The heat kernel in one dimension for the real line is given by the usual gaussian density function:
$$g(t,x,y)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{(x-y)^2}{2t}}\, .$$
In particular, by differentiating this function, one finds that for $|x-y|\leq \sqrt{T}$,
$$\sup\_{t\in [0,T]} g(t,x,y) =\frac{C}{|x-y|}\, ,$$
for some constant $C$. 
My question is about the heat kernel for a bounded interval. More precisely, in the case of the interval $[0,\pi]$ instead of the real line, the heat kernel is given by
$$g(t,x,y)=\frac 2 \pi \sum\_{k\geq 1}\sin(kx) \sin(ky) e^{-k^2 t}\, .$$
Then do we have a similar estimate on the suppremum in time of the heat kernel, i.e. do we have an inequality like
$$\sup\_{t\in [0,T]} g(t,x,y) \geq \frac{C}{|x-y|}\, ,$$
valid for $|x-y|$ small enough?



Singularity of the heat kernel
Yes, away from the boundary: the heat kernel for the interval is given by $$\tag{1}g(t,x,y)=(2\pi t)^{-1/2}\sum\_{n\in\mathbb{Z}} (-1)^n \exp\left(-\frac{(x-y-n\pi)^2}{2t}\right),$$ and it is not difficult to show that the term corresponding to $n = 0$ is dominating for small time. It is in fact known that $$g(t,x,y) \approx C\_1 t^{-1/2} \exp\left(C\_2 \frac{(x-y)^2}{2t}\right) \min\left\{1,\frac{xy(\pi-x)(\pi-y)}{t}\right\}$$ with different constants $C\_1$ and $C\_2$ in upper and lower bound (a similar estimates holds for any bounded smooth domain). A sharper estimate can be obtained from (1) with some effort (and a similar bound for balls in $\mathbb{R}^n$ was [proved recently](https://arxiv.org/abs/1612.01833) by my colleagues Jacek Małecki and Grzegorz Serafin).


