Let f : R → R be a continuous function that is convex on (-infinity, 0] and [0, ∞) and has a local maximum at the point 0. Prove that the function f is not differentiable at the point 0.

Answer 1

We will prove that

lim(h->0+)(f(0) - f(x+h))/(0 - h) < 0 and lim(h->0-)(f(0) - f(x+h))/(0 - h) > 0

if those limits exist at all.

Actually I will prove only the first part, as the second has a symmetrical argument.

But once both are proven then they imply that

lim(h->0)(f(0) - f(x+h))/(0 - h) does not exist, i.e., f is not differentiable at 0.

Lemma:

For any h > 0

f(0) - f(h/2) > (f(0) - f(h))/2

Proof:

By definition of convex function

f(h/2) < (f(0) + f(h))/2

From that

-f(h/2) > (-f(0) - f(h))/2

f(0) - f(h/2) > f(0) + (-f(0) - f(h))/2

f(0) - f(h/2) > (f(0) - f(h))/2

Now we continue with the main proof.

By definition of local maximum there exits h₀>0 such that

f(0) > f(x+h₀)

and hence

(f(0) - f(x+h₀))/(-h₀) < 0

Consider the following sequence s₀, s₁, ..., where

s_i = (f(0) - f(h₀/2ⁱ))/(-h₀/2ⁱ)

Intuitively we are approaching 0 from the right by halving the distance to 0 at each step.

We show that s_i+1 < s_i.

Using the Lemma and the fact that -h₀ < 0

s_i+1 = (f(0) - f(h₀/2ⁱ⁺¹))/(-h₀/2ⁱ⁺¹) < (f(0) - f(-h₀/2ⁱ))/2(-h₀/2ⁱ⁺¹) = s_i

So {s_i}is a decreasing sequence with each s_i < s₀ = (f(0) - f(x+h₀))/(-h₀) < 0

This shows that if

lim(h->0+)(f(0) - f(x+h))/(-h)

exists then it must be negative.

And this completes the proof.