Prove that the limit of the inner product is equal to the inner product of the limits in $L^2$ I'm working on the following problem.

Let $\{f_n\},f,\{g_n\},g$ be functions from $[-\pi,\pi]$ to $\mathbb{C}$ that are all Riemann integrable. Prove that if $\{f_n\}$ converges in the mean to $f$ and $\{g_n\}$ converges in the mean to $g$ then $$\Vert f\Vert=\lim_{n\rightarrow\infty}\Vert f_n \Vert$$ and $$(f,g)=\lim_{n\rightarrow\infty}(f_n,g_n)$$ where $(f,g):=\int_{-\pi}^\pi f(x)\overline{g(x)}dx$ and $\Vert f \Vert:=\left[\int_{-\pi}^\pi |f(x)|^2dx\right]^{1/2}$

For the first one I have got something which seems too trivial...
$$(\lim_{n\rightarrow \infty}\Vert f_n\Vert)^2=\lim_{n\rightarrow\infty}\int_{-\pi}^\pi|f(x)|^2dx=(f(x))^2$$ and taking the square root of both sides gives the result. But I feel like I am missing some subtlety.
The second one I'm sure sure even how to get started.

I am assuming that by convergence in mean you mean the $L^2$ norm.
For the first, any norm satisfies $| \|x\|- \|y\| | \le \|x-y\|$. Hence $| \|f_n\|- \|f\| | \le \|f-f_n\|$ and the result follows.
For the second, use the polarization identity. That is $\langle x , y \rangle = \frac{1}{4} \sum_{k=0}^3 i^k \|x+i^ky \|^2 $. We have $\| \alpha (f-f_n) + \beta(g-g_n)\| \le |\alpha| \|f-f_n\| + |\beta| \|g-g_n \|$, hence $\alpha f_n + \beta g_n $ converges in mean to $\alpha f + \beta g $ for any $\alpha, \beta$. Applying the first result gives $\|\alpha f_n + \beta g_n \| \to \| \alpha f + \beta g \| $. It follows from the polarization identity that $\langle f_n , g_n \rangle = \frac{1}{4} \sum_{k=0}^3 i^k \|f_n+i^k g_n \|^2 \to \frac{1}{4} \sum_{k=0}^3 i^k \|f+i^k g \|^2 = \langle f , g \rangle$.
