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. $x \in \mathbb{R}$, $\lim_{y \to 0} f_y(x) = f(x)$?
  • 3
    $\begingroup$ This is a convlution of two $L^1$ function thus is also $L^1$. $\endgroup$
    – user99914
    Dec 10 '15 at 8:49

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


is bounded by


  • $\begingroup$ For the first question, you should probably detail why you may permute the integrals. $\endgroup$
    – Alex M.
    Dec 17 '15 at 22:09
  • $\begingroup$ I think part 3 follows even more straightforwardly from Chebyshev inequality and the result you proved in part 1. No? $\endgroup$
    – bartgol
    Dec 17 '15 at 22:11
  • $\begingroup$ @Alex M. Thanks, I will try to come up with something readable after I get some sleep :-) $\endgroup$ Dec 17 '15 at 22:15
  • $\begingroup$ @bartgol It looks like Chebyshev, but isn't the supremum getting in the way? $\endgroup$ Dec 17 '15 at 22:16
  • $\begingroup$ Uhm, I don't think so. Without the supremum it would be plain Chebyshev. But then, since the inequality would hold for every $y>0$, it would also hold for the supremum. No? $\endgroup$
    – bartgol
    Dec 17 '15 at 22:21

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.


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