Real analysis - show convex functions are left & right differentiable

[quasar987]

[SOLVED] Real analysis - show convex functions are left & right differentiable

Homework Statement

Let f:R-->R be convex. Show f admits in every point a left derivative and a right derivative.

Homework Equations

A function f:R-->R is convex if x1 < x < x2 implies

$$f(x)\leq \frac{x_2-x}{x_2-x_1}f(x_1)+\frac{x-x_1}{x_2-x_1}f(x_2)$$

Or equivalently, if whatever x, y, and $\lambda$ in [0,1],

$$f(\lambda x + (1-\lambda)y\leq \lambda f(x) + (1-\lambda)f(y)$$By left derivative at x0, we mean the limit

$$D_lf(x_0)\lim_{x\rightarrow x_0^-}\frac{f(x)-f(x_0)}{x-x_0}$$

and by right derivative at x0, we mean the limit

$$D_rf(x_0)\lim_{x\rightarrow x_0^+}\frac{f(x)-f(x_0)}{x-x_0}$$

The Attempt at a Solution

Let's stick to the left derivative.

I know convex functions are Lip****z, so the differential quotient is bounded.

I have also proven in an earlier exercise that the differential quotient is increasing as x increases:

"If x1 < x < x2, then $$\frac{f(x)-f(x_1)}{x-x_1}\leq \frac{f(x_2)-f(x_1)}{x_2-x_1} \leq\frac{f(x_2)-f(x)}{x_2-x}$$"

But this does not give the conclusion because I must show the differential quotient converges for any sequence, monotonous or not, converging to x0.

Can we show limsup=liminf? Can we show it is Cauchy? I don't see how.

 

[morphism]

Keep in mind that you're taking the limit as x approaches x_0 from the left. So out of the sequences converging to x_0, we need only consider those whose tails increase to x_0.

 

[quasar987]

how come? I thought limit from the left only meant that we only consider sequences whose points are lesser than x_0

 

[morphism]

I worded that very badly!

What I was trying to get across is that if L = lim(x->a-) g(x) exists, and g(x) is increasing, then this limit is going to be sup{g(x) : x < a} (by uniqueness of limits).

 

[quasar987]

this allows us to conclude that the limit of the differential quotient exists?

 

[morphism]

Let g be the differential quotient. Why does sup{g(x) : x < x_0} exist?

 

[quasar987]

I see your point! Now I can try to show directly that the limit is sup{g(x) : x < x_0}.

And this is easy! you rock :D