Let $f \in \mathcal{C}^{\infty}_{K}(\mathbb{R})$ be a smooth function with compact support 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.
I am really struggling to show this and would be grateful for hints or a solution.
Let $\eta_{x,t,z}$ be such that \begin{align*} f(x+\sqrt{t}z)=f(x)+\sqrt{t}zf'(x)+\dfrac{tz^{2}}{2}f''(x)+\dfrac{t\sqrt{t}z^{3}}{6}f^{(3)}(\eta_{x,t,z}). \end{align*} Note that $\int_{\mathbb{R}}ze^{-\frac{1}{2}z^{2}}dz=0$, so we obtain that \begin{align*} g_{t}(x)=\dfrac{1}{2}f''(x)\int_{\mathbb{R}}(z^{2}-1)e^{-\frac{1}{2}z^{2}}dz+\dfrac{\sqrt{t}}{6}\int_{\mathbb{R}}f^{(3)}(\eta_{x,t,z})z^{3}e^{-\frac{1}{2}z^{2}}dz, \end{align*} so \begin{align*} \left|g_{t}(x)-\dfrac{1}{2}f''(x)\int_{\mathbb{R}}(z^{2}-1)e^{-\frac{1}{2}z^{2}}dz\right|\leq\dfrac{\sqrt{t}\cdot\max|f^{(3)}|}{6}\cdot\int_{\mathbb{R}}|z|^{3}e^{-\frac{1}{2}z^{2}}dz\rightarrow 0 \end{align*} uniformly in $x$ as $t\rightarrow 0$.
I do not see why the term $\frac{1}{2}f''(x)\int_{\mathbb{R}}(z^{2}-1)e^{-\frac{1}{2}z^{2}}dz$ has to be vanished.