Prove Constant Function Cannot Satisfy Condition for r>1

[Treadstone 71]

"Let r be a rational number greater than 1. Let f:R->R be a function that satisfies the condition that for all real numbers x and y,

|f(x)-f(y)|\leq (x-y)^r

Prove that f is a constant."

Perhaps there's some subtlety which I misunderstood, but even a constant function fails to satisfy the conditions for some choice of r. For instance, let r=3 and f be a constant. Let x<y, and the inequality fails, i.e.,

|f(x)-f(y)|=0\leq (x-y)^3 &lt;0

 

[AKG]

The question says that IF f satisfies this condition THEN it is a constant, not IF f is a constant THEN it satisfies the condition. If the thing you're asked to prove is true, then given what you just showed when r = 3, you can only conclude that NO function f (constant or otherwise) satisfies the condition.

 

[Treadstone 71]

But since x and y are arbitrary, x-y could be negative. In that case, if r is some rational number with an odd numerator and an even denominator, this whole thing fails, regardless of what kind of function it is.

 

[Muzza]

Probably just a typo, it should probably say

|f(x) - f(y)| <= |x - y|^r.

 

[AKG]

If r is the type you suggested, then since x-y could be negative, the condition wouldn't hold, as you point out. Again, I repeat, it says IF the condition holds THEN f is a constant.