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Consider $\varphi \in S(R^n)$ (space of rapidly decreasing functions). Consider the heat kernel

$$ K_t(x) = \displaystyle\frac{1}{{(4\pi t)}^{n/2}} \displaystyle e^{- \displaystyle\frac{|x|^2}{4t}}, t>0 , x \in R^n$$

I want to show that $$\displaystyle\lim_{t \rightarrow 0^{+}}\displaystyle\int_{R^n} e^{- \displaystyle\frac{|x|^2}{4t}} \varphi(x) \ dx = \varphi(0).$$

My try: we have $\displaystyle\int_{R^n} K_t(x) \ dx = 1$ and $\lim_{t \rightarrow 0^{+} }\displaystyle\int_{|x|\geq \epsilon}\displaystyle\frac{1}{{(4\pi t)}^{n/2}}e^{- \displaystyle\frac{|x|^2}{4t}}=0$ for all $\epsilon >0 $. Then

$$ \displaystyle\lim_{t \rightarrow 0^{+} } |\displaystyle\int_{R^n} e^{- \displaystyle\frac{|x|^2}{4t}} \varphi(x) \ dx - \varphi(0)| = \displaystyle\lim_{t \rightarrow 0^{+} } | \displaystyle\int_{R^n} \displaystyle\frac{1}{{(4\pi t)}^{n/2}}e^{- \displaystyle\frac{|x|^2}{4t}} \varphi(x) \ dx -\displaystyle\int_{R^n}\displaystyle\frac{1}{{(4\pi t)}^{n/2}} e^{- \displaystyle\frac{|x|^2}{4t}}\varphi(0) \ dx|$$

$$ \leq \displaystyle\lim_{t \rightarrow 0^{+} } || \varphi - \varphi(0)||_{\infty}. \displaystyle\lim_{\epsilon \rightarrow 0^{+} } \displaystyle\int_{|x|\geq \epsilon}\displaystyle\frac{1}{{(4\pi t)}^{n/2}}e^{- \displaystyle\frac{|x|^2}{4t}} \ dx$$

$$ =\displaystyle\lim_{\epsilon \rightarrow 0^{+} } || \varphi - \varphi(0)||_{\infty} \displaystyle\lim_{t \rightarrow 0^{+} }\displaystyle\int_{|x|\geq \epsilon}\displaystyle\frac{1}{{(4\pi t)}^{n/2}}e^{- \displaystyle\frac{|x|^2}{4t}} \ dx = 0$$

I dont know if my solution is correct (i am not sure about the last line). Someone can give me a hint to this exercise ?

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  • $\begingroup$ Why is this convergence true, $\lim_{t \rightarrow 0^{+} }\displaystyle\int_{|x|\geq \epsilon}\displaystyle\frac{1}{{(4\pi t)}^{n/2}}e^{- \displaystyle\frac{|x|^2}{4t}}=0$. Does the sequence of functions, $\displaystyle\frac{1}{{(4\pi t)}^{n/2}}e^{- \displaystyle\frac{|x|^2}{4t}}$, converge uniformly to 0? $\endgroup$
    – alpastor
    Commented Dec 1, 2018 at 2:29

1 Answer 1

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For the last line to be true, you need show that the limits can be interchanged. I will propose her another solution. Note

\begin{eqnarray} \left|\int_{\mathbb{R}^n}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}\varphi(x)-\int_{\mathbb{R}^n}\frac{1}{{(4\pi t)}^{n/2}} e^{- \frac{|x|^2}{4t}}\varphi(0) \right| &\leq& \nonumber \\ \int_{\mathbb{R}^n}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}|\varphi(x)-\varphi(0)| &=& \nonumber \\ \int_{|x|\leq\delta}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}|\varphi(x)-\varphi(0)| +\int_{|x|>\delta}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}|\varphi(x)-\varphi(0)| \tag{1}\end{eqnarray}

For $\epsilon>0$, choose $\delta>0$ such that $|\varphi(x)-\varphi(0)|\leq\epsilon$ for $\|x\|\leq\delta$. We conclude from $(1)$ that

$$\left|\int_{\mathbb{R}^n}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}\varphi(x)-\int_{\mathbb{R}^n}\frac{1}{{(4\pi t)}^{n/2}} e^{- \frac{|x|^2}{4t}}\varphi(0) \right| \leq \\ \epsilon+\int_{|x|>\delta}\frac{1}{{(4\pi t)}^{n/2}} e^{-\frac{|x|^2}{4t}}|\varphi(x)-\varphi(0)| $$

Now you can conclude.

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  • $\begingroup$ nice solution. Thanks !!! ! $\endgroup$ Commented Nov 8, 2013 at 23:25
  • $\begingroup$ You are welcome @LeandroTavares $\endgroup$
    – Tomás
    Commented Nov 8, 2013 at 23:45
  • $\begingroup$ Why $\lim_{t\to 0^+}\int_{|x|>\delta}K_t(x)|\varphi(x)-\varphi(0)|dx=0$? $\endgroup$
    – eraldcoil
    Commented Jun 3, 2022 at 19:28

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