Showing an orthogonal complement is closed using Cauchy-Schwarz

Question

Given the space $H_0 = \{f \in L^2(P) : f(\omega) = f(T\omega)\}$ where $T$ is a bijection from $\Omega \rightarrow \Omega$, the space in question is the orthogonal complement: $$ H_0^\perp := \{g \in L^2(P) : E[gf] = 0 \forall f \in H_0\}. $$ My goal is to show that $H_0^\perp$ is closed. If we take a sequence $g_1,g_2,\dots \in H_0,$ which is convergent to $g$ in $L^2$, we need to show that $E[gf] = 0$. My idea is to apply Cauchy-Scwarz on the scalar product, which in this case is the expectation. This gives us $E[g_nf] \leq \sqrt{E[g_n^2]E[f^2]}$ for every $n$ and for $g$. This upper bound however does not seem very helpful given that $E[g_nf] = 0$. Can this bound somehow be used to show $E[gf] = 0$ when paired with how we defined $f$, or is there a better way to go about this problem? Perhaps I incorrectly applied Cauchy-Schwarz?

Answer 1

It turns out to be quite straightforward once you find the trick:

$$E[f(g - g_n)] \leq \sqrt{E[f^2]E[(g - g_n)^2]} \rightarrow 0$$

since $g_m \rightarrow g$. The left hand side is equal to $E[fg] - E[fg_n] = E[fg]$.