Show $|u(t, x)|\leq\sup_{y\in\mathbb{R}}|u_0(y)|$ and $|\partial_xu(t,x)|\leq\frac1{\sqrt{\pi t}}\sup_{y\in\mathbb{R}}|u_0(y)|$ for $1$D heat equation Let $u_0=u_0(x): \mathbb{R} \to \mathbb{R}$ be a bounded $C^\infty$ class function and let $u = u(t, x): [0, \infty) \times \mathbb{R} \to \mathbb{R}$ be the solution to the initial value problem for the one-dimensional heat equation:
$$
\partial_t u = \partial_{xx} u \ \ \ in \ (0, \infty) \times \mathbb{R}\\
u(0, x) = u_0(x) \ \ \ \ \ \ \ \ \ \ \ \ in \  \mathbb{R}.
$$
Show that the following holds.
For any $(t, x) \in (0,\infty) \times \mathbb{R}$,
$$
|u(t, x)| \leq \sup_{y\in \mathbb{R}} |u_0(y)|
$$
and
$$
|\partial_x u(t,x)| \leq \frac{1}{\sqrt{\pi t}} \sup_{y\in\mathbb{R}}|u_0(y)|
$$
The latter permits an orderly exchange of derivatives and integrals.
I would appreciate it if you could tell me how to prove it, as I have no policy at all...
What I know:
Heat kernel$\ K = K(t, x): (0,\infty) \times \mathbb{R}, C^\infty,$
$$
K(t, x) = \frac{1}{\sqrt{4\pi t}}e^{-\frac{|x|^2}{4t}}.
$$
Solutions are
$$
\int_{-\infty}^\infty K(t, x-y) u_0(y)dy\ (t>0),\\ u_0(x)\ (t=0).
$$
i.e.
$$
\int_{-\infty}^\infty \frac{1}{\sqrt{4\pi t}}e^{-\frac{|x-y|^2}{4t}} u_0(y)dy\ (t>0),\\ u_0(x)\ (t=0).
$$
When $t>0$
$$
\partial_x u(t,x) = \frac{\partial}{\partial x} \int_{-\infty}^\infty K(t, x-y) u_0(y)dy\\
=\int_{-\infty}^\infty \frac{\partial}{\partial x} K(t, x-y) u_0(y)dy.
$$
 A: In the comments, we have proved $|u(t,x)|\leq \sup_{y\in\mathbb R}|u_0(y)|$. Now, we  use a similar argument to prove
$$|\partial_x u(t,x)| \leq \frac{1}{\sqrt{\pi t}} \sup_{y\in\mathbb{R}}|u_0(y)|.$$
As you already wrote in OP, we need to use
$$
\partial_x u(t,x) =\int_{-\infty}^\infty \frac{\partial}{\partial x} K(t, x-y) u_0(y)dy.
$$
Since $K(t,x)=\frac{1}{\sqrt{4\pi t}}e^{-\frac{|x|^2}{4t}}$, we have
$$\frac{\partial}{\partial x} K(t, x-y)=\frac{1}{\sqrt{4\pi t}}e^{-\frac{(x-y)^2}{4t}}\left(-\frac{2(x-y)}{4t}\right).$$
Hence
\begin{align*}
\int_\mathbb R\left|\frac{\partial}{\partial x} K(t, x-y)\right|\,dy&=\frac{1}{\sqrt{4\pi t}}\int_\mathbb R\left|\frac{2(x-y)}{4t}e^{-\frac{(x-y)^2}{4t}}\right|\,dy\\
&=\frac{1}{\sqrt{4\pi t}}\int_\mathbb R\left|\frac{2z}{4t}e^{-\frac{z^2}{4t}}\right|\,dz\\
&=\frac1{\sqrt{\pi t}}\int_0^\infty\left(\frac{2z}{4t}\right)e^{-\frac{z^2}{4t}}\,dz\\
&=\frac1{\sqrt{\pi t}}\int_0^\infty e^{-\frac{z^2}{4t}}\,d\left(\frac{z^2}{4t}\right)\\
&=\frac1{\sqrt{\pi t}}\int_0^\infty e^{-u}\,du=\frac1{\sqrt{\pi t}}.
\end{align*}
Therefore, we have
$$|\partial_x u(t,x)|\leq \int_\mathbb R\left|\frac{\partial}{\partial x} K(t, x-y)\right||u_0(y)|\,dy\leq \sup_{z\in\mathbb R}|u_0(z)|\int_\mathbb R\left|\frac{\partial}{\partial x} K(t, x-y)\right|\,dy=\frac{1}{\sqrt{\pi t}}\sup_{y\in\mathbb R}|u_0(y)|.$$
