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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4 Answers4



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: $|f(y)| \leq |x||y|.$

Jesse Madnick
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    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. – Jesse Madnick Sep 13 '10 at 04:53

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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Arturo Magidin
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  • 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. – Rudy the Reindeer Jul 31 '12 at 12:13

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.

Asaf Karagila
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    Asaf... You and analysis? – JT_NL Aug 20 '11 at 13:38
  • If the thread was already bumped... correction past mistakes sounds like a good plan. :-) – Asaf Karagila Dec 21 '12 at 15:58
  • Why is is sufficient to analyze the unit circle? – mavavilj Feb 02 '19 at 18:18
  • @mavavilj: Things scale up/down to the unit circle. And scaling is continuous. – Asaf Karagila Feb 02 '19 at 18:54
  • @AsafKaragila Do you further assume that the vector space is finite? – mavavilj Feb 02 '19 at 19:01
  • @mavavilj: No... – Asaf Karagila Feb 02 '19 at 19:42
  • @mavavilj: I'm not sure which RHS you mean. – Asaf Karagila Feb 02 '19 at 20:02
  • @mavavilj: Also, where do I mention "dimension" in the whole story? – Asaf Karagila Feb 02 '19 at 20:06
  • Then how is the R.H.S. finite? In 2nd paragraph. If it's bounded, it must be finite? – mavavilj Feb 02 '19 at 20:07
  • @mavavilj: Do you understand what does it mean for a linear operator to be bounded on the unit circle? Perhaps you should look up the relevant definitions. Finiteness is irrelevant. – Asaf Karagila Feb 02 '19 at 20:21
  • @AsafKaragila My notes have a lemma: Linear operator $T:E \rightarrow F$ is bounded iff its norm is finite. In the C-S the norm of the operator is on the left. But is the right finite so that the left can be? Further the lemma says to find a $C < \infty$ s.t. $\|Tx\|_F \leq C\|x\|_E$ $\implies$ $\|T\| \leq C$. So maybe you did $C=1$? However, since it seems like you fixed $x$, rather than $v$, then should $\|x\|=1$? Or which one is your parameter? – mavavilj Feb 02 '19 at 20:38
  • @mavavilj: Yes, the norm is finite. What does that have to do with the dimension? – Asaf Karagila Feb 02 '19 at 22:04
  • The norm definition I have allows the norm to attain values in $[0,\infty)$. I read that this includes $\infty$, no? It's open in the right end, but this is often the case when marking for "goes to infinity", since the infinity cannot be really attained. On the other hand, another reference says that in normed spaces $\| \cdot \|$ is always assumed to not attain infinity, it always has finite value. – mavavilj Feb 02 '19 at 22:19
  • @mavavilj: Where in my answer do you see the word "finite" or "infinite"? – Asaf Karagila Feb 02 '19 at 23:38
  • @AsafKaragila I'm asking, how would one distinguish between them, given your answer. Or how your answer relates to the notes that I have, which say that norms are always finite. While vector spaces might be infinite dimensional. Your answer leaves open as to whether the R.H.S. could be infinite? So I just want to check, what definitions your answer uses. – mavavilj Feb 03 '19 at 11:34
  • @mavavilj: The RHS is never infinite. It's a real number and all real numbers are finite. I'm still not sure what that has to do with the dimension of the space. – Asaf Karagila Feb 03 '19 at 11:42
  • @AsafKaragila Well since the inputs to norms are vectors, then if the vector space is infinite dimensional, then the norm could be infinite? Particularly e.g. in $\|\cdot\|_2$? Since one'd have an infinite positive sum inside the square root? – mavavilj Feb 03 '19 at 12:04
  • @mavavilj: Norms and inner products are abstractly defined. – Asaf Karagila Feb 03 '19 at 12:11
  • @AsafKaragila What does that mean? Surely one can do your proof for finite and infinite cases separately. Also, what use is an "abstract definition" if it would fall apart, if one has to consider specific cases? – mavavilj Feb 03 '19 at 12:12
  • @mavavilj I think it's time to end this discussion. It's not the point of the comments. – Asaf Karagila Feb 03 '19 at 12:45

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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