For $E$ a finite-dimensional Banach Space, there exists an orthonormal basis.

Question

Let $E$ be a Banach space over the reals with finite dimension $k$. Let $\{ e_{1}, ..., e_{k} \}$ be a basis such that $||e_{i}|| = 1$, for all $i$.

Then every $x \in E$ can be decomposed into $\sum^{k}_{i=1} x_{i} e_{i}$. Define the maps $f_{i} : x \rightarrow x_{i}$. We claim that this $f_{i}$ is a linear continuous functional on $E$.

For continuity, I would like to say that "clearly" the norm $||x_{i}e_{i}|| \leq ||x||$ for all $i$, but I am unsure how I can actually prove this without orthonormality. Of course, to say that all of the $e_{i}$ are orthonormal, I would need an inner product on $E$, which I am not given.

Is there a theorem to show that any finite-dimensional real Banach space has an easily defined inner-product which we can use to define my desired orthonormal basis?

If not, is there an alternative way to show the function is continuous?

Answer 1

For $x= \sum_{k=1}x_ke_k$ define $||x||_1:= \sum_{k=1}|x_k|$. Then $|| \cdot||_1$ is a norm on $E$. Since $E$ is finite-dimensional, $|| \cdot||_1$ and $|| \cdot||$ are equivalent norms on $E$.

We have $|f_i(x)|=|x_i| \le ||x||_1.$

This shows that $f_i$ is continuous.

No inner producct is needed.

Answer 2

Let $\Delta =\{(a_1,a_2,..,a_n)\in \mathbb R^{n}:\sum a_i^{2} \leq 1\}$. This is a compact set in $\mathbb R^{n}$ with it usual norm. Consider the function $(a_1,a_2,..,a_n) \to \|\sum a_ie_i\|$. It is easy to see, from the definition of a norm, that this function is continuous. Also this function is strictly positive at every point (by linear independence of $e_i$'s). Hence it has a positive infimum $r$. This means $\|\sum a_ie_i\| \geq r$ for all $(a_1,a_2,..,a_n) \in \Delta$. Given any nonzero vector $(b_1,b_2,..,b_n)$ we can apply this inequality to $(a_1,a_2,..,a_n)$ defined by $a_i=b_i/\sqrt {\sum b_i^{2}}$ to get $\|\sum b_ie_i\| \geq r\sqrt {\sum b_i^{2}}$ for all $(b_1,b_2,..,b_n)$. Continuity if $f_i$'s is obvious from this.