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.
 A: 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?.
A: 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.
