TITLE: For $f \in L^1(\mathbb{R})$ and $y > 0$, we have ${1\over{\sqrt{y}}} \int_\mathbb{R} f(x - t)e^{{-\pi t^2}\over{y}}dt \in L^1(\mathbb{R})$?
QUESTION [6 upvotes]: For $f \in L^1(\mathbb{R})$ and $y > 0$, let$$f_y(x) := {1\over{\sqrt{y}}} \int_\mathbb{R} f(x - t)e^{{-\pi t^2}\over{y}}dt.$$

Do we have $f_y \in L^1(\mathbb{R})$ for every $ y > 0$?
Do we have $\lim_{y \to 0} \int_\mathbb{R} |f(x) - f_y(x)|\,dx = 0$?
Does there exist $C > 0$ such that, for every $f \in L^1(\mathbb{R})$,$$\left\{x \in \mathbb{R} : \sup_{y > 0} \left|f_y(x)\right| > \lambda\right\} \le {C\over\lambda} \int_\mathbb{R} \left|f(t)\right|\,dt?$$
If $f \in L^1(\mathbb{R}$, then do we have that for a.e. $x \in \mathbb{R}$, $\lim_{y \to 0} f_y(x) = f(x)$?

REPLY [3 votes]: This is an attempt to answer all 4 questions by elementary means.
Write $T_y$ for the linear operator that maps $f\in L^1$ to $f_y.$
For the first question we have
$$\eqalignno{
\|T_y f\|_1&=\frac1{\sqrt y}\int_x\left|\int_t f(x-t)\exp(-\frac{\pi t^2}y)dt\right|dx\\
&\leq\frac1{\sqrt y}\int_x\int_t\left|f(x-t)\right|\exp(-\frac{\pi t^2}y)dtdx\\
&=\frac1{\sqrt y}\int_t\int_x|f(x-t)|dx\exp(-\frac{\pi t^2}y)dt&(*)\\
&=\|f\|_1\left(\frac1{\sqrt y}\int_t\exp(-\frac{\pi t^2}y)dt\right)\\
&=\|f\|_1.
}$$
where in (*) we use the Fubini-Tonelli theorem.
The second question is the assertion that $\lim_{y\to0}T_yf=f$ in $L^1.$ This is easy to verify for $f$ the indicator function of an interval, and from there to finite linear combinations (block functions). But block functions are dense in $L^1,$ so if $f_b$ is a block function close to $f$ then
$$\eqalign{
\|T_yf-f\|_1&\leq\|T_y f-T_y f_b+T_y f_b-f_b+f_b-f\|_1\\
&\leq\|T_y(f-f_b)\|_1+\|T_yf_b-f_b\|_1+\|f_b-f\|_1\\
&\leq2\|f-f_b\|_1+\|T_yf_b-f_b\|_1
}$$
in view of the inequality in the answer to question 1.
Question 4 is about convergence almost everywhere, which follows from the stronger question 2 about convergence in $L^1.$
For question 3 note that the function
$$\mathbb R^+\to\mathbb R:y\mapsto\frac1{\sqrt y}\exp(-\frac{\pi t^2}y)$$
reaches its maximum when $y^2=\pi t^2,$ so we have
$$\sup_y\frac1{\sqrt y}\exp(-\frac{\pi t^2}y)=\frac1{\pi^{1/4}\sqrt{|t|}}\exp(-\sqrt\pi|t|).$$
Let us denote the expression on the right hand side by $\phi(t),$ then $\phi$ is an integrable function.
Now we estimate
$$\eqalignno{
\int_x\sup_y|T_yf(x)|dx&\leq\int_x\sup_y\frac1{\sqrt y}\int_t|f(x-t)|\exp(-\frac{\pi t^2}y)dt\ dx\\
&\leq\int_x\int_t|f(x-t)|\sup_y\frac1{\sqrt y}\exp(-\frac{\pi t^2}y)dt\ dx\\
&=\int_t\int_x|f(x-t)|\sup_y\frac1{\sqrt y}\exp(-\frac{\pi t^2}y)dx\ dt&(*)\\
&=\|f\|_1\int_t\phi(t)dt
}$$
from which we deduce that $x\mapsto\sup_y|T_yf(x)|$ is an integrable function, and for any $\lambda>0$ the measure of
$$\left\{x\in\mathbb{R}:\sup_y\left|f_y(x)\right|>\lambda\right\}$$
is bounded by
$$\frac{\|\phi\|_1}\lambda\|f\|_1.$$