# If $B$ is a continuous bilinear function such that $B(h,k) = o(\lVert(h,k)\rVert^2)$, then $B=0$.

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)

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