- #1

Government$

- 87

- 1

## Homework Statement

Let f be defined for all real x and suppose that,

[itex]\left|f(x) - f(y)\right|\leq(x-y)^2[/itex]

for all real x and y. Prove that f is constant.

## The Attempt at a Solution

First of all, is following allowed. Since f is constant then [itex]\left|f(x) - f(y)\right|=0[/itex], and form here to build my proof?

Here is what i don't understand. If i suppose that [itex]\left|f(x) - f(y)\right|\leq(x-y)^2[/itex] then does that imply that [itex]f(x)[/itex] is always greater then [itex]f(y)[/itex]?

Since [itex](x-y)^2>0[/itex] when [itex]x≠y[/itex] then that means that [itex]\left|f(x) - f(y)\right|> 0[/itex] because if it were to be [itex]\left|f(x) - f(y)\right|= 0[/itex] we would violate [itex]x≠y[/itex] . Onliy way that [itex]\left|f(x) - f(y)\right|> 0[/itex] is that [itex]f(x)[/itex] is always greater then [itex]f(y)[/itex]

But this leads me to contradiction since [itex]f[/itex] should be constant i.e. [itex]\left|f(x) - f(y)\right|=0[/itex].

On the other hand [itex](x-y)^2=0[/itex] iff [itex]x=y[/itex] then that means that [itex]\left|f(x) - f(y)\right|= 0[/itex].

So perhaps my question boils down to is can i take that [itex]x=y[/itex] or am i only allowed to that [itex]x≠y[/itex]?Also i am trying to think of all things that will prove that some function is constant. First thing that comes to mind is that first derivative is 0. Can that be a starting point?