Conditions for concavity/convexity - multivariable case
Here's the problem:
Give an informal argument to show that a function F which satisfies F(0) = F(1) = 0
and F''(t) >=0 must satisfy F(t) <= 0 for 0 <= t <= 1.
The function F is defined for fixed x1 and x2 by
F(t) = f(tx1 + (1 - t)x2) - tf (x1) - (1 + t)f(x2);
where f is a smooth function which satisfies f '' (x) >= 0. Show that F satisfies the conditions of the first paragraph and deduce that f is convex.
How can this result be extended to the case where f is a function of many variables (i.e. where x1 is replaced by the vector x1, etc)?
For the first part you can give a geometric argument or alternatively use Rolle's Theorem to formalise it. The second part I simply have to use the chain rule and I get the definition of convexity for f(.)
I have a bit of difficulty with the last part - generalizing the result.
I am not sure how to get the second derivative of F''(t). If i am correct i get
F''(t) = Σ_i [ (X1_i-X2_i) Σ_j (X1_j-X2_j) (d^2f/dX_i dX_j) ]
I need conditions for F''(t0)<= 0 to be equivalent to "the hessian matrix is positive definite"
i am not sure that the above expression is right. Even if it is, there are some simplifications which can be made i think. Anyways i am not sure how to get "the hessian is positive definite" as a condition for convexity, from this expression.
Thanks a lot!
Re: Conditions for concavity/convexity - multivariable case
I figured it out.
F''(t) can be written as X'HX where X=(X1_i -X_2_i) For i = 1,2.....n
and H is the hessian
Therefore F''(t) >= 0 is equivalent to H being positive definite
which in turn implies that F(t)<= 0 for 0<=t<=1... This gives us convexity of f(.)