[proofplan] We first prove the identity up to the stopping time at which Brownian motion exits a compact interval. On that stopped interval, $f'$ and $f''$ are bounded and $f''$ is uniformly continuous, so [Taylor's theorem](/theorems/827) along a partition of $[0,t]$ has a remainder controlled by the Brownian mesh and the quadratic variation sum. The first-order sums converge to the stochastic integral, the second-order sums converge to the time integral by weighted quadratic variation, and the remainder vanishes. Letting the compact interval expand gives the stated localized identity. [/proofplan] [step:Localize Brownian motion to a compact range] Fix $t\geq 0$ and, for each integer $n\geq 1$, define the stopping time \begin{align*} \tau_n &= \inf\{u\geq 0: |W_u|\geq n\}. \end{align*} The stopped path $(W_{u\wedge\tau_n})_{0\leq u\leq t}$ takes values in the compact interval $[-n,n]$. Since $f\in C^2(\mathbb R)$, the functions $f'$ and $f''$ are bounded on $[-n,n]$, and $f''$ is uniformly continuous on $[-n,n]$. It is therefore enough to prove the formula with $t$ replaced by $t\wedge\tau_n$ and then let $n\to\infty$, because $\tau_n\uparrow\infty$ almost surely by continuity of Brownian paths. [/step] [step:Apply Taylor's theorem on a partition before the stopping time] Let $\pi=\{0=t_0<t_1<\cdots<t_m=t\}$ be a partition of $[0,t]$, and set \begin{align*} Y_u &= W_{u\wedge\tau_n}. \end{align*} For each $k$, [Taylor's theorem](/theorems/827) gives \begin{align*} f(Y_{t_{k+1}})-f(Y_{t_k}) &= f'(Y_{t_k})(Y_{t_{k+1}}-Y_{t_k}) +\frac{1}{2}f''(Y_{t_k})(Y_{t_{k+1}}-Y_{t_k})^2 +R_k, \end{align*} where \begin{align*} |R_k| &\leq \frac{1}{2}\omega_n\left(|Y_{t_{k+1}}-Y_{t_k}|\right)(Y_{t_{k+1}}-Y_{t_k})^2. \end{align*} Here $\omega_n:[0,\infty)\to[0,\infty)$ is the modulus of continuity of $f''$ on $[-n,n]$, defined by \begin{align*} \omega_n(r)&=\sup\{|f''(x)-f''(y)|: x,y\in[-n,n], |x-y|\leq r\}. \end{align*} Summing over $k$ telescopes the left-hand side: \begin{align*} f(Y_t)-f(Y_0) &= \sum_{k=0}^{m-1}f'(Y_{t_k})(Y_{t_{k+1}}-Y_{t_k}) \\ &\quad+\frac{1}{2}\sum_{k=0}^{m-1}f''(Y_{t_k})(Y_{t_{k+1}}-Y_{t_k})^2 +\sum_{k=0}^{m-1}R_k. \end{align*} [/step] [step:Pass the three sums to their limits] Let $|\pi|=\max_k(t_{k+1}-t_k)$ denote the mesh of the partition, and let $|\pi|\to0$ along deterministic partitions. Write $[W]_u$ for the [quadratic variation of Brownian motion](/theorems/3543) at time $u$. Since $f'(Y_u)$ is bounded and adapted, the definition of the Itô integral gives \begin{align*} \sum_{k=0}^{m-1}f'(Y_{t_k})(Y_{t_{k+1}}-Y_{t_k}) &\to \int_0^t f'(W_{s\wedge\tau_n})\,dW_{s\wedge\tau_n} \end{align*} in probability. The weighted quadratic variation convergence for Brownian motion gives \begin{align*} \sum_{k=0}^{m-1}f''(Y_{t_k})(Y_{t_{k+1}}-Y_{t_k})^2 &\to \int_0^{t\wedge\tau_n} f''(W_s)\,d\mathcal L^1(s) \end{align*} in probability. This weighted convergence follows from the quadratic variation identity $[W]_{u}=u$ and [uniform continuity](/page/Uniform%20Continuity) of the continuous adapted weight $s\mapsto f''(Y_s)$ on $[0,t]$. For the remainder, Brownian paths are uniformly continuous on $[0,t]$ almost surely, so \begin{align*} \max_{0\leq k<m}|Y_{t_{k+1}}-Y_{t_k}|&\to0 \end{align*} almost surely. Also the quadratic sums $\sum_k(Y_{t_{k+1}}-Y_{t_k})^2$ converge in probability to $t\wedge\tau_n$ and are therefore bounded in probability. Since $\omega_n(r)\to0$ as $r\downarrow0$, the estimate on $R_k$ gives \begin{align*} \sum_{k=0}^{m-1}R_k&\to0 \end{align*} in probability. Passing to the limit in the telescoping identity yields \begin{align*} f(W_{t\wedge\tau_n}) &= f(W_0) + \int_0^t f'(W_{s\wedge\tau_n})\,dW_{s\wedge\tau_n} +\frac{1}{2}\int_0^{t\wedge\tau_n} f''(W_s)\,d\mathcal L^1(s) \end{align*} almost surely. [/step] [step:Remove the localization] The stopped stochastic integral satisfies \begin{align*} \int_0^t f'(W_{s\wedge\tau_n})\,dW_{s\wedge\tau_n} &= \int_0^{t\wedge\tau_n} f'(W_s)\,dW_s. \end{align*} Since $\tau_n\uparrow\infty$ almost surely, for almost every outcome there is an index $n_0$ such that $t<\tau_n$ for all $n\geq n_0$. For those $n$, \begin{align*} W_{t\wedge\tau_n} &= W_t, & \int_0^{t\wedge\tau_n} f''(W_s)\,d\mathcal L^1(s) &= \int_0^t f''(W_s)\,d\mathcal L^1(s), \end{align*} and the stopped stochastic integral agrees with the localized stochastic integral $\int_0^t f'(W_s)\,dW_s$. Letting $n\to\infty$ gives \begin{align*} f(W_t) &= f(W_0) + \int_0^t f'(W_s)\,dW_s +\frac{1}{2}\int_0^t f''(W_s)\,d\mathcal L^1(s) \end{align*} almost surely. [/step]