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Let $$\Phi(x,t)=\frac{1}{\sqrt{4\pi t}}e^{-x^2/4t}$$ be the fundamental solution of the heat equation (or the heat kernel).

What is the supremum of $\Phi$ over $x$: $$\sup_x \Phi(x,t)?$$

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    $\begingroup$ It's not hard to see that it occurs at $x=0$. $\endgroup$ – Ian Oct 21 '15 at 1:24
  • $\begingroup$ First, $|\Phi(x,t)|\ne \frac{1}{\sqrt{4\pi t}}$. Rather, $|\Phi(x,t)|=\Phi(x,t)$ since $\Phi(x,t)$ is non-negative. Second, the least upper bound over $x$ is simply $\frac{1}{\sqrt{4\pi t}}$. $\endgroup$ – Mark Viola Oct 21 '15 at 1:28
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    $\begingroup$ Yes you are right. I just realised...so in conclusion, $\sup_x\Phi(x,t)=\frac{1}{\sqrt{4\pi t}}$? $\endgroup$ – math101 Oct 21 '15 at 1:32
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Note that, we have $\exp(\xi) < 1$ for $\xi < 0$ and $\exp(0) = 1$. Hence, for every $x \ne 0$, we have $$ \Phi(x,t) = \frac 1{(4\pi t)^{1/2}}\exp\left(-\frac{x^2}{4t}\right) < \frac 1{(4\pi t)^{1/2}} $$ As $$ \Phi(0, t) = \frac 1{(4\pi t)^{1/2}}$$ we have that $$ \sup_{x\in \mathbf R} \Phi(x,t) = \frac 1{(4\pi t)^{1/2}} $$

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