How to prove that a convex function $f(x)$ with $f(0) = 0$ satisfes $f'(x)x > f(x)$?
I am not sure how to use the property of convexity to solve this.
How to prove that a convex function $f(x)$ with $f(0) = 0$ satisfes $f'(x)x > f(x)$?
I am not sure how to use the property of convexity to solve this.
On
Straight-forward approach:
We are interested in the expression $xf'(x) > f(x) \Leftrightarrow xf'(x) - f(x) > 0$.
In such cases, in Calculus, a good way of working our way is naming a function $F(x) = xf'(x) - f(x)$ be the expression that we are interested in and studying that.
Note that $F'(x) = xf''(x) + f'(x) - f'(x) = xf''(x) > 0 $ for $ x>0$. But, that means that $F(x)$ is an increasing function for $x > 0$. Can you conclude now ?
Alternative via MVT:
If $f$ is convex, then $f''(x) > 0$. This means that $f'(x)$ is an increasing function.
Apply the Mean Value Theorem for $f$ now, over an interval $[0,x]$. Then, there exists an element $\xi \in (0,x)$ such that :
$$f'(\xi) = \frac{f(x) - f(0)}{x - 0} \Leftrightarrow f'(\xi) = \frac{f(x)}{x}$$
But, as we said the function $f'(x)$ is increasing, thus for any $x > \xi$ it is $f'(x) > f'(\xi)$.
That means that $f'(x) > \frac{f(x)}{x} \Leftrightarrow f'(x)x > f(x)$ for $x >0$.
Note that when $x=0$ the strict inequality does not hold, thus : $xf'(x) \geq f(x)$.
On
You can easily construct a convex function that is not differentiable and attains $$f(0) = 0$$ The easiest example is probably $f(x) = |x|$ which has derivative everywhere except at $x=0$.
Now to take this one step further, just split line segments (for example already approximating some convex function) into smaller line segments. The derivative will have a jump at every knot between these lines. Now do that systematically, perhaps in a dyadic fashion similar to how discrete wavelets are constructed. You will get a function with as many discontinuities in the derivative as you want.
So your question needs to be attached with the big IF it is differentiable.
This proof uses only the definition of a convex function and existence of $f'(x)$.
If $h>0$ then $x=(t)(0)+(1-t)(x+h)$ where $t =\frac h {x+h}$. Hence $f(x) \leq tf(0)+(1-t) f(x+h)$. I will leave it to you to re-write this inequality as $\frac {f(x+h) -f(x)} h \geq \frac {f(x)-f(0)} x$. Letting $h \to 0$ gives $f'(x) \geq \frac {f(x)} x$ for $x>0$. The proof for $x<0$ is similar.
Taking $f\equiv 0$ we see that strict inequality need not hold.