Prove that $V$ and $\mathbb{C}_{2}^{n}$ are isometric
Let $V$ a vector space over $\mathbb{C}$ with dimention $n$ and let $\mathcal{B}=\{x_{1},\cdots,x_{n}\}$ be the basis of $V$. For eery $p\geq 1$ fixed and $x=\sum_{j=1}^{n}\alpha_{j}x_{j}$ we define the function $\varphi(x)=\{\sum_{j=1}^{n}|\alpha_{n}|^{p}\}^{\frac{1}{p}}$
a)Prove that $\varphi$ is a norm of $V$
b) Suppose that $V$ is a space with inner product $\left \langle \cdot \right \rangle$ and $||\cdot||$ the norm generated by the inner product, also suppose that $\mathcal{B}$ is a orthonormal basis of $V$. Prove that $V$ and $\mathcal{C}_{2}^{n}$ are isometric where $\mathbb{C}_{2}^{n}=(\mathbb{C}^{n},||\cdot||_{2})$
I already proved that $\varphi$ is a norm of $V$, I basically had to use raised the p-norm $(||\alpha+\beta||_{p}^{p})$ to prove that $||\alpha+\beta||_{p}\leq ||\alpha||_{p}+||\beta||_{p}$ but I don't know how to make the second one, thanks for your help
- If $\|\cdot\|_p$ is the $p$-norm on $\mathbb C^n$, and $l \colon \mathbb C^n \to V$ is the map $(\alpha_1,\dots,\alpha_n) \mapsto \alpha_1x_1+\cdots+\alpha_nx_n$, note that $l$ is a linear isomorphism and that $\varphi(x) = \|l^{-1}(x)\|_p$ for all $x \in V$. Thus, (a) follows from the following basic fact:
Given two vector spaces $E$ and $F$, and an injective linear map $T \colon E \to F$, if $\|\cdot\|$ is a norm on $F$, then $\|x\|' := \|Tx\|$ for all $x \in E$ defines a norm $\|\cdot\|'$ on $E$.
- Observe that for any $(\alpha_1,\dots,\alpha_n) \in \mathbb C^n$ we have that $$ \begin{align*} \left\| \sum_{k=1}^n \alpha_kx_k \right\|^2 &= \left\langle \sum_{i=1}^n \alpha_ix_i, \sum_{j=1}^n \alpha_jx_j \right\rangle \\ &= \sum_{i=1}^n \sum_{j=1}^n \alpha_i \overline{\alpha_j} \langle x_i,x_j \rangle \\ &= \sum_{k=1}^n |\alpha_k|^2 \quad (\textrm{since $\mathcal B$ is orthonormal}). \end{align*} $$ Hence, if $l$ is the map defined above, $\|l(\alpha)\| = \|\alpha\|_2$ for every $\alpha \in \mathbb C^n$, in other words, $l$ is an isometry between $(V,\|\cdot\|)$ and $(\mathbb C^n,\|\cdot\|_2)$.