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I'm trying to compute the operator norm of the multiplication operator $\mathcal{M}_{\hat{g}}:L^2(\mathbb{R}^d)\to L^2(\mathbb{R}^d)$ given by $\mathcal{M}_{\hat{g}}(f)=f\cdot \hat{g}$, where $\hat{g}$ is the Fourier transform of $g\in L^1(\mathbb{R}^d)$. I think that the operator norm is equal to $\Vert g\Vert_{L^1}$, but I was only able to show it is a bound for $\Vert \mathcal{M}_{\hat{g}}\Vert_{op}$.

I know that $\Vert \mathcal{M}_{\hat{g}}\Vert_{op}=\Vert \hat{g}\Vert_{L^\infty}$ and that $\Vert \hat{g}\Vert_{L^\infty}\leq \Vert \mathcal{F}\Vert_{op}\cdot \Vert g\Vert_{L^1}=\Vert g\Vert_{L^1}$, where $\mathcal{F}$ is the Fourier transform. I want to find a sequence of $L^2$ functions such that

$$ \Vert f_n\cdot \hat{g}\Vert_{L^2} \overset{n\to \infty}{\to}\Vert g\Vert_{L^1} \quad \text{and} \quad \Vert f_n\Vert_{L^2}=1 \quad \text{for all} \quad n, $$

but I'm not sure how to go about this. This might be a trivial question but I'm not sure how to solve this point. I'm also not sure whether this is indeed true.

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  • $\begingroup$ Very rarely you will find a single vector where the operator norm is attained. You should try to find a sequence of functions $f_n$ of norm $1$ such that $M_{\hat g} f \to \|g\|_{L^{1}}$. $\endgroup$ Commented Feb 16, 2021 at 9:33
  • $\begingroup$ @KaviRamaMurthy I also tried doing that but I was unsure about how to go about this. I thought of normalized indicators, but got stuck. $\endgroup$ Commented Feb 16, 2021 at 9:41
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    $\begingroup$ This may help: math.stackexchange.com/questions/550669/… $\endgroup$ Commented Feb 16, 2021 at 9:46
  • $\begingroup$ @Keen-ameteur I am not sure to understand the question. You know that $\Vert \mathcal{M}_{\hat{g}}\Vert_{op}=\Vert \hat{g}\Vert_{L^\infty}$, and you want to know if $\Vert \mathcal{M}_{\hat{g}}\Vert_{op}=\Vert g\Vert_{L^1}$ ? If so, the question can be reformulated as "do we have $\Vert g\Vert_{L^1} = \Vert \hat{g}\Vert_{L^\infty}$ for $g \in L^1$ ?". $\endgroup$ Commented Feb 16, 2021 at 9:48
  • $\begingroup$ @TheSilverDoe I think the question can be reformulated as such. $\endgroup$ Commented Feb 16, 2021 at 9:58

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As you remarked, for $g∈ L^1$, we have $\|\mathcal{M}_{\widehat g}\|_{L^2\to L^2} = \|\widehat g\|_{L^\infty} ≤ \|g\|_{L^1}$.

$\bullet$ If $g≥ 0$, then $\|g\|_{L^1} = ∫ g = \widehat{g}(0) ≤ \|\widehat g\|_{L^\infty}$.

$\bullet$ However, in the general case, this reverse inequality is false. See for example the answer of Giuseppe Negro here: Estimate the $L^1$-norm of the Fourier transform, or the answer of David C Ullrich here: Is the inverse of the Fourier transform $L^1(\mathbb R)\to (C_0(\mathbb R),\Vert \cdot \Vert_\infty)$ bounded?.

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Just wanted to add a relatively easy counter-example.

Let $M>0$ and suppose $g(x)=1_{[-M,M]}\cos(x)$.

Then, using that $\sin$ and $\cos$ is odd, you get that

$$ \hat{g}(\xi)=2\int_0^M \cos(x)\cos(\xi x)\textrm{d}x $$ And applying the age-old double integration by parts trick, we get that $$ \int_0^M \cos(x)\cos(\xi x)\textrm{d}x=[\sin(x)\cos(\xi x)]_0^M-\xi\left([-\cos(x)\sin(\xi x)]_0^M +\xi\int_0^M \cos(x)\cos(\xi x)\textrm{d}x\right) $$ Gathering everything, we get that $$ \hat{g}(\xi)=\frac{2}{1+\xi^2}\left(\sin(M)\cos(\xi M)+\xi\cos(M) \sin(\xi M\right)), $$ which is uniformly bounded by 4, independently of $M$.

However, $\|g\|_{L^1}\geq \lfloor\frac{M}{2\pi}\rfloor \frac{\pi}{\sqrt{2}}$, so indeed, it is not true that $\|g\|_{L^1}=\|\hat{g}\|_{L^{\infty}}$ for $M$ sufficiently large.

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  • $\begingroup$ I mean... I guess this example is sort of analogous to Ullrich's answer referenced below. Take it for what it's worth. $\endgroup$ Commented Feb 16, 2021 at 11:56

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