Let $\{B_t\}$ be Brownian motion with $B_0=x\in\mathbb{R}$. Let $f\in C_c^2(\mathbb{R})$. According to page 11 of this document, we have that
$$ \lim_{t\to 0}\frac{\mathbb{E}^x[f(B_t)]-f(x)}{t}=\frac{1}{2}f''(x).$$ If you follow the reasoning, there is a point where the author obtains through 2 change of variables that $$ \frac{\mathbb{E}^x[f(B_t)]-f(x)}{t}=\int_\mathbb{R}\frac{1}{\sqrt{2\pi }}\frac{f(x+\sqrt{t}\cdot y)-f(x)}{t}e^{-y^2}dy.$$ He then uses Taylor's formula to get that $$f(x+\sqrt{t}y)-f(x)=f'(x)\sqrt{t}\cdot y+\frac{1}{2}f''\big(x+\xi\sqrt{t}\cdot y\big)ty^2.$$ for some $\xi\in[0,1]$ and concludes by dominated convergence that $$\lim_{t\to 0 }\int_\mathbb{R}\frac{1}{\sqrt{2\pi }}\frac{f(x+\sqrt{t}\cdot y)-f(x)}{t}e^{-y^2}dy =\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}\lim_{t\to 0}f''\big(x+\xi\sqrt t y\big)y^2e^{-y^2}dy.$$ I don't get the last step. First of all, why was he able to use dominated convergence in the end and not from the start since the integrand is the same? Second, how did he get rid of the term $f'(x)\sqrt{t} y$ in the Taylor expression?
Either something is weird about this whole argument or I'm missing a main point.
Let's do this slowly. We start with $$ \frac{\mathbb{E}^x[f(B_t)]-f(x)}{t}=\mathbb{E}^x\left[\frac{f(B_t)-f(x)}{t}\right] =\int_{\mathbb{R}}\frac{f(x+\sqrt{t}\cdot y)-f(x)}{t}\frac{e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}} $$ and we expand $f(x+\sqrt{t}\cdot y)$ to get $$ \dots=\int_{\mathbb{R}}\frac{f'(x)\sqrt{t}\cdot y}{t}\frac{e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}} +\int_{\mathbb{R}}\frac{\frac12f''(x+\xi\sqrt{t}\cdot y)ty^2}{t}\frac{e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}}. $$ The first integrand is an odd function of $y$ so its integral over $\mathbb{R}$ is 0 and we are left with the second integral $$ \frac{\mathbb{E}^x[f(B_t)]-f(x)}{t}=\int_{\mathbb{R}}\frac12f''(x+\xi\sqrt{t}\cdot y)\frac{y^2e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}}. $$
Now comes the DCT part. Since $f\in C^2_c(\mathbb{R})$, we have $\sup\lvert f''\rvert=M<\infty$ and so the integrand on the RHS is dominated by the integrable function $\frac12 M e^{-y^2}$. Hence we have $$ \lim_{t\to 0}\int_{\mathbb{R}}\frac12f''(x+\xi\sqrt{t}\cdot y)\frac{y^2e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}} =\int_{\mathbb{R}}\lim_{t\to 0}\frac12f''(x+\xi\sqrt{t}\cdot y)\frac{y^2e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}} $$ i.e., $$ \lim_{t\o 0}\frac{\mathbb{E}^x[f(B_t)]-f(x)}{t} =\int_{\mathbb{R}}\lim_{t\to 0}\frac12f''(x+\xi\sqrt{t}\cdot y)y^2\frac{e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}}=\frac12f''(x). $$