This is a question from the public notes,Page3
Prove that $\langle X,Y\rangle:= \mathbb{E}[XY] $ makes $\mathcal{H}$ into a real inner product space
$$\mathcal{H}:=\{X:X\, \text{is a real-valued random variable with}\,\mathbb{E}[X^2]<\infty\}$$
Q1. I know how to prove the Cauchy inequality, but I don't know how does Cauchy hint related to this question
Q2 I think $\langle X,Y\rangle=Cov(X,Y)$ also make $\mathcal{H}$ into a real inner product?
Q3 Here is how I approach this question for proving vector space, is this valid?
$$X,Y\in\mathcal{H}\,\text{so that}\,\mathbb{E}[X^2]<\infty,\mathbb{E}[Y^2]<\infty\\(X-Y)^2\geq0\\XY\leq2XY\leq X^2+Y^2\\2\mathbb{E}[XY]\leq\mathbb{E}[X^2]+\mathbb{E}[Y^2]<\infty\\ \mathbb{E}[X^2]+2\mathbb{E}[XY]+\mathbb{E}[Y^2]<\infty\quad \text{finite plus finite is finite}\\\mathbb{E}[(X+Y)^2]<\infty\implies (X+Y)\in\mathcal{H}$$
Regarding closed under multiplication, the argument is similar, finite times finite is also finite for any $X\in\mathcal{H}$ and $c\in\mathbb{R}$
Appreciate for any comment and help for those three questions
re-edit:
$$\require{cancel}\xcancel{XY\leq}\left|2XY\right|\leq X^2+Y^2$$
Q1: C-S inequality is usually used to prove that $E|XY| <\infty$ but your argument (with the changes sugggested in my comment above) allows you to avoid C-S inequality. Q2: No, $cov(X,Y)$ does not work because $cov (1,1)=0$. You need $ \langle X, X \rangle=0$ implies $X=0$ for $ \langle ., .\rangle$ to be an inner product.