Does the inner product $\langle \cdot, \cdot \rangle$ induce any other norms other than the 2 norm?

Question

In the lecture my professor wrote that the standard inner product on $R^n$ is given by

$\langle x, y \rangle = x^Ty = \sum\limits_{i=1}^n x_i y_i$

which induces a norm $\sqrt{\langle x,x \rangle} = \|x\|_2$

My question is do inner products induce other types of norms...or rather are norms such as the 1-norm or the $\infty$-norm induces by some inner product?

Answer 1

This is a really interesting question, and here is a partial answer. The $1$ and $\infty$ norms do not come from inner products. For a norm to have an associated inner product actually gives you a lot of structure. For example (if the scalars are real for convenience), $$\left\| x - y \right\|^2 = \langle x - y, x -y \rangle = \langle x, x \rangle - 2 \langle x, y \rangle + \langle y, y \rangle = \left\|x \right\|^2 - 2 \langle x, y \rangle + \left\| y \right\|^2$$ In fact it turns out that there is an identity called the parallelogram law $$2 \left\|x\right\|^2 + 2\left\|y\right\|^2 = \left\|x + y \right\| + \left\| x - y\right\|$$ A norm obeys this identity iff it has an associated inner product. You can verify that the $1$ and $\infty$ norms do not obey this identity (by finding examples), and therefore cannot have an inner product. In fact the $p$-norms on $\mathbb{R}^n$ only obey this identity when $p=2$.

Thanks to the comments for some additions. For a proof of the "iff" claim, see this related question. If you have a norm which obeys the parallelogram law, you can actually express the inner product directly in terms of the norm by (again real case for convenience) $$\langle x, y \rangle = \frac{1}{4} \left( \left\| x + y \right\|^2 - \left\| x - y \right\|^2 \right)$$ See here for more information.

Answer 2

Every inner product on $\mathbb{R}^n$ can be written as $\langle x, y \rangle = x^t A y$, where $A$ is a (symmetric) positive definite matrix. These matrices can be orthogonally diagonalized, i.e. there is an orthogonal matrix $M$ so that $A = M^t \operatorname{diag}(\lambda_1, \ldots, \lambda_n) M$. This means in particular $\langle x, y \rangle = (Mx)^t \operatorname{diag}(\lambda_1, \ldots, \lambda_n) (Mx)$.

Now note that the open balls with respect to the inner product $\langle x, y \rangle = x^t \operatorname{diag}(\lambda_1, \ldots, \lambda_n) y$ are ellipsoids and the map $x \mapsto Mx$ is essentially the composition of reflections and rotations [of course this can only be visualized for $n \le 3$]. So the open balls of an arbitrary inner product in $\mathbb{R}^n$ are rotated and reflected ellipsoids.