# Norm of the integral operator in $L^2(\mathbb{R})$.

Suppose we have an integral operator $A$ such that $$Af(x) = \frac{1}{\sqrt{2\pi}}\int\limits_{\mathbb{R}}e^{-\frac{(x-y)^2}{2}}f(y)dy$$

To find $\|A\|$ we can use the unitary Fourier transform $F$, as it saves the norm of a function and hence $\|A\|=\|FA\|$. Due to the convolution theorem we have: $$FAf(x) = e^{-\frac{x^2}{2}}Ff(x)$$

So $FAf(x)$ is a composition of the unitary Fourier transform $F$ and the multiplication operator $M$ by the function $e^{-\frac{x^2}{2}}$. Norm of the latter in $L^2(\mathbb{R})$ equals to $\mathrm{ess\,sup}\left(e^{-\frac{x^2}{2}}\right) = 1$. And the norm of $F$ is surely also $1$. That means that $\|A\| \leq 1$

The quastion is how one proves that $\|A\|=1$ (or is it so)?

My guess of $\|A\|=1$ bases on the fact that if $f\equiv 1$, than $Af \equiv 1$. But we can't use it straightforward, since $f \equiv 1$ does not lie in $L^2(\mathbb{R})$. Even if we consider sequance $f_n=\chi[-n,n]$ - indicators of $[-n,n]$, that doesn't do much.

While finding such sequince $\{f_n\}$ that $\|Af_n\|/\|f_n\| \to 1$ is one way to prove the equility, I wonder if it somehow follows from the fact that $F$ is unitary operator and not just have unity norm. And so it somehow is evident that oparator $FAf(x) = e^{-x^2}Ff(x)$ has a unity norm and not just $\leq 1$.

Any suggestion is appreciated.

Yes, it is true that $\|A\|=1$. The operator $A$ is sometimes called a Fourier multiplier or simply a multiplier because it acts by pointwise multiplication of the Fourier transform. (Alternatively, multipliers are the same as convolution operators). So anyway, we can prove the following:
Let $m\in L^\infty(\mathbb{R})$ be a bounded and measurable function, and let $T_m:L^2(\mathbb{R})\to L^2(\mathbb{R})$ be the operator defined by $F(T_m f) = m(F(f))$, where $F$ is the Fourier transform. Then $\|T_m\|_{L^2\to L^2} = \|m\|_{L^\infty}$.
Your argument already shows that $\|T_m\|\leq \|m\|_{L^\infty}$, so we will show that $\|T_m\|\geq \|m\|_{L^\infty}$ by showing that $\|T_m\| \geq \|m\|_{L^\infty} - \varepsilon$ for any $\varepsilon>0$.
Indeed, let $\varepsilon > 0$ (small enough that $\varepsilon < \|m\|_{L^\infty}$) and consider the set $$E_\varepsilon = \{x\in\mathbb{R}; |m(x)| \geq \|m\|_{L^\infty} - \varepsilon\}$$ where $m$ is close to its maximum value. This set has positive measure. It may in fact have infinite measure, so intersect it with a ball of large radius if necessary so that $0<|B_R\cap E_\varepsilon|<\infty$. Now consider the function $f$ defined by $F(f)(\xi) = \frac{m(\xi)}{|m(\xi)|}\chi_{B_R \cap E_\varepsilon}(\xi)$. Then we can compute $$\|f\|_{L^2}^2 = \|F(f)\|_{L^2}^2 = |B_R\cap E_\varepsilon|$$ and $$\|T_m f\|_{L^2}^2 = \|m F(f)\|_{L^2}^2 = \int_{B_R\cap E_\varepsilon} |m(\xi)|^2 \,d\xi \geq |B_R\cap E_\varepsilon| (\|m\|_{L^\infty} - \varepsilon)^2.$$ Thus $$\|T_m f\|_{L^2} \geq (\|m\|_{L^\infty} - \varepsilon) \|f\|_{L^2},$$ which shows $\|T_m\| \geq \|m\|_{L^\infty} - \varepsilon$.
Notice that if we try to do this procedure with your operator, we would apply $A$ to functions whose Fourier transform has support in smaller and smaller balls around the origin. Such functions tend to spread out and become nearly constant (so that their limit is the function $f=1$ that you suggested). Also by slightly modifying this proof you could find that your suggestion of using indicator functions for large intervals would work, because most of the mass of their Fourier transform lies near the origin.
You can approximate the constant function with Gaussians $\phi_{\sigma^2}(x)=\frac{1}{\sqrt{2\pi \sigma^2}} \exp(-x^2/2\sigma^2)$. If I calculated correctly you have $\|\phi_{\sigma^2}\|^2= \dfrac{1}{2\sqrt{\pi \sigma^2}}$ and since $A\phi_{\sigma^2} = \phi_{\sigma^2+1}$ (the convolution of two Gaussians is Gaussian with the sum of the variances) you get $$\|A\|^2\ge \frac{\|A \phi_{\sigma^2}\|^2}{\|\phi_{\sigma^2}\|^2} = \frac{\|\phi_{\sigma^2+1}\|^2}{\|\phi_{\sigma^2}\|^2} = \frac{\sigma^2}{\sigma^2+1} \to 1$$ for $\sigma \to \infty$.