Find derivative of convolution with gaussian Let $A(\sigma)$, $\sigma > 0$ be an operator that acts on bounded continuous functions $f$ on $\mathbb{R}$ by the rule
$$
   (A(t)f)(x) = \int\limits_{\mathbb{R}} f(y)\frac{1}{\sqrt{2 \pi t}}\exp\left( -\frac{(x-y)^2}{2t} \right)dy,
$$
and let $(A(0)f)(x) = f(x)$. Then $(A(t)f)(x) \to f(x)$ when $t \to +0$. But is it true in general that
$$
   \left(\frac{ A(t) - A(0)}{t}f\right)(x) \to \frac{1}{2}\frac{d^2}{dx^2}f(x), \;\;\; t \to +0
$$
for any $C^2(\mathbb{R})$ bounded function $f$? 
If $f(x)$ is bounded and continuous on $\mathbb{R}$ then $u(t,x)=(A(t)f)(x)$ is a solution of Cauchy problem for the heat equation:
$$
   \left\{ \begin{array}{l} u_{t}   =  \frac{1}{2} u_{xx}, \;\;\; t>0, \; x \in \mathbb{R} \\
   u(0,x)  =  f(x), \;\;\; x \in \mathbb{R} \end{array} \right.
$$
Then $\frac{1}{t}(A(t)-A(0))f(x) = \frac{1}{t}(u(t,x)-u(0,x))$. So the question is equivalent to a question: is it true that for bounded $f(x) \in C^2(\mathbb{R})$ there exists a right derivative
$$
   \lim\limits_{t\to +0}\frac{u(t,x)-u(0,x)}{t} = u_{t}(+0,x) = \frac{1}{2}u_{xx}(0,x)
$$
so that the equation $u_t = \frac{1}{2}u_{xx}$ is valid for limit value of $t$.
 A: Applying a Fourier Transform to your first equation and taking the Derivative gives
$$ \frac{d}{dt} F(\omega) e^{-2\pi^2\omega^2 t} = F(\omega) e^{-2\pi^2\omega^2 t} (-2\pi^2\omega^2)
$$
which is transformed back
$$ \left(\frac{1}{2}\frac{d^2}{dx^2} f(x)\right) \ast N(x,t) \xrightarrow{t\rightarrow +0} \frac{1}{2}\frac{d^2}{dx^2} \ ,$$
where $N(x,t)$ is the Gaussian and '$\ast$' the convolution.
A: Let $N(t,x) = \frac{1}{\sqrt{2 \pi t}} \exp \left( -\frac{x^2}{2t} \right)$, $t>0$. For any bounded $f(x) \in C(\mathbb{R})$ we have
$$
   \int f(y)N(t,x-y)dy \to f(x), \;\;\; t \to +0
$$
Indeed, fix $x \in \mathbb{R}$, $t>0$ and $\varepsilon > 0$ and take $\delta = \delta(\varepsilon) > 0$ such  that $|f(x)-f(y)|< \varepsilon$ when $|x-y|<\delta$, then
$$
\int f(y)N(t,x-y)dy = \int\limits_{|x-y|< \varepsilon} + \int\limits_{|x-y| \geqslant \varepsilon}
$$
We have
$$
   f(x)-\varepsilon \leqslant \int\limits_{|x-y|< \varepsilon}f(y)N(x-y)dy \leqslant f(x)+\varepsilon,
$$
because $\int N(t,x-y) = 1$ and $N(t,x-y) > 0$ for any $y$. Next let $M$ be such constant that $|f(x)| \leqslant M$ for any $x$. Then we can write
$$
  \left| \int\limits_{|x-y|\geqslant \varepsilon}f(y)N(t,x-y) dy \right| \leqslant M \int\limits_{|y| \geqslant \varepsilon} N(t,y) dy \leqslant \frac{M}{\varepsilon^2} \int\limits y^2 N(t,y) dy = \frac{M}{\varepsilon^2} t \to 0
$$
when $t \to +0$. Then
$$
   \varlimsup\limits_{t \to +0} \int f(y) N(t,x-y)dy \leqslant f(x) + \varepsilon \\
   \varliminf\limits_{t \to +0} \int f(y) N(t,x-y)dy \geqslant f(x) - \varepsilon
$$
for an arbitrary $\varepsilon > 0$. From this we can derive that
$$
   \int f(y) N(t,x-y) \to f(x), \;\;\; t \to +0
$$
for any bounded continuous $f$. 
Now let $f$ be a $C^2$ bounded function but not necessary with bounded second derivative. Consider a cut-off function $\eta \in C^2_{c}(\mathbb{R})$ such that $\eta(y) = 1$ for such $y$ that $|x-y|<1$ and $\eta(y) = 0$ for $|x-y|>2$. Then it can be shown that $|f(\cdot)*N(\cdot,y) - \eta(\cdot)f(\cdot)*N(t,\cdot)| = o(t)$ when $t \to +0$. Then we can deal with $\eta f$ instead of $f$ if second derivative of $f$ is not bounded. Without loss of generality we consider $f$ with bounded second derivatives.
We have $u(t,x) = f(\cdot)*N(t,\cdot) (x)$ and for any $t>0$ we have
$$
   u_{t}(t,x) = \frac{1}{2}u_{xx}(t,x) = \frac{1}{2}f''(\cdot)*N(t,\cdot) (x) \to \frac{1}{2}f''(x), \;\;\; t \to + 0
$$
Function $t \mapsto u(t,x)$ is continuous on $[0,T]$, $u_{t}(t,x)$ exists on $(0,T)$ and exists limit $u_{t}(+0,x)$. Then
$$
   u_{t}(+0,x) = \lim\limits_{t\to+0} \frac{u(t,x)-u(0,x)}{t}
$$
(for this statement see Demidovich, Problems in Mathematical Analysis). Hence
$$
   \lim\limits_{t \to +0} \frac{u(t,x) - u(0,x)}{t} = \frac{1}{2} f''(x).
$$
