Jul 7, 2017 at 0:41
• If I'm not mistaken I think such an inequality would say that the inverse fourier transform is continuous on $L^{\infty}$ when $k=1$. I don't think that is true. Commented Jul 7, 2017 at 1:13

I did not read carefully the hp before. What I said here is true but is not an answer to the question.

It look like false. Take $f_n(x)$ the characteristic function of the >interval $[0,n]$. For any $n\in\mathbb{N}$ this function is in $L^1$ but >$\lim_{n\rightarrow\infty}\|f_n\|_{L^1}=\infty$. On the other hand, if my >calcula are correct, one has $$\hat{f_n}(\xi)=\frac{1-e^{-in\xi}}{i\xi}.$$ The $L^{\infty}$ norm of the latter quantity stays bounded in $n$.

Now I try to answer the question. Let me prove that $$\|u\|_{\infty}\leq\|\hat{u}\|_{L^1}.$$ If $\hat{u}$ is not in $L^1$ then there is nothing to prove. Moreover to give a meaning to $\hat{u}$ one has to require that $u$ is in $L^1$. Therefore I am working on the space $$X:=\{u\in L^1:\hat{u}\in L^1\}.$$ It is classical, but I do not have a reference and the proof is quite long (but not hard), that if $u$ is in $X$ then $$2\pi u(x)=\int \hat{u}(\xi)e^{ix\xi}d\xi$$ for a. e. $x$. Therefore one has $$\|u(x)\|_{\infty}\leq\|\int\hat{u}(\xi)e^{ix\xi}d\xi\|_{\infty}\leq\int|\hat{u}(\xi)|d\xi=\|\hat{u}\|_{L^1}.$$ Then you can interpolate this inequality with $\|u\|_{L^2}=\|\hat{u}\|_{L^2}$ to get the intermediate numerology.

• Remember that $1 \le m \le 2$. So you cannot take $L^\infty$ there.
– user428573
Commented Jul 7, 2017 at 12:19
• oh! you're totally right! Commented Jul 7, 2017 at 12:32
• Maybe now the proof is done Commented Jul 7, 2017 at 13:16
• Thank you. Can you add the details on the interpolation? That's where I'm not clear.
– user428573
Commented Jul 8, 2017 at 7:17

Set $u=\mathcal F v$ in Hausdorff-Young, using the fact that $$\mathcal F \mathcal F v = \overline{v}(-x),$$ so in particular $\| \mathcal F \mathcal F v \|_{L^p}=\|v\|_{L^p}$ for all $p\in [1, \infty]$.