Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that $B: H \times K \Rightarrow F$ is a continuous bilinear function, where $H,K$ and $F$ are real normed spaces.

I have to prove (not as homework) that if $B(h,k) = o(\lVert(h,k)\rVert^2)$, then $B=0$.

Since $B$ is bilinear and continuous, we have that $\lVert B(h,k)\rVert \leq \lVert B\rVert \lVert h \rVert \lVert k\rVert \leq \lVert B\rVert \lVert(h,k)\rVert^2$, where $\lVert B\rVert$ is the operator norm.

Hence we have $$0=\lim_{(h,k)\rightarrow 0} \frac{\lVert B(h,k)\rVert}{\lVert (h,k) \rVert^2} \leq \lim_{(h,k)\rightarrow 0} \frac{\lVert B\rVert \lVert(h,k)\rVert^2}{\lVert (h,k) \rVert^2}= \lVert B\rVert.$$

If there is some way for me to get that the last limit is also $0$, I have what has to be proven. But I don't see a way to this. Could anyone provide me with a tiny hint? (No full answers please)

share|cite|improve this question
Could you please edit in some examples of the kind of $B$ you are talking about? I am unable to combine things in the way you indicate. Mostly I have no clue what $\parallel B \parallel$ should mean here. Anyway, please type in a few actual $B(h,k),$ which would appear to be a function taking real values. Plus, if this is what I think, the word continuous is superfluous. – Will Jagy Oct 30 '12 at 0:29
@WillJagy, Is this ok? – sxd Oct 30 '12 at 0:36
That's more informative, certainly. I'll think about it. – Will Jagy Oct 30 '12 at 0:46
@WillJagy, thank you! – sxd Oct 30 '12 at 0:47
up vote 3 down vote accepted

Let $f, g$ be arbitrary and take $\epsilon > 0$ some parameter. Then look at the behaviour $\epsilon^{-2} B(\epsilon f, \epsilon g)$ as $\epsilon \to 0$.

share|cite|improve this answer
I'm sorry, but towards which step in my proof is this advice? – sxd Oct 30 '12 at 0:02
This is a hint how you can prove your proposition, i.e. that $B = 0$ if $B$ is bilinear and satisfies $B(h,k) = o(\| (h,k) \|^2 )$. I'm afraid your attempt just shows $\| B \| \geq 0$ which you know anyway. To get $\| B \| = 0$ try what I suggested. – DanielM Oct 31 '12 at 20:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.