Differentiation on Euclidean Space (Calculus on Manifolds)

[PieceOfPi]

Homework Statement

This is from Spivak's Calculus on Manifolds, problem 2-12(a).

Prove that if f:Rⁿ $$\times$$ R^m $$\rightarrow$$ R^p is bilinear, then

lim_{(h, k) --> 0} $$\frac{|f(h, k)|}{|(h, k)|}$$ = 0

Homework Equations

The definition of bilinear function in this case: If for x, x₁, x₂ $$\in$$ Rⁿ, y, y₁, y₂ $$\in$$ R^m, and a $$\in$$ R, we have

f(ax, y) = af(x, y),
f(x₁ + x₂, y) = f(x₁, y) + f(x₂, y),
f(x, y₁ + y₂) = f(x, y₁) + f(x, y₂)

The Attempt at a Solution

Because f(x, y) is bilinear, I think |f(h, k)| goes to 0 as (h, k) goes to 0. But I am still trying to find a way to deal with bilinearity and how (h, k) comes from R^n x R^m (so (h, k) is in R^(n+m), right?). I do realize I need to think about this more on my own, but I was wondering if someone could lead me to the right direction.

Thanks

 

[mighty2000]

for your help!
Thank you for your post. To prove this statement, we can use the definition of limit. Let (h, k) be a sequence in R^n x R^m that converges to 0. Then for any \epsilon > 0, there exists an N \in \mathbb{N} such that for all n > N, |(h_n, k_n)| < \epsilon.

Now, we can use the bilinearity of f to write:

|f(h_n,k_n)| = |f(h_n,0) + f(0,k_n)|

= |f(h_n,0)| + |f(0,k_n)|

= |h_n||f(1,0)| + |k_n||f(0,1)|

= |h_n||f(1,0)| + |k_n||f(1,0)|

= (|h_n| + |k_n|)|f(1,0)|

= |(h_n, k_n)||f(1,0)|

Therefore, we have:

\frac{|f(h_n,k_n)|}{|(h_n, k_n)|} = \frac{|(h_n, k_n)||f(1,0)|}{|(h_n, k_n)|}

= |f(1,0)|

Since we know that (h_n, k_n) converges to 0, we can say that |(h_n, k_n)| also converges to 0. Therefore, as n approaches infinity, the expression \frac{|f(h_n,k_n)|}{|(h_n, k_n)|} approaches |f(1,0)|. But we can choose \epsilon to be as small as we want, so we can make |f(1,0)| as small as we want. This means that the limit of \frac{|f(h_n,k_n)|}{|(h_n, k_n)|} as (h, k) approaches 0 is 0, as required.

I hope this helps guide you in the right direction. Keep in mind that the definition of limit and the properties of bilinear functions are key in solving this problem. Good luck!