I only know basic $L^p$ theory (nothing about distributions) and am trying to prove the following:

Let $t>0$, $f\in L^{p}(\mathbb{R}^n,m)$, $\Gamma(t,x)=(4\pi t)^{-n/2}e^{-|x|^2/4t}$ and $$ u(t,x)=\int_{\mathbb{R}^n}\Gamma(t,x-y)f(y)dy. $$

If $1\leq p<\infty$, then $u(t,\cdot)\to f$ in $L^p$ as $t\to 0$.

(EDIT: I've asked the original parts 2 and 3 as separate questions.)

I've shown that $||u(t,\cdot)||_p\leq ||f||_p$ for $1\leq p\le\infty$ but haven't gotten anywhere from here. Could someone please give some hints or suggest a reference (these seem to be very well-known and widely applicable results).

  • $\begingroup$ Are you sure it is $e^{-x^2/(4t)}$ not $e^{-|x|^2/(4t)}$ ? $\endgroup$ – science Mar 9 '15 at 5:35
  • $\begingroup$ I'm sure you're right, thanks for the correction! $\endgroup$ – Aubrey Mar 9 '15 at 5:40
  • $\begingroup$ You are welcome! $\endgroup$ – science Mar 9 '15 at 5:42
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    $\begingroup$ (Just getting the facts straight: I was NOT the one who downvoted science's answer below, but I'm trying to do this without any use of distribution theory.) Any hints would be helpful, thanks! $\endgroup$ – Aubrey Mar 9 '15 at 17:12

(1) We notice first that

$$\int_{\mathbb{R}^n} \Gamma(t,x) dx = 1\; \forall t \in \mathbb{R}_{+} $$

And, as $t \rightarrow 0$, $\Gamma(t,x) \rightarrow 0, \; x \not=0$, so that, by the Dominated Convergence Theorem, $\int_{\mathbb{R}^n\backslash B(0,\delta)} \Gamma(t,x) dx \rightarrow 0 \;\forall \delta > 0$ as $t \rightarrow 0$. This will help us now:


$$\| u(t,\cdot) - f\|_p^p = \int_{\mathbb{R}^n} \left|\int_{\mathbb{R}^n}\Gamma(t,y)[f(x-y)-f(x)]dy\right|^pdx $$

By Minkowski's inequality for integrals, we've that

$$\|u(x,\cdot)-f\|_p \le \int_{\mathbb{R}^n}\left(\int_{\mathbb{R}^n} |f(x-y)-f(x)|^p dx\right) ^{1/p}\Gamma(t,y) dy\;\;\;\;(1) $$

Now we'll need a

Lemma: The translation is continuous in $L^p$, i.e., for each $f \in L^P$, we've that

$$\|\tau_hf - f \|_p \rightarrow 0$$

As $h \rightarrow 0$

The proof of this Lemma is a standard one: frist, we show that it is true for $f \in C^{\infty}_c(\mathbb{R}^n)$, then approximate any $L^p$ functions by functions of this kind. I won't provide details for this proof here.

So, now we've that

$$\|u(t,\cdot)-f\|_p \le \int_{\mathbb{R}^n} \|\tau_y f - f\|_p \Gamma(t,y)dy \\ \le \int_{B(0,\delta)}\|\tau_y f - f\|_p \Gamma(t,y)dy + 2\|f\|_p \int_{B(0,\delta)^{c}}\Gamma(t,y)dy $$

By the observations and the Lemma, we've that we may choose $\delta$ such that $\|\tau_y f - f \|_p \le \epsilon \;\forall |y|<\delta$, and, for this $\delta$, if we make $t$ sufficiently small, we'll have that the second integral is littler than or equal to $\epsilon$, so that, for small enough $t$,

$$ \|u(t,\cdot) - f\|_p \le \epsilon + 2\|f\|_p\;\epsilon$$

This Clearly finishes the proof of (1).

