I konw the Itô isometry states
\begin{equation} \mathbb{E} \left(\int_{0}^{\infty} \alpha_{s} dB_{s} \right)^{2} = \mathbb{E} \int_{0}^{\infty} \alpha_{s}^{2} ds\end{equation}
Can I obtain the following result:
\begin{equation} \mathbb{E}\left[\int_{0}^{T} \int_{0}^{t} f(s)dB_{s}g(t)dB_{t}\right]=\int_{0}^{T} \int_{0}^{t}f(s)g(t)dsdt\end{equation}
Recall the polarization identity, from which it follows that if $T\colon \mathcal{H_1}\to\mathcal{H_2}$ is any isometry between Hilbert spaces then necessarily $\langle Tf,Tg\rangle = \langle f,g\rangle$. Applying this to the Itô isometry, we obtain that $$ \mathbb E\int_0^\infty\int_0^\infty \alpha_s\ dB_s\ \beta_t\ dB_t=\mathbb E\int_0^{\infty}\alpha_s\beta_s\ ds. $$
Finally, choose $\alpha_s=1_{s<t}f(s)$ and $\beta_t=1_{t<T}g(t)$ to conclude.