- Thread starter
- #1

#### CaptainBlack

##### Well-known member

- Jan 26, 2012

- 890

\( f(x_1)-f(x_2) \le (x_1-x_2)^2 \)

CB

- Thread starter CaptainBlack
- Start date

- Thread starter
- #1

- Jan 26, 2012

- 890

\( f(x_1)-f(x_2) \le (x_1-x_2)^2 \)

CB

- Moderator
- #2

- Feb 7, 2012

- 2,707

Since nobody has responded so far, here's my approach to the problem. First, if $f(x)$ is a solution then so is $f(x) + k$ for any constant $k$. So by adding a constant to $f$ we may as well assume that $f(0)=0.$ Also, if $f(x)$ is a solution then so is $-f(x).$ So anything that we can prove about $f$ will also be true for $-f.$

\( f(x_1)-f(x_2) \le (x_1-x_2)^2 \)

CB

Thus $f(x) = f(x)-f(0) \leqslant (x-0)^2 = x^2$, and also $-f(x)\leqslant x^2$. Therefore $|f(x)| \leqslant x^2$.

Next, $f(x) = \bigl(f(x) - f(\frac12x)\bigr) + \bigl(f(\frac12x) - f(0)\bigr) \leqslant \bigl(\frac12x\bigr)^2 + \bigl(\frac12x\bigr)^2 = \frac12x^2.$ As before, the same applies to $-f(x)$ and so $|f(x)| \leqslant \frac12x^2.$

Now repeat this argument inductively to show that $|f(x)| \leqslant \dfrac1{2^n}x^2$ for all $n\geqslant1.$ Then let $n\to\infty$ to conclude that $f(x)=0.$

Therfore the only functions with this property are the constant functions.

- Thread starter
- #3

- Jan 26, 2012

- 890

My solution:

\( f(x_1)-f(x_2) \le (x_1-x_2)^2 \)

CB

The inequality implies that:

\(| f(x_1) - f(x_2) |\le (x_1-x_2)^2\)

so:

\( \displaystyle \left| \frac{f(x_1) - f(x_2)}{x_1-x_2} \right| \le |x_1-x_2| \)

Which implies that \(f(.)\) is differentiable everywhere on \( \mathbb{R} \) and that its derivative is everywhere \(=0\). Hence any such function is a constant, and any constant function satisfies the conditions of the problem ...

CB