Finite dimensional normed vector space, projection continuous?

Question

Let $X$ be an $n$ -dimensional normed $\mathbb{R}$-vector space. Let $\{e_i \mid i=1,...,n\}$ be a basis of $X$. Is it true that the function $f:X\rightarrow \mathbb{R}$, $\sum \lambda_i e_i \mapsto \lambda_i$ is continuous? I thought it would be easy to prove this but I wasn't able to.

Could you help me? Thanks!

Answer 1

Here is a proof that does not use any theorem on general normed linear spaces. It uses only the Bolzano Weierstras Theorem for the usual norm on $\mathbb R^{n}$. Suppose $\sum a_i^{(m)}e_i \to 0$. We have to show that $a_i^{(m)} \to 0$ for each $i$. We claim that $\{(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})\}$ is bounded. Suppose, if possible, $\left\Vert (a_{1}^{(m_{j})},a_{2}^{(m_{j})},...,a_{n}^{(m_{j})})\right\Vert_2 \to \infty $. Here $\|.\|_2$ denotes the usual norm on $\mathbb R^{n}$. Let $b_{i}^{(m_{j})}=\frac{a_{i}^{(m_{j})}}{\left\Vert (a_{1}^{(m_{j})},a_{2}^{(m_{j})},...,a_{n}^{(m_{j})})\right\Vert_2 },1\leq i\leq n.$ Then $\sum_{i=1}^{n}b_{i}^{(m)}x_{i}\to 0$ and $% \left\Vert (b_{1}^{(m_{j})},b_{2}^{(m_{j})},...,b_{n}^{(m_{j})})\right\Vert_2 =1$ $\forall j$. Since unit vectors form a compact set in $\mathbb R^{n}$ we can take limit of $\sum_{i=1}^{n}b_{i}^{(m)}x_{i}$ through a subsequence to get an equation of the type $ \sum_{i=1}^{n}c_{i}x_{i}=0$ where $\left\Vert (c_{1},c_{2},...,c_{n})\right\Vert_2 =1.$ This is impossible because $% \{x_{1},x_{2},...,x_{n}\}$ be a basis. We have proved that $% \{(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})\}$ is bounded. We now extract a convergent subsequence of this sequence to get $ \sum_{i=1}^{n}d_{i}x_{i}=0$ forcing each $d_{i}$ to be $0$. This shows that every limit point of $\{(a_{1}^{(m)},a_{2}^{(m)},...,a_{n}^{(m)})\}$ is $(0,0,,,,0)$ and hence the claim is true.

Answer 2

Hint: Let $P_1 v = v_1$. Observe \begin{align} |v_1-w_1| = |P_1v-P_1w| \leq \|P_1\|\|v-w\|. \end{align}

Edit: A upon the request of @KaviRamaMurthy, I have decided to include a quick proof that all linear maps $A:\mathbb{R}^n \rightarrow \mathbb{R}^n$ are bounded.

Using matrix representation, we see that \begin{align} \|Av\|^2=&\ \sum^n_{j=1}\left|\sum^n_{j=1} a_{ij}v_j \right|^2 \leq \sum^n_{j=1}\left( \sum^n_{j=1} |a_{ij}|^2\right)\left( \sum^n_{j=1} |v_j|^2\right) \\ =&\ \left(\sum^n_{i,j=1}|a_{ij}|^2 \right)\|v\|^2 \end{align} which means $\|Av\| \leq C\|v\|$ where $C$ is independent of $v$.

Answer 3

Since any two norms on a finite dimensional space are equivalent, and $\|\sum_i \lambda_i e_i\|_* = \sum_i |\lambda_i|$ defines a norm, there exists a constant $C>0$ such that $$ \|\sum_i \lambda_i e_i\|_*\leq C\|\sum_i \lambda_i e_i\|,\quad\forall (\lambda_i)_{i=1}^n. $$It follows $$ \left|f(\sum_i \lambda_i e_i)\right |= |\lambda_j|\le \|\sum_i \lambda_i e_i\|_*\leq C\|\sum_i \lambda_i e_i\|.$$

EDIT: Here's a sketch of the proof that any two norms on a f.d.v.s. $V$ on $\mathbb{F}$($=\mathbb{R}$ or $\mathbb{C}$) are equivalent. Let $(e_i)_{i=1}^n$ be a basis of $V$, and endow $V$ a norm defined by $\|\sum_i x_i e_i\|_1 =\sum_i |x_i|$. Then, $(V,\|\cdot\|_1)$ and $(\mathbb{F}^n,\|\cdot\|_1)$ are isometric and hence homeomorphic. Notice that this implies $\{\sum_i x_ie_i\;|\;\sum_i |x_i| = 1\}$ is compact in $V$.
Now, let $f(\sum_i x_ie_i) =\|\sum_i x_i e_i\|$ where $\|\cdot\|$ denotes the given norm of $V$. If we show $\|\cdot\| \sim\|\cdot\|_1$, then the claim follows since $\sim$ is an equivalence relation. Observe that $$ |f(\sum_i x_ie_i)-f(\sum_i y_ie_i)| = |\|\sum_i x_ie_i\|-\|\sum_i y_ie_i\||\leq \|\sum_i (x_i-y_i)e_i\| \leq \sum_i |x_i-y_i|\cdot \max_i \|e_i\|. $$ This shows $f:(V,\|\cdot\|_1)\to [0,\infty)$ is continuous. Since $E=\{\sum_i x_ie_i\;|\;\sum_i |x_i| = 1\}$ is compact, by extremum value theorem, there is $0<m\leq M<\infty$ such that $$ m\leq f(x) =\|x\|\le M, \quad \forall x\in E. $$ This gives $m \|x\|_1 \le \|x\|\le M\|x\|_1$ for all $x\in V$.