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


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*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. $x \in \mathbb{R}$, $\lim_{y \to 0} f_y(x) = f(x)$?


Question (1)
We shall use the theorem: $f, g \in L^1 (\Bbb R) \implies f * g \in L^1 (\Bbb R)$.
Choose $g_y (t) = \frac 1 {\sqrt y } \Bbb e ^{- \frac {\pi t^2} y}$ and verify that $\int \limits _{\Bbb R} |g_y| \ \Bbb d t = 1$, so the aforementioned theorem gives us that $f_y = f * g_y \in L^1 (\Bbb R)$.
Question (2)
We shall use the theorem: let $f, g \in L^1(\Bbb R)$ with $\int \limits _{\Bbb R} g \ \Bbb d t = 1$ and let $g_r (t) = \frac 1 r g(\frac t r)$; then $\| f * g_r - f \| _1 \to 0$.
Choose $g(t) = \Bbb e ^{- \pi t^2}$; choose $r = \sqrt y$ and note that $g_y (t) = \frac 1 {\sqrt y} g(\frac t {\sqrt y})$, so we may apply the aforementioned theorem to get that $\int \limits _{\Bbb R} |f_y(x) - f(x)| \Bbb d x = \| f * g_y - f \|_1 \to 0$.
The theorems can be found, for instance, in lecture 4 from the course on Real Analysis by Cristopher Heil from Georgia Tech (theorem 1.12 for (1), and the bottom of p20 and top of p21 for (2)). Alternatively, you may want to take a look at the Lecture 2 of a course given by Hart Smith from the University of Washington (pages 2-6 for (1) and page 11 for (2)). In general, a search on the web for "convolution lp" or "convolution lp spaces" will return even more results.
