= Solution
The bounded classical solution is the <heat-kernel convolution>
$$
\boxed{u(x,t)=\frac1{4\pi t}\int_{\mathbb R^2}\exp\!\left(-\frac{|x-y|^2}{4t}\right)g(y)\,dy,\qquad t>0.}
$$
The two-dimensional <heat kernel> has integral one. Boundedness of $g$ permits differentiation under the integral for $t>0$, proving $u_t=\Delta u$ and $|u|\le\|g\|_\infty$. Its <approximate identity> property gives $u(x,t)\to g(x)$, locally uniformly since $g$ is <continuous>. This bounded solution satisfies the required Gaussian-growth bound.
For <heat-equation uniqueness under Gaussian growth>, let $w$ be the difference of two solutions, so $w(x,0)=0$ and $|w|\le M_0e^{a|x|^2}$ on a finite time interval. Choose $b>a$ and a slab $0\le t\le\tau<1/(4b)$. The positive comparison solution
$$
\Phi_b(x,t)=\frac1{1-4bt}\exp\!\left(\frac{b|x|^2}{1-4bt}\right)
$$
satisfies $(\Phi_b)_t=\Delta\Phi_b$. For any $\varepsilon>0$, its faster spatial growth makes $|w|\le\varepsilon\Phi_b$ on a sufficiently large lateral cylinder boundary. At the initial boundary, $w=0$. The <heat equation maximum principle> applied to $\pm w-\varepsilon\Phi_b$ proves the same inequality inside. First allow the cylinder radius to increase, then let $\varepsilon\downarrow0$. Repeating these slabs proves uniqueness on every finite interval with the stated uniform growth bound. No spatial integrability of $g$ is required.
At time $T$, this is <Gaussian filtering> with covariance $2T I$, hence \b[$\sigma=\sqrt{2T}$ per coordinate]. With the angular-frequency <Fourier transform> convention $\widehat f(\xi)=\int e^{-ix\cdot\xi}f(x)\,dx$, the <Gaussian filtering Fourier multiplier> is
$$
\boxed{\widehat u(\xi,T)=e^{-T|\xi|^2}\widehat g(\xi)=e^{-\sigma^2|\xi|^2/2}\widehat g(\xi).}
$$
For merely bounded $g$, this identity is understood through <tempered distributions>. The multiplier leaves the zero frequency unchanged and attenuates large <frequencies> exponentially: it suppresses rapid noise fluctuations but blurs <image edges>. In cycles-per-length frequency $k$, the multiplier is $e^{-4\pi^2T|k|^2}$.
For the printed <Perona-Malik equation>, retain the supplied conductance $c(s)=s e^{-s^2/(2\lambda^2)}$, which includes an extra factor $s$. In one dimension write $p=u_x$ and $F(p)=c(|p|)p=p|p|e^{-p^2/(2\lambda^2)}$. Then $u_t=F'(u_x)u_{xx}$, with
$$
F'(p)=|p|e^{-p^2/(2\lambda^2)}\left(2-\frac{p^2}{\lambda^2}\right)\quad(p\ne0),\qquad F'(0)=0.
$$
The <forward-backward threshold for gradient-weighted exponential diffusion> is therefore
$$
\boxed{0<|u_x|<\sqrt2\lambda:\ \text{forward diffusion};\quad |u_x|>\sqrt2\lambda:\ \text{backward diffusion}.}
$$
At $|u_x|=0$ and $\sqrt2\lambda$ the coefficient vanishes, giving degenerate diffusion. Increasing $\lambda$ increases the forward-diffusion range. Small nonzero slopes smooth, while large slopes formally sharpen; a negative coefficient produces short-wave growth and <ill-posedness>, so this is a dynamics explanation, not a general existence theorem for arbitrary data. Replacing the printed conductance by $e^{-s^2/(2\lambda^2)}$ would give threshold $\lambda$, but that is a different equation.
The <four-neighbour mean expansion> follows by <Taylor expansion>: opposing first and third <derivatives> cancel, so
$$
\operatorname{mean}_h(u)(x)=u(x)+\frac{h^2}{4}\Delta u(x)+\frac{h^4}{48}\bigl(u_{xxxx}(x)+u_{yyyy}(x)\bigr)+O(h^6).
$$
If $u(x_0)=\operatorname{mean}_h(u)(x_0)$ for every sufficiently small $h$, division by $h^2$ and passage to the <limit> give \b[$\Delta u(x_0)=0$]. This is a pointwise conclusion; it does not assert harmonicity in a whole neighbourhood.
For the area <median> over the disk, the <disk-median curvature expansion>, at a point where $|\nabla u|\ne0$, is
$$
\operatorname{median}_{B_h(x_0)}u=u(x_0)+\frac{h^2}{6}\left(\Delta u-\frac{\nabla u^T(D^2u)\nabla u}{|\nabla u|^2}\right)(x_0)+o(h^2).
$$
Consequently the analogous <median> fixed-point condition yields
$$
\boxed{\Delta u(x_0)-\frac{\nabla u(x_0)^T(D^2u(x_0))\nabla u(x_0)}{|\nabla u(x_0)|^2}
=|\nabla u(x_0)|\operatorname{div}\!\left(\frac{\nabla u}{|\nabla u|}\right)(x_0)=0.}
$$
This is the vanishing of the <level-line curvature> at $x_0$: to second order the level line is straight there. It need not be a straight segment. For example $u(x,y)=y+x^3$ has zero disk <median> at the origin for every radius, by odd symmetry, but its zero level line is the cubic $y=-x^3$. Vanishing <curvature> on an entire connected regular level arc would force that arc to be straight. The factor $1/6$ is for the disk's uniform area measure; a circle-boundary <median> has a different scale factor, while producing the same zero-curvature equation.
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