Proving Limit of Bilinear Function to 0

[yifli]

Prove that $$\mbox{lim}_{(h,k) \rightarrow 0} \frac{|f(h,k)|}{|(h,k)|} = 0$$, where f is a bilinear function from R^n X R^m -> R^p

I don't know where to start only given the information that f is bilinear.

 

[micromass]

Hi yifli!

Start by putting

$$h=h_1e_1+...+h_ne_n$$

and

$$k=k_1e_1+...+k_me_m$$

where the e_i form a basis.

Now, how can you write f(h,k) now?

 

[yifli]

Hi yifli!

Start by putting

$$h=h_1e_1+...+h_ne_n$$

and

$$k=k_1e_1+...+k_me_m$$

where the e_i form a basis.

Now, how can you write f(h,k) now?

so f(h,k) can be written as follows:
$$f(h,k)=h_1(k_1f(e_1,\overline{e_1})+\cdots+k_mf(e_1,\overline{e_m}))+h_2(k_1f(e_2,\overline{e_1})+ \cdots +k_mf(e_2,\overline{e_m}))+\cdots+h_n(k_1f(e_n,\overline{e_1})+\cdots+k_mf(e_n,\overline{e_m}))$$

e_i and \bar{e_i} represents the basis of R^n and R^m respectively

when both h_i and k_i tend to zero, I guess the numerator gets closer to zero faster than denominator because of the product h_ik_i.
but how do you argue this rigorously?

 

[micromass]

so f(h,k) can be written as follows:
$$f(h,k)=h_1(k_1f(e_1,\overline{e_1})+\cdots+k_mf(e_1,\overline{e_m}))+h_2(k_1f(e_2,\overline{e_1})+ \cdots +k_mf(e_2,\overline{e_m}))+\cdots+h_n(k_1f(e_n,\overline{e_1})+\cdots+k_mf(e_n,\overline{e_m}))$$

e_i and \bar{e_i} represents the basis of R^n and R^m respectively

when both h_i and k_i tend to zero, I guess the numerator gets closer to zero faster than denominator because of the product h_ik_i.
but how do you argue this rigorously?

Indeed, let me illustrate this on an example. I hope the general method will be clear from that:

Take

$$f:\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}:(x,y)\rightarrow xy$$

Then we need to calculate

$$\lim_{(h,k)\rightarrow 0}{\frac{|hk|}{|(h,k)|}}$$

The easiest thing to do is pick the norm (it works without this too).

$$|(h,k)|=\max\{h,k\}$$

Then we know that $h\leq \max\{h,k\}$, thus

$$\frac{|h|}{|(h,k)|}\leq 1$$

Hence

$$0\leq \lim_{(h,k)\rightarrow 0}{\frac{|hk|}{|(h,k)|}}\leq \lim_{(h,k)\rightarrow 0}{|k|}=0$$

Try to do this in the general case...