(2) EDIT: There was a mistake in this part, as @robjohn mentioned. I think this one is right: Let $f \in L^1\cap L^p$ at first. Then, by the Young's inequality,

$$ \|u(t,\cdot)\|_{\infty} \le \|\Gamma(t,\cdot)\|_{p'} \|f\|_p $$


$$ \|u(t,\cdot)\|_{1} \le \|f\|_1 $$

By $L^p$-Interpolation, we've that

$$ \|u(t,\cdot)\|_p \le \|u(t,\cdot)\|_1^{1/p}\|u(t,\cdot)\|_{\infty}^{(p-1)/p}\le C(f,p)\|\Gamma(t,\cdot)\|_{p'} ^{(p-1)/p} $$

Where $p'^{-1} + p^{-1} = 1$. But

$$\|\Gamma(t,\cdot)\|_{r}^r = \frac{1}{(4\pi t)^{nr/2}}\int e^{\frac{-r|x|^2}{4t}}dx = \frac{1}{(4\pi t)^{nr/2}} c(n) \int_{0}^{\infty} s^{n-1}e^{-\frac{rs^2}{4t}}ds = \\ \frac{c(n)}{(4\pi t)^{nr/2}} \sqrt{\frac{4t}{r}}^n \int_0^{\infty} w^{n-1}e^{-w^2} dw $$

Which shows that, for $r>1$, $\|\Gamma(t,\cdot)\|_r \rightarrow 0$ as $t \rightarrow \infty$. Thus, we've proved the Theorem for $f \in L^1 \cap L^p$.

For the general case, let $g \in L^1\cap L^p$ be such that $\|g-f\|_p \le \varepsilon$, and then

$$ \|u_f(t,\cdot)\|_p \le \|u_g(t,\cdot)-u_f(t,\cdot)\|_p + \|u_g(t,\cdot)\|_p \le \varepsilon + \varepsilon$$

If $t$ is big enough, where we've used once again that $\|u(t,\cdot)\|_p\le \|f\|_p$. So, the Theorem is proved.

Part (3) is lengthier than the previous one. The shortest proof I know to this fact uses that the convolution type Operator associated to the Gaussian is of Weak type 1-1 and dominated by a multiple of the Hardy-Littlewood Maximal Operator, but, as you said your background is of only $L^p$ spaces, then I suppose this is too complicated for this discussion.

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    $\begingroup$ The continuity lemma for $L^p$ only holds for $p\lt\infty$. I know it's a prerequisite for (1), but it should probably be mentioned. $\endgroup$ – robjohn Mar 9 '15 at 18:29
  • $\begingroup$ Yeah, Yeah, I should note that I was assuming $p < \infty$. $\endgroup$ – João Ramos Mar 9 '15 at 18:36

You need the fact

$$ \lim_{t\to 0} \frac{1}{(4\pi t)^{n/2}} e^{-\frac{|x|^2}{4t}} = \delta(x) $$

where $\delta(x)$ is the Dirac delta function.

  • $\begingroup$ Thanks, but this is supposed to be an exercise not assuming any knowledge of distribution theory; and you may well assume that in fact I don't know any yet. So basic $L^p$ theory and inequalities are my only tools, unfortunately. $\endgroup$ – Aubrey Mar 9 '15 at 5:43
  • $\begingroup$ @Aubrey: I can see the tag of distribution theory! $\endgroup$ – science Mar 9 '15 at 5:45
  • $\begingroup$ Would you care explaining what is the down vote for? The OP changed the title and the tag after I submitted my answer! $\endgroup$ – science Mar 9 '15 at 7:06
  • $\begingroup$ Still though, it might be better if you could perhaps remove the answer, since the question hasn't really been answered? $\endgroup$ – Aubrey Mar 9 '15 at 13:38
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    $\begingroup$ @Aubrey: A friendly piece of advice from a friendly moderator. You have some false impressions about how the site works. It often happens here that somebody answers a question using tools that the original asker is unfamiliar with. BECAUSE THE ANSWERS ARE NOT ONLY FOR YOU BUT ALSO FOR FUTURE VISITORS WHO HAVE A SIMILAR QUESTION, we view this as one of the strengths of the site. Just wait for another answer! $\endgroup$ – Jyrki Lahtonen Mar 9 '15 at 18:08

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