In my Real Analysis book I found a very interesting problem. Let f be a function on R such that: |f(x-y)| <= (x-y)^2 for all x,y in R. Show that, f is a constant function on R.
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Originally Posted by ThePerