$l^p$ norm on an inner product space

Question

On a finite-dimensional inner product space $X$, can we define its $l^p$ norm
with $p \geq 1$, as the $l^p$ norm on its coordinate space under an orthonormal basis of $X$, and the $l^p$ norm doesn't depend on the choice of the orthonormal basis of $X$? Why is it? Thanks.

Answer 1

The $l^p$-norm does depend on the choice of the orthonormal basis also in finite dimensions ($> 1$).

Consider $\mathbb{R}^2$ and $x = (1,1)$. For orthonormal bases in $\mathbb{R}^2$ with the standard inner product, consider

1. $e_1,e_2$,
2. $\frac{1}{\sqrt{2}}(e_1+e_2), \frac{1}{\sqrt{2}}(e_1-e_2)$.

With respect to the first ONB, we have

$$\lVert x\rVert_p = (1^p + 1^p)^{1/p} = 2^{1/p},$$

while with respect to the second we have

$$\lVert x\rVert_p = \left(\sqrt{2}^p + 0^p\right)^{1/p} = \sqrt{2}.$$

The two coincide only for $p = 2$.

You can embed this counterexample in every inner product space of dimension $> 1$. Take an ONB $e_1,e_2,e_3,\dotsc$ [ignore $e_3$ etc. if the dimension is exactly $2$], and let $x = e_1 + e_2$. Then look at the ONB $f_1,f_2,e_3,\dotsc$, where $f_1 = \frac{1}{\sqrt{2}}(e_1+e_2)$ and $f_2 = \frac{1}{\sqrt{2}}(e_1-e_2)$.