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Let $ S $ be a normed vector space (over $\mathbb{R}$, say, for simplicity) with norm $ \lVert\cdot\rVert$.

Can this norm be induced from inner product only through $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$ ?

As to prove the if part of "A norm is induced by inner product iff the norm satisfies parallelogram equality", $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$ is used.

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  • $\begingroup$ Wouldn't any other norm you define be equivalent to the standard induced norm? $\endgroup$ – KyleW Aug 13 '15 at 6:07
  • $\begingroup$ This is the Jordan-von Neumann Theorem. Here's a proof in the complex case: math.dartmouth.edu/archive/m113w10/public_html/… . $\endgroup$ – DisintegratingByParts Aug 13 '15 at 7:01
  • $\begingroup$ @KyleW I didn't get you. If you mean like L_3 norm to be equivalent to the standard induced norm, then it's not true. $\endgroup$ – vivkul Aug 13 '15 at 12:50
  • $\begingroup$ In a finite dimensional inner product space, any two norms you induce from the inner product will be equivalent. $\endgroup$ – KyleW Aug 13 '15 at 13:10
  • $\begingroup$ Your sentence beginning "As to prove .." is unintelligible. Please clarify it: saying "can be induced" rather than "is induced" makes a significant difference to the meaning. $\endgroup$ – Rob Arthan Aug 14 '15 at 21:46
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Well, this depends entirely on what you mean by a norm being induced by an inner product. The way this expression is usually used, the norm induced by the inner product $\langle \cdot, \cdot \rangle$ is by definition the norm $\lVert \cdot \rVert = \sqrt{\langle \cdot, \cdot \rangle}$. If you were to use the expression more loosely, you might for instance say that $\lVert \cdot \rVert = 2\sqrt{\langle \cdot, \cdot \rangle}$ is also "a norm induced by $\langle \cdot, \cdot \rangle$". You wouldn't be following standard usage, but it would be understandable. In any case, this definitional question is the only one at issue here; as far as your question seems to imply that there is anything to prove here, I believe this is mistaken. In the theorem you quote at the end of the question, the above standard definition of a norm being induced by an inner product is presupposed.

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In an inner product space $$ \langle v, w \rangle = \frac{1}{2}(\|v + w\|^2 -\|v\|^2 - \|w\|^2) $$ So the norm determines the inner product. The norm on a normed vector space is induced by an inner product iff the function $\langle \cdot, \cdot \rangle$ defined by the above formula satisfies the axioms for an inner product. The Jordan-von Neumann theorem says that this is so iff $\|\cdot\|$ satisfies the parallelogram identity: $$ \|v + w\|^2 + \|v - w\|^2 = 2(\|v\| + \|w\|)^2 $$ Normed vector spaces with this property are called Euclidean.

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