Defining the $L^2$ norm of a vector valued function I am considering a collection of function of the type, $ f:[0,2\pi]\rightarrow \mathbb{R^2}$. I want to define the $L^2$ norm of the function in that space.
I am defining the a norm of $f(=(f_1,f_2)')$ as $\rho(f)=\int_0^{2\pi}(f^2_1(x)+f^2_2(x))^{\frac{1}{2}}dx$. 
Could anyone please suggest me how to show whether the above norm is an $L^2$-norm or not? Or any suggestion about any $L^2$ norm in such a space. 
Thanks in advance.
 A: If $(X,\mu)$ is a measure space and $H$ is a Hilbert space, then an $L^2$ space of $H$-valued functions on $X$ is usually defined by taking $L^2(X,H)=\{f:X\to H: \int_X \|f(x)\|^2\,d\mu<\infty\}$, with norm $\|f\|=\left(\int_X \|f(x)\|^2\,d\mu\right)^{1/2}$.  This norm comes from the inner product $\langle f,g\rangle = \int_X \langle f(x),g(x)\rangle\,d\mu$, with which $L^2(X,H)$ is a Hilbert space.
The norm you defined is more of an $L^1$ norm on a space of $H$ valued functions; you are taking $\|f\|=\int_X \|f(x)\|\,d\mu$.  This will not be a Hilbert space. 
If the Hilbert space dimension of $H$ is $n$, then $L^2(X,H)$ is isomorphic to the Hilbert space direct sum of $n$ copies of $L^2(X)$, which is also isomorphic to the Hilbert space tensor product $L^2(X)\otimes H$.  In particular, for $X=[0,2\pi]$ with Lebesgue measure and $H=\mathbb R^2$ with standard inner product, $L^2(X,H)=L^2([0,2\pi],\mathbb R^2)\cong L^2([0,2\pi])\otimes \mathbb R^2\cong L^2([0,2\pi])\oplus L^2([0,2\pi])$.
A: Well, I'm not sure if we could define it, but let's try to check if $L^2 $ is a vector space and $||.||_2$ is a norm for any $1\leq p \leq\infty$,
The cases $p=1$ and $p=\infty$ are clear, so let's check $1< p <\infty$ and let $f,g \in L^2$
we have $$|f(x)+g(x)| \leq (|f(x)|+|g(x)|)^p \leq 2^{p}(|f(x)|^p + |g(x)|^p)$$
consequently, $f+g \in L^p$. Moreover
$||f+g||_p^{p}=\int |f+g|^{p-1}|f+g| \leq \int |f+g|^{p-1}|f|+\int |f+g|^{p-1}|g| $
But $|f+g|^{p-1} \in L^{p'}$, and by Hölder's inequality we obtain
$$||f+g||_{p}^p \leq ||f+g||_{p}^{p-1}(||f||_p+||g||_p)^{p-1}$$, in other words
$||f+g||_{p}\leq ||f||_p+||g||_p$
