Proving linearity of a subspace of $L^2$

Question

I'm reading p.79 of Steele s Stochastic Calculus and Financial Applications. It defines a space of functions $\mathcal{H}^2$ as the space of all measurable adapted functions such that $\mathbb{E}[\int_0^T f^2(w, t) dt] < \infty$. It also claims that $\mathcal{H}^2$ is a closed linear space of $\mathcal{L}^2(dP \times dt)$. I'm trying to verify that $\mathcal{H}^2$ is a linear space. Showing that $0 \in \mathcal{H}^2$ is trivial, but showing $f, g \in \mathcal{H}^2 \implies \alpha f + \beta g \in \mathcal{H}^2$ is giving me trouble. So far I've done \begin{align} \mathbb{E}\bigg{[}\int_0^T (\alpha f(t) + \beta g(t))^2 dt\bigg{]} \leq& \alpha^2\mathbb{E}\bigg{[}\int_0^T f^2(t) dt\bigg{]} + \beta^2\mathbb{E}\bigg{[}\int_0^T g^2(t) dt\bigg{]} + 2\alpha\beta \mathbb{E}\bigg{[}\int_0^T f(t)g(t) dt\bigg{]} \end{align} The first two terms are finite by definition of $\mathcal{H}^2$, but how do we know that the third term is finite? Cauchy-Schwarz inequality on $\int_0^T f(t)g(t) dt$ doesn't seem helpful here because it gives a term of order 4 inside the expectation (unless I'm using it the wrong way).

Answer 1

Plain old C-S gives us the following

$$\int f(t) g(t) dt \le \left(\int f^2(t) dt \right)^{1/2} \left(\int g^2(t) dt \right)^{1/2} \tag{1} $$

Let $X = \left(\int f^2(t) dt \right)^{1/2}$, and $Y = \left(\int g^2(t) dt \right)^{1/2}$

Now, Cauchy-Schwarz again gives us that

$$ \mathbb{E}[XY]^2 \le \mathbb{E}[X^2] \mathbb{E} [Y^2] $$ $$\Rightarrow \mathbb{E} \left[\left( \int f^2(t) dt \right)^{1/2} \left(\int g^2(t) dt \right)^{1/2}\right] \le \left(\mathbb{E} \left[ \int f^2(t) dt \right]\right)^{1/2} \left(\mathbb{E}\left[ \int g^2(t) dt \right]\right)^{1/2} \tag{2}$$

Now, note that \begin{align} \mathbb{E} \left[ \int f(t)g(t) dt \right] &\overset{(a)}{\le} \mathbb{E} \left[ \left(\int f^2(t) dt \right)^{1/2} \left(\int g^2(t) dt \right)^{1/2} \right] \\ &\le \left(\mathbb{E} \left[ \int f^2(t) dt \right]\right)^{1/2} \left(\mathbb{E}\left[ \int g^2(t) dt \right]\right)^{1/2} \end{align}

Where $(a)$ follows from taking expectation of $(1)$, and the second inequality is just $(2)$.