I'm working on an exercise to better-understand vector spaces. In particular, I'm trying to show that the space of square-integrable real-valued functions $\mathcal{F}$ is closed under addition.
$$ \mathcal{F} = \{ f ~|~ f : \mathbb{R} \rightarrow \mathbb{R} ~,~ \int^\infty_{-\infty} |f(x)|^2 dx < \infty \} $$
Proof: Take two functions $f, g \in \mathcal{F}$ and let $h = f + g$.
We then have;
\begin{align} \int^\infty_{-\infty} |h(x)|^2 dx &= \int^\infty_{-\infty} |f(x) + g(x)|^2 dx \\ &\le \int^\infty_{-\infty} \left(|f(x)| + |g(x)|\right)^2 dx \\ &\le \int^\infty_{-\infty} |f(x)|^2 + |f(x)||g(x)| + |g(x)|^2 dx \\ &\le \int^\infty_{-\infty} |f(x)|^2 + \int^\infty_{-\infty} |f(x)||g(x)| + \int^\infty_{-\infty} |g(x)|^2 dx \end{align}
By definition, the first and last terms are finite, as $f, g \in \mathcal{F}$. How can I show that the middle term is also finite (and thus that $\mathcal{F}$ is closed under addition?).
Thank you.
$|a+b|^2 \leq 2(|a|^{2}+|b|^{2})$. Apply this to $a =f(x),b=g(x)$ and integrate.
[$|a+b| \leq |a|+|b|$. Now, $(|a|-|b|)^{2} \geq 0$. Expanding this we get $|a|^{2}+|b|^{2}-2|a||b| \geq 0$. In other words $2|a||b|\leq |a|^{2}+|b|^{2}$. Adding $|a|^{2}+|b|^{2}$ to both sides we get $2|a||b|+ |a|^{2}+|b|^{2} \leq 2|a|^{2}+2|b|^{2}$. This gives $|a+b|^{2}\leq (|a|+|b|)^{2}=2|a||b|+ |a|^{2}+|b|^{2} \leq 2|a|^{2}+2|b|^{2}$].