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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?

Dominik
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Fraïssé
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2 Answers2

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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.

Community
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Jason Knapp
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    It bears mentioning that an inner product can be "recovered" from a norm obeying the parallelogram law via the polarization identity – Ben Grossmann Sep 16 '15 at 13:24
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    See this question plus answers: http://math.stackexchange.com/questions/21792/norms-induced-by-inner-products-and-the-parallelogram-law. There it is shown that the validity of the parallelogram law suffices to induce an inner product. The converse is even easier, though. – tomglabst Sep 16 '15 at 13:25
  • In fact, take $||x+y||^2$ in the first formula and add them up to receive the parallelogram law. – tomglabst Sep 16 '15 at 13:46
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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.

Dominik
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  • Could you go into detail how to note that the open balls are ellipsoids? This is the crucial point, isn't it? – tomglabst Sep 16 '15 at 13:49
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    @tomglabst This pretty much follows directly from the definition of an ellipsoid. Note that the induced norm satisfies $| x |^2 = \sum \limits_{k = 1}^n \lambda_k x_k^2$. – Dominik Sep 16 '15 at 13:56