Consider the Cauchy problem
$$\left\{ \begin{array}{l l} u_{t} - \kappa u_{xx}=0 & \quad \mbox{$x \in \mathbb{R}, t>0$,}\\ \quad u(x,0) = \psi(x), \end{array} \right. $$
where $\kappa > 0$ is a constant, and $\psi (x)$ is an ${even}$ function. Show that the solution $u$ is even in $x$ for all $t>0$. i.e., $u(x,t) = u(-x,t)$.
The discussion in the comments insists on the fact that uniqueness is obtained provided that we have two constants $A, a > 0$ such that $$ |u(x,t)| \leq A e^{a |x|^2} \, , \qquad x\in\Bbb R,\, t\in [0,T] \, . $$ If this is satisfied, then one can speak of the solution.
That being said, if we set $\xi = -x$, we have $u_\xi = -u_x$, and $u_{\xi\xi} = u_{xx}$. Since $\psi$ is an even function, we have $\psi(\xi) = \psi(x)$. Therefore, $$ \left\lbrace \begin{aligned} u_t - \kappa u_{\xi\xi} &= 0 \, ,\\ u(\xi,0) &= \psi(\xi) \, , \end{aligned} \right. \qquad \xi\in\Bbb R,\, t\in [0,T] \, . $$ The bound on the solution is unchanged too: $|u(\xi,t)| \leq A e^{a|\xi|^2}$. Therefore, the PDE problem for $u(\xi,t)$ is the same as for $u(x,t)$. From uniqueness of the solution, we conclude that $u(\xi,t) = u(x,t)$.