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In a inner product space with inner product $\langle\ ,\ \rangle$ and real or complex line as its base field, for each point $x$ in the space, is $\langle x,-\rangle$ continuous function on the second argument, and is $\langle - ,x\rangle$ continuous function on the first argument? "Continuous" is defined respect to the topology induced by the inner product.

Thanks and regards!

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Yes.

Fix x in the inner product space, and let $f(y) = \langle y, x \rangle$ denote the inner product function. Note that this is a linear functional -- that is, it is linear in y, and maps vectors to scalars.

It is a well-known theorem that linear functionals are continuous (on the entire space) if and only if they are bounded. Here, "bounded" means that there exists a constant M such that $|f(y)| \leq M|y|$ for all y in the space.

That the inner product functional is bounded now follows from the Cauchy-Schwarz Inequality: $|f(y)| \leq |x||y|.$

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    $\begingroup$ In general, if X and Y are real/complex normed vector spaces and T: X -> Y is a linear operator, then the following are equivalent: (1) T is continuous on X; (2) T is continuous at 0; (3) T is bounded; (4) T maps bounded subsets to bounded subsets; (5) T is Lipschitz continuous on X. $\endgroup$ – Jesse Madnick Sep 13 '10 at 4:53
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Fix $x$. We have $|\langle x,y\rangle - \langle x,z\rangle| = |\langle x,y-z\rangle|\leq ||x||||y-z||$ by the Cauchy-Schwarz inequality. This gives you easily that $\langle x,-\rangle\colon \mathbf{V}\to\mathbb{F}$ is not only continuous, but uniformly continuous. Similarly for $\langle -,x\rangle$.

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  • $\begingroup$ Great, +1, thank you. I like this proof : ) Gives me an application of Cauchy Schwarz and answers my question about whether the inner product is continuous. $\endgroup$ – Rudy the Reindeer Jul 31 '12 at 12:13
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If we define $T_x v = \langle v,x\rangle$ then $T_x$ is a linear functional. Since an inner product induces a norm, and thus it is continuous if and only if it is bounded on the unit circle.

And so $|T_x v | = |\langle v,x\rangle | \le \|v\|\cdot \|x\|$ and for $v$ on the unit circle, i.e. $\|v\| = 1$ we have that $T_x$ is indeed bounded and therefore continuous.

The proof is similar if you fix the left argument, however it may not be a linear functional if it is a complex vector space, but rather antilinear (that is linear up to conjugation). The norm is unaffected by that, though, so it's not a big step to overcome.

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    $\begingroup$ Asaf... You and analysis? $\endgroup$ – Jonas Teuwen Aug 20 '11 at 13:38
  • $\begingroup$ If the thread was already bumped... correction past mistakes sounds like a good plan. :-) $\endgroup$ – Asaf Karagila Dec 21 '12 at 15:58
  • $\begingroup$ Why is is sufficient to analyze the unit circle? $\endgroup$ – mavavilj Feb 2 at 18:18
  • $\begingroup$ @mavavilj: Things scale up/down to the unit circle. And scaling is continuous. $\endgroup$ – Asaf Karagila Feb 2 at 18:54
  • $\begingroup$ @AsafKaragila Do you further assume that the vector space is finite? $\endgroup$ – mavavilj Feb 2 at 19:01
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I think the easiest way is to show that it is a convex function, and then use the theorem that says that if a convex function defined on a convex set, then the function is continuous in every interior point. Since we are talking about $\Bbb{R}^n$, then it is true for all points because $\Bbb{R}^n$ is a convex and open set.

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