Estimate for the heat equation inside and outside the self-similar region

Question

Let us consider the 2 dimensional heat equation with zero initial condition \begin{cases} \partial_t \psi = \Delta \psi + f & \text{in } \mathbb R^2 \times (0, \infty),\\ \psi(\cdot, 0) = 0 &\text{in } \mathbb R^2. \end{cases} I would like to prove that if $f$ satisfies $$|f(x, t)| \le \begin{cases} 1/(1 + t)^\nu & \text{for } |x| \le \sqrt t,\\ 1/(1 + |x|)^{2\nu} & \text{for } |x| \ge \sqrt t, \end{cases}$$ for $\nu \in (1, 2)$, then $$|\psi(x, t)| \le C \begin{cases} 1/(1 + t)^{\nu-1} & \text{for } |x| \le \sqrt t,\\ 1/(1 + |x|)^{2(\nu-1)} & \text{for } |x| \ge \sqrt t. \end{cases}$$ I feel that I managed to prove it using a barrier argument, that is, constructing explicitly a super solution $\bar \psi (x, t)$ with the desired behaviour such that $$|\psi(x, t)| \le |\bar \psi(x, t)| \quad \forall (x, t) \in \mathbb R^2 \times (0, \infty).$$ However, I do not completely trust my argument and I would feel more comfortable using the explicit representation of the solution with the heat kernel, i.e. $$\psi(x, t) = \int_0^t \int_{\mathbb R^2} \frac{1}{4\pi (t - s)} \exp\left(-\frac{|x - y|^2}{4(t - s)}\right)f(y, s) dyds.$$ But I struggle doing it this way, as we are dealing with a convolution of the right hand side $f$ with the heat kernel, and therefore it isn't obvious that if $f$ has different behaviour inside and outside the self-similar region, so does $\psi$. I can prove that if $$f(x, t) = \frac{1}{(1 + t)^\nu} \mathbb 1_{|x| \le \sqrt t} ~\Rightarrow~ |\psi(x, t)| \le C\frac{1}{(1 + t)^{\nu-1}}$$ for all $(x, t)$, using direct computation with the heat kernel. But this is not quite what I want, I need a different behaviour for $|x| \ge \sqrt t$ and this is not obvious. Any help would be more than welcome.