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I am lost with the following exercise that is posed in Steve Rosenberg's book "The Laplacian on a Riemannian Manifold":

Show that for a continuous function f,

\begin{equation} \lim_{t \to 0} \quad \frac{1}{\sqrt{4 \pi t}} \, \int_\mathbb{R} \exp \left(-\frac{(x - y)^2}{4t}\right) f(y) \, dy = f(x) \end{equation}

I am not sure how to tackle this - using L'Hopital's Rule doesn't really simplify the limit .. any hint would be a huge help, many thanks !!

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    @QiaochuYuan: Hm .. by looking at the integral I thought the kernel becomes small as $t \to 0$, since the t stands in the denominator of a negative fraction of the exponential. Another thing that confuses me is we are integrating$y$over the whole line, so what do you exactly mean by saying that a fixed$x$is close to$y$? Sorry for these dumb questions, and thanks for your patience! Your comment sounds very interesting as it might help me understanding the impact of the kernel better, if you could elaborate a bit more that would be great !2012-04-17

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I like the probabilistic approach.

The integral on the left is $\mathbb{E}(f(X_t))$, where $X_t$ is a normal random variable with mean $x$ and variance $2t$. Since $X_t\to x$ in distribution as $t\to0$, we get $\mathbb{E}(f(X_t))\to f(x).$ We have assumed that $f$ is a bounded, continuous function.$ $ $ $ $ $

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    What could be said, if the function is not bounded, but merely $\mathbb E(f(X_t))$ is finite for small enough t?2015-02-27
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From the distribution point of view let's observe that $\delta_a(d)=\dfrac 1{a\sqrt{\pi}}e^{-\dfrac {d^2}{a^2}}$ admits the limit $\delta(d)$ as $a \to 0$ (see the animation at the right).

Let's set $a:=2\sqrt{t}$ and $d:=x-y$ then $\delta_t(x-y)=\dfrac 1{\sqrt{4\pi t}}e^{-\dfrac {(x-y)^2}{4t}}$ will have the limit $\delta(x-y)$ as $t\to 0$ and

$\lim_{t \to 0} \quad \frac{1}{\sqrt{4 \pi t}} \, \int_\mathbb{R} \exp \left(-\frac{(x - y)^2}{4t}\right) f(y) \, dy =\int_\mathbb{R} \delta(x-y) f(y) \, dy = f(x)$

Of course justifications are required in the previous steps. You may find them in introductions to distributions like Zemanian's 'Distribution Theory and Transform Analysis' or in the chap. 6 of Olver's 'Generalized Functions and Green's Functions'.

Hoping this helped,

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If we assume $f(x)$ to be analytic around $x$, then we can expand it in Taylor series at $x$ and we obtain $ \lim_{t \to 0} \quad \frac{1}{\sqrt{4 \pi t}} \, \int_\mathbb{R} \exp \left[-\frac{(x - y)^2}{4t}\right] \sum_n f^{(n)}(x) (y-x)^n \, dy = f(x) . $

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    I completely forgot to specifiy that f can only assumed to be continuous. I have now added this to the post. But in any case, thanks for this initial answer!2012-04-16