Let $S(t)$ be the semi-group generated by the Dirichlet Laplacian in $L^2(0,1)$, which is given, for $y\in L^2(0,1)$, by 


$$S(t)y=\displaystyle\sum\_{n=1}^\infty e^{-n^2\pi^2 t} \langle y,\sin(n\pi x) \rangle \sin(n\pi x).$$ 


For $t>0, y\in L^1(0,1),$ we have $\|S(t)y\|\_{L^\infty(0,1)}\le C t^{-\frac{1}{2}} \|y\|\_{L^1(0,1)}$.


My question is about an **explicit value** of the constant $C$.


Thanks!



A Laplacian semi-group estimation
A rough estimate is obtained as follows. The $L^{+\infty}$−norm of $S(t)y$ is majorized by $\sum\_{n=1}^\infty e^{-n^2\pi^2 t}|\langle y,\sin(n\pi x)\rangle|$ which in turn is majorized by $\|y\|\_{L^1}\sum\_{n=1}^\infty e^{-n^2\pi^2 t}$ *assuming that* $\langle y,\sin(n\pi x)\rangle$ is intended to mean $\int\_0^1y(s)\sin(n\pi s){\kern.7mm\rm d\kern.7mm}s$. So it remains to estimate $\sum\_{n=1}^\infty e^{-n^2\pi^2 t}$. This is majorized by $\int\_0^{+\infty}e^{-s^2\pi^2 t}{\kern.7mm\rm d\kern.7mm}s$. Recalling that $\int\_0^{+\infty}e^{-u^2}{\kern.7mm\rm d\kern.7mm}u=\frac12\sqrt\pi$, and making the substitution $u=s\pi\sqrt t$, we obtain $\pi\sqrt t\int\_0^{+\infty}e^{-s^2\pi^2 t}{\kern.7mm\rm d\kern.7mm}s=\frac12\sqrt\pi$ whence further $\int\_0^{+\infty}e^{-s^2\pi^2 t}{\kern.7mm\rm d\kern.7mm}s=t^{-\frac12}\frac1{2\sqrt\pi}$. So $C=\frac1{2\sqrt\pi}$ is one possibility.


By the way, if one wants to have $S(0)y=y$ for at least $y=\langle\kern.6mm\sin(n\pi s):0<s<1\kern.6mm\rangle$ with $n\in\mathbb Z^+$, one should instead have 
$$
(S(t)y)(s)=\sum\_{n=1}^{+\infty}e^{-n^2\pi^2 t}\left(2\int\_0^1y(s)\sin(n\pi s){\kern.7mm\rm d\kern.7mm}s\right)\kern.7mm\sin(n\pi s)\text{ .}
$$ 
Then my argument above gives $C=\pi^{-\frac12}\kern.6mm$.


