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It is an easy question, but i want to make it clear :)

Let $(V,\langle -,- \rangle)$ be an inner product space over $\mathbb{K}$.

Then, is the inner product $\langle -,- \rangle:V\times V\rightarrow \mathbb{K}$ continuous?

Indeed, I have proven it, but I want to make it sure :)

Is this true?

EDIT:

Here's what i tried.

I tried to show that $\langle -,- \rangle$ is continuous at a fixed point $(\alpha,\beta)$.

Assume that for a given $\epsilon >0$, there is a neighborhood $N$ of $(\alpha,\beta)$ such that for all $(x,y)\in N, |\langle x,y \rangle - \langle \alpha,\beta \rangle |<\epsilon$.

Let $\| \cdot \|$ be the norm induced by the inner product.

Then, it is only sufficient to show that $\| x-\alpha \| \| y \| + \| \alpha \| \| y - \beta \|$ is smaller than $\epsilon$.

Consider separate cases when either $\alpha$ or $\beta$ is zero.

Mathemagic
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    May you please add what work you have done then? If you want to see another answer, then still, add your work and state that. – ShakesBeer Aug 26 '14 at 10:15
  • @coolydudey60 I just wanted to know whether this is really true or not. I'll add what i tried anyway. – Mathemagic Aug 26 '14 at 10:18
  • Indeed it is, and you'll only waste more time than you already have by writing it up because this is a duplicate. http://math.stackexchange.com/questions/4501/is-inner-product-continuous-when-one-arg-is-fixed. – ShakesBeer Aug 26 '14 at 10:18
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    @cooleydudey60 I have checked that post before i post my question. That post is asking whether an inner product is continuous when one argument is fixed. Isn't it? – Mathemagic Aug 26 '14 at 10:26
  • Yes, well look more carefully and read the comments! As far as your post is concerned (is this statement true?), the answer is given in that question (if not proved), so this is a duplicate. – ShakesBeer Aug 26 '14 at 10:29
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    @coolydudey60 I think you mean a comment by Robin. He said "an inner product is continuous of both arguments". This is really confusing since he may mean that it is continuous when left argument is fixed or right argument is fixed. I guess this comment also confuses the OP there, so he commented there, is an inner product continuous under the product topology on $V\times V ?$ , but then there is no reply to that comment. That's the reason why I posted this question. I'm sorry if it is only me who got confused. – Mathemagic Aug 26 '14 at 10:38
  • Alright ok, I was quick to assume you hadn't done your research properly and hadn't looked much, and I must apologise. However, do try to make your questions as complete as possible. And yes, it is true. – ShakesBeer Aug 26 '14 at 11:07

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