# Norm of multiplication operator by a Fourier transform of an $L^1$ function

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.

• 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 $M_{\hat g} f \to \|g\|_{L^{1}}$. Commented Feb 16, 2021 at 9:33
• @KaviRamaMurthy I also tried doing that but I was unsure about how to go about this. I thought of normalized indicators, but got stuck. Commented Feb 16, 2021 at 9:41
• This may help: math.stackexchange.com/questions/550669/… Commented Feb 16, 2021 at 9:46
• @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 $g \in L^1$ ?". Commented Feb 16, 2021 at 9:48
• @TheSilverDoe I think the question can be reformulated as such. Commented Feb 16, 2021 at 9:58

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?.

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.

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