Let $f \in \mathcal{C}_{0}(\mathbb{R})$ s.t. $f''$ exists and $f'' \in \mathcal{C}_{0}(\mathbb{R})$ and $t \geq0$. Then
$ g_{t}(x) =\frac{1}{t}\int_{\mathbb{R}}(f(x+\sqrt{t}z)-f(x)-\frac{t}{2}f^{''}(x))e^{-\frac{1}{2}z^2}dz \rightarrow 0 $ uniformly for $t \rightarrow 0$ holds.
One sees the pointwise convergence with Taylor expansion for $f$ but I don't know how to show the uniform convergence. Here Uniform convergence of integral function is the solution to the assertion with stronger assumptions. Sorry for the questions being so similar.
I would be grateful for hints or a solution.
We have from Taylor's Theorem, \begin{align*} f(x+\sqrt{t}z)=f(x)+\sqrt{t}zf'(x)+\dfrac{tz^{2}}{2}f''(x)+tz^2R(x,\sqrt{t}z) \end{align*} where $R(x,h)=\int_0^1(1-t)(f''(x+th)-f''(x))dt$. Note that $\int_{\mathbb{R}}ze^{-\frac{1}{2}z^{2}}dz=0$, so we obtain that \begin{align*} g_{t}(x)=\int_{\mathbb{R}}R(x,\sqrt{t}z)z^{2}e^{-\frac{1}{2}z^{2}}dz. \end{align*} $R(x,h)$ is uniformly continuous in $(x,h)$ for bounded $h$ as $f''$ vanishes at infinity. As $R(x,0)=0$ we thus have for $h<\delta, R(x,h)<\epsilon$. Also we have that $R(x,h)\leq \max(|{f''}|)$ Putting all this together we can bound $g_t(x)$ as follows, \begin{align*} |g_{t}(x)|\leq \max{|f''|}\int_{|z|\geq\frac{\delta}{\sqrt{t}}}z^{2}e^{-\frac{1}{2}z^{2}}dz+\epsilon\int_{|z|<\frac{\delta}{\sqrt{t}}}z^{2}e^{-\frac{1}{2}z^{2}}dz \end{align*} This bound is independent of $x$ and approaches $\epsilon$ as $t\to 0$