Krdytkyu

Let $u(x,t)$be the solution for the Cauchy Problem

$$u_t-u_{xx} = 0 mbox{ in $mathbb{R}times ]0, infty[$}$$ $$u(x,0)
= u_0(x) mbox{ in $mathbb{R}$} $$

where $u_0in S(mathbb{R})$(schwartz space in $mathbb{R})$. Conclude that, for each fixed $t_0>0$,
the function $xto u(x,t_0)$ belongs to the Gevrey class of order
$1/2$ in $mathbb{R}$

I already defined the Gevrey class here: $sup_K |partial^{alpha}u|le C^{|alpha|+1}alpha!^s$ then $u$ is analytic for $sle 1$ but here is the definition again:

A function $uin C^{infty}$ belongs to the Gevrey Class of order $s$
if for every compact $K$ of $Omega$ there is a constant $C$ such that

$$sup_K |partial^{alpha}u|le C^{|alpha|+1}alpha!^s, alphainmathbb{Z}_+^N$$

this exercise comes after this one:

define, for $xinmathbb{R}$,

$$v(x) = int_{mathbb{R}}e^{ixlambda-alambda^2}dlambda$$

Show that $v$ belongs to the Gevrey cass of order $1/2$ in
$mathbb{R}$

So maybe they have something in common. The last exercise above looks like a fourier transform.

UPDATE

$$||partial_x^n phi||_{L^2} = (frac{1}{sqrt{4pi t}}int|partial_{x}^n| e^{-(xsqrt{4t})^2}dx)^{1/2} = (frac{1}{sqrt{pi}}int|partial_x^n e^{-x^2}|dx)^{1/2} = ||partial_x^n phi_{1/4}||L^2$$

By your second box, we have the result that

If $v_a$ is defined by
$$v_a(x) = int_{mathbb{R}} e^{ixlambda -alambda^2};dlambda$$
the $v_a$ is in the Gervy class of order $frac{1}{2}$.

In fact, if you look at the result from Finding $sup_{lambda ge 0}{lambda^k e^{−alambda^2/2}}$, you see that we actually have something better:

$$sup_{x in mathbb{R}} partial^alpha v(x) le C^{|alpha| + 1} alpha!^frac{1}{2}$$

Moreover, $v_t$ is actually (up to a scaling factor) the fundamental solution of the heat equation. We can obtain it by taking the Fourier transform in $x$ with frequency variable $lambda$, solving the resulting ODE, and inverting the transform.

Thus, it would suffice to prove the following result:

$$sup_{x} partial^alpha v_t * g (x) le C^{alpha + 1}alpha!^frac{1}{2}$$ for any Schwartz class $g$.

Note that
$$begin{align*}
sup_{x} |partial^alpha (v_t * g)(x)| &=leftlVert(partial^alpha v_t) * g (x)rightrVert_{L^infty}\
&le lVert partial^alpha v_trVert_{L^infty} lVert g rVert_{L^1}\
&le C^{|alpha|+1}lVert g rVert_{L^1} alpha!^frac{1}{2}
end{align*}$$
by Holder's inequality. It follows that $u(x,t) = v_t * g$ is in the Gervy class of order $frac{1}{2}$ for any fixed $t > 0$.

In this answer, all norms are spatial norms (i.e. there is no integration/supremum wrt the parameter $t_0$ that was already fixed in the question, and I have dropped the subscript $0$).

The unique solution to the heat equation with nice initial data is well known as the convolution with the Gaussian kernel,
$$ u(x,t) = phi_t * u_0(x) = frac1 {sqrt{4pi t}}int_{mathbb R} e^{-|x-y|^2/4t} u_0(y) dy.$$

Then by Cauchy-Schwarz ($|int fg|le |f|_{L^2}|g|_{L^2}$ ),
begin{align}
&|partial_x^n u(x,t)|_{L^infty} \
& = |(partial_x^n phi_t) * u_0|_{L^infty} \
& = left |int_{mathbb R}partial^n_xphi_t(x-y) u_0(y) dyright |_{L_x^infty} \
& le left | |partial^n_xphi_t(x-y)|_{L^2_y} |u_0|_{L^2}right |_{L_x^infty} \
& = left | |partial^n_xphi_t|_{L^2} |u_0|_{L^2}right |_{L_x^infty} \
& = |partial^n_xphi_t|_{L^2} |u_0|_{L^2} \
&le C^{n+1} |partial_x^n phi_{1/4}|_{L^2} |u_0|_{L^2} \
&le C_1^{n+1} |partial_x^n phi_{1/4}|_{L^2} end{align}
Above, we used Chain rule,
$$partial^n_xphi_t(x) = C(t) partial^n_x exp(-x^2/(4t)) = frac{1}{(sqrt{4t})^{n}} (partial_x^nphi_{1/4})(x/sqrt{4t}) $$
and also $ int_{mathbb R} f(lambda x)^2 dx = frac1lambda int_{mathbb R} f^2(y) dy $, so that
$$ |partial^n_xphi_t|_{L^2} le C^{n+1}|partial_x^n phi_{1/4}|_{L^2} $$
Note that $$partial_x^n phi_{1/4}(x) = (-1)^nH_n(x) phi_{1/4}(x),$$ where $H_n$ is the $n$th (physicist) Hermite polynomial.
The Hermite polynomials have the well known $L^2$ norm when weighted with $phi_{1/4}$ (which I quote from that same wikipedia page),
$$int H_n^2 phi_{1/4}^2 le |phi_{1/4}|_{L^infty} int H_n^2 phi_{1/4} le C sqrt{2pi} n! $$
Thus
$$ |partial_x^n u(x,t)|_{L^infty} le C_2^{n+1} sqrt{n!}.$$

Response to update in OP: note that you seem to have avoided chain rule by using
$$ int |partial_x^n [phi_{1/4} (lambda x) ]|^2 dx overset{?}{=} int |[partial_x^n phi_{1/4}] (lambda x)|^2 dx $$
Note the very intentional bracketing. This isn't true, try writing it out explicitly for $n=1$.