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

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

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  • $\begingroup$ It's H-Y if you assume $Fu\in L^m$. For general $u\in L^k$ it doesn't even make sense because the FT is not guaranteed to be in $L^m$. Questions of this type would be much better suited for MSE, though. $\endgroup$
    – Christian Remling
    Commented Jul 7, 2017 at 1:29
  • $\begingroup$ Or I guess you could declare the RHS to be $\infty$ if $Fu\notin L^m$, and then you're good no matter what. $\endgroup$
    – Christian Remling
    Commented Jul 7, 2017 at 1:33
  • $\begingroup$ @ChristianRemling Yes, that is the convention I'm using. However, why it is H-Y is not clear to me. $\endgroup$
    – user89890
    Commented Jul 7, 2017 at 1:35
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    $\begingroup$ If you use the correct normalization, you have $\mathcal{F}^2 u(x) = u(-x)$ which is clearly an $L^p$ isometry. $\endgroup$
    – Willie Wong
    Commented Jul 7, 2017 at 2:02
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    $\begingroup$ The "proof", assuming the Fourier inversion formula, is one line long. The statement appears at least two times in the link I showed you (once written in the form $\mathcal{F}^2 = \mathcal{P}$). These statements all hold on Schwartz space, which is dense in any $L^p$ with $1\leq p \leq 2$. If you can't work out the details: ask on Math.SE; this is not the place for it. $\endgroup$
    – Willie Wong
    Commented Jul 8, 2017 at 1:41

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