Any finite dimensional vector space is isomorphic to its dual. My problem is showing that it is isometric. If $\{e_i\}_{1\leq i\leq n}$ is a basis, let $\{f_i\}_{1\leq i\leq n}$ be the corresponding dual basis. My naive attempt at setting up an isometry would be to map $v = \sum_{i=1}^n\lambda_ie_i$ to $f = \sum_{i=1}^n\lambda_if_i$. However, without concrete knowledge of what the norm is, I have problems computing the norm of $f$. For example, in an attempt to establish a lower bound for $f$, I have $||f|| \geq \frac{\sum_{i=1}^n\lambda_if_i(\sum_{i=1}^n\lambda_ie_i)}{||\sum_{i=1}^n\lambda_ie_i||}=\frac{\sum_{i=1}^n\lambda_i^2}{||\sum_{i=1}^n\lambda_ie_i||}$, which I have no idea how to proceed. Help appreciated!
Show that a finite dimensional banach space is isometrically isomorphic to its dual
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linear-algebra
functional-analysis
1 Answers
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This is not true. Think about $X:={\mathbb R}^3$ with the norm $\|x\|:=\max\bigl\{|x_1|,|x_2|,|x_3|\bigr\}$. The unit ball is the cube $[-1,1]^3$. On the other hand the unit ball of the dual $V^*$ is the octahedron $\bigl\{(y_1,y_2,y_3)\,\bigm|\,|y_1|+|y_2|+|y_3|\leq1\bigr\}$.
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0I guess that means that the bijection I have set up is the wrong one. I remember my lecturer mentioning something like if two norms are equivalent they are essentially isometric as we will always be able to find a map which maps each vector in the domain to the 'correct one' in the range. Is this not true? – 2017-02-25