I am slightly confused by Evans' proof of this theorem (Theorem 8 in Section 2.4 in the second edition of the book "Partial Differential Equation". By heat equation it is meant the homogeneous heat equation $u_t - \Delta u = 0$).
Theorem. Suppose $u \in C^2_1(U_T)$ solves the heat equation in $U_T$. Then $u \in C^\infty(U_T)$.
Here, $U$ is a bounded open set in $\mathbb{R}^n$, $U_T := U \times (0,T]$ and $u \in C^2_1(U_T)$ means that $u : U_T \to \mathbb{R}$ is s.t. $u \in C(U_T)$ and $u_t$, $D_x u$ and $D^2_x u$ exist and are continuous in $U_T$ (thus, up to the top $U \times \{t=T\}$). Although the notation $C^\infty(U_T)$ is not explicitly explained for this kind of domains, it should mean the obvious thing that $u$ is smooth in the interior of $U_T$ and every derivative is continuous up to the top $U \times \{t=T\}$. Let us introduce another piece of notation: for $(x^0,t^0) \in U_T$, let $$C(x^0,t^0) = \{ (y,s) \in \mathbb{R}^n \times \mathbb{R} : |x^0-y| \leq r \mbox{ and } t^0-r^2 \leq s \leq t^0 \}$$ be the cylinder of height $r^2$. Notice tht $C$ is closed and $(x^0,t^0)$ is the point at the center of the top part $U \times \{t=T\}$.
The proof, in few words, consists of two steps. In the first step, one assumes $u \in C^\infty(U_T)$, fix a point $(x^0,t^0) \in U_T$ and $r$ so small that $C := C(x^0,t^0;r) \subset U_T$. Letting $C'$, $C''$ be the cylinders defined as $C$ but with $r$ replaced by $3r/4$ and $r/2$ resp., one shows that $$\forall (x,t)\in C'', \quad u(x,t) = \iint_C K(x,y,t,s) u(y,s)\,\mbox{d}y\,\mbox{d}s, \tag{$*$}$$ where $K(x,t,y,s)$ is a smooth function on $C \setminus C'$ which is zero for all points $(y,s) \in C'$ ($K$ is a combination of the fundamental solution and pf certain derivatives of a cut off function that has been used to localize to $C$).
In the second step, one extends ($*$) to any solution of class $C^2_1(U_T)$. Once this has been done, the smoothness of $u$ in $C''$ is trivial. Evans says:
If $u$ satisfies only the hypotheses of the theorem, we derive ($*$) with $u^\varepsilon = \eta_\varepsilon *u$ replacing $u$, $\eta_\varepsilon$ being the standard mollifier in the variables $x$ and $t$, and let $\varepsilon \to 0$.
Now, it is clear to me that one gets ($*$) for $u^\varepsilon$ and that the moral is: "since ($*$) does not depend on the derivatives of $u$, we can pass to the limit inside the representation formula for $u^\varepsilon$ to get the one for $u$". However, I'm puzzled because, according to the standard properties of mollification, $u^\varepsilon$ is smooth not in $U_T$ but in $$U_T^\varepsilon := \{ (y,s) \in U_T : {\rm dist}((y,s),\partial U_T)) > \varepsilon \},$$ and moreover $u^\varepsilon \to u$ uniformly on compact subsets of $U_T \setminus (U \times \{t=T\})$ as $\varepsilon \to 0^+$. Thus, one gets without problems $u \in C^\infty( U_T \setminus (U \times \{t=T\}))$ but I don't see how the continuity of derivatives up to the boundary follows.