Is there a function $f\in C^1$ but $E \left(\int_0^T f^2(W_t)\,dt\right) = \infty$ for $W_t$ being a Brownian Motion? Can you give an example?
The question arrose because I am considering the limit $$\lim_{n\to\infty} \sum_n f(W_{t_{i-1,n}}) (W_{t_{i,n}}-W_{t_{i-1,n}})$$ for $t_{i,n}$ being an appropriate partition of the interval $[0,T]$.
If $E \left(\int_0^T f^2(W_t)\,dt\right)$ is finite, this limit should converge to $\int_0^T f(W_t)\,dW_t$.
Let $f(x) = e^{x^2}$ and $T > \frac 14$. For $t > \frac 14$ we have \begin{align*}E[f(W_t)^2] = \frac{1}{\sqrt{2\pi t}}\int_{-\infty}^\infty e^{2x^2}e^{-\frac 1{2t} x^2}dx = \infty \end{align*} so $$E\left[\int_0^T f(W_t)^2dt\right] \ge E\left[\int_{1/4}^T f(W_t)^2 dt\right] = \int_{1/4}^T E[f(W_t)^2]dt = \infty.$$
If we want to remove the condition on $T$, we can simply choose $f(x) = e^{\frac{x^2}{2T}}$ so that for $t > \frac T2$ we have \begin{align*}E[f(W_t)^2] = \frac{1}{\sqrt{2\pi t}}\int_{-\infty}^\infty e^{\frac{x^2}{T}}e^{-\frac 1{2t} x^2}dx \ge\frac{1}{\sqrt{2\pi t}}\int_{-\infty}^\infty 1dx = \infty \end{align*} and apply the same argument.