In page 377 of the article Diffusion processes with continuous coefficients I (Stroock Varadhan - 1969), one reads:
I am having trouble understanding why $g$ is a fundamental solution of the heat equation.
Attempt We are trying to show that $g$ is a fundamental solution, according to the wikipedia we are looking for a distribution $g$ such that $$Lg(s,x) = \delta_{s,x} $$
where $L u = \frac{\partial u}{\partial s } + \frac{1}{2} \sum_i \frac{\partial^2 u}{\partial x_i^2 }$
This means that for every test function $$Lg(s,x) f = \int_{0}^\infty\int_{\Bbb{R}^d} L g_{s,x}(t,y)f(t,y)\, dy\,dt $$
First let's take the derivatives of $g(s,x;t,y) = 1_{t > s}[2 \pi (t-s)]^{-d/2}\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\}$ for $t > s$
First the time derivative
$$\frac{\partial g}{\partial s} = -\frac{d}{2}[2 \pi (t-s)]^{-d/2-1}(-2 \pi)\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\} +\\ \bigg(\frac{\vert y - x \vert^2}{(2(t-s))^2}\bigg)(-2) [2 \pi (t-s)]^{-d/2}\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\} \\ = d \pi[2 \pi (t-s)]^{-d/2-1}(-2 \pi)\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\} +\\ \bigg(\frac{\vert y - x \vert^2}{(t-s)}\bigg)(- \pi) [2 \pi (t-s)]^{-d/2-1}\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\}$$
now the spatial derivatives $$\frac{\partial g}{\partial x_i} = [2 \pi (t-s)]^{-d/2}\bigg(\frac{y_i - x_i}{t - s}\bigg)\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\}\\ \frac{\partial^2 g}{\partial x_i^2} = [2 \pi (t-s)]^{-d/2}\bigg(-\frac{1}{t - s}\bigg)\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\}\\ +[2 \pi (t-s)]^{-d/2}\bigg(\frac{(y_i - x_i)^2}{(t - s)^2}\bigg)\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\} \\ $$
So summing over $i$ we obtain $$\sum_i \frac{\partial^2 g}{\partial x_i^2 } = d[2 \pi (t-s)]^{-d/2}\bigg(-\frac{1}{t - s}\bigg)\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\}\\ +[2 \pi (t-s)]^{-d/2}\bigg(\frac{\vert y - x\vert^2}{(t - s)^2}\bigg)\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\} $$ so $$\sum_i \frac{\partial^2 g}{\partial x_i^2 } = -2 \pi d[2 \pi (t-s)]^{-d/2-1}\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\}\\ +[2 \pi (t-s)]^{-d/2 -1} 2\pi\bigg(\frac{\vert y - x\vert^2}{(t - s)}\bigg)\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\} $$
wich means that
$$\frac{\partial g}{\partial s} + \frac{1}{2}\sum_i \frac{\partial^2 g}{\partial x_i^2 } = 0 $$
Now define $$H(s,x) = \int_0^\infty\int_{\Bbb{R}^d} g(s,x; t, y) f(t,y) \, dy\, dt\\ = \int_s^\infty\int_{\Bbb{R}^d} [2 \pi (t-s)]^{-d/2}\exp \bigg\{-\frac{\vert y - x \vert^2}{2(t-s)}\bigg\} f(t,y) \, dy\, dt\\.$$ We then compute:
$$\frac{ \partial}{\partial s} H (s,x) = \lim_{h \to 0} \frac{1}{h} \Bigg( \int_s^\infty\int_{\Bbb{R}^d} g(s,x; t, y) f(t,y) \, dy\, dt - \int_{s + h}^\infty\int_{\Bbb{R}^d} g(s+h,x; t, y) f(t,y) \, dy\, dt\Bigg) \\ =\lim_{h \to 0} \frac{1}{h} \Bigg( \int_s^{s + h}\int_{\Bbb{R}^d} g(s,x; t, y) f(t,y) \, dy\, dt \\ + \int_{s + h}^\infty\int_{\Bbb{R}^d} g(s,x; t, y) - g(s+h,x; t, y) f(t,y) \, dy\, dt\Bigg) \\ = f(s,x) + \int_s^{\infty}\int_{\Bbb{R}^d} \frac{\partial}{\partial s} g(s,x; t, y) f(t,y) \, dy\, dt$$
now $$ \frac{ \partial^2}{\partial x_i^2} H (s,x) =\int_s^{\infty}\int_{\Bbb{R}^d} \frac{\partial^2}{\partial x_i^2} g(s,x; t, y) f(t,y) \, dy\, dt $$
therefore
$$ \frac{ \partial}{\partial s} H (s,x) + \frac{1}{2}\sum_i \frac{ \partial^2}{\partial x_i^2} H (s,x) = f(s,x) + \int_s^{\infty}\int_{\Bbb{R}^d} \big(\frac{\partial}{\partial s} + \frac{1}{2} \sum_i \frac{\partial^2}{\partial x_i^2}\big) g(s,x; t, y) f(t,y) \, dy\, dt \\ = f(s,x)$$
Are those computations correct? are there any hypothesis needed on $f$ (other that $f$ being bounded and continuous) for this to work out?
why do the authors use the function $G_\lambda f$ and why do they focus on the relation $$\lambda h - \partial_s h - \frac{1}{2} \Delta h = f ?$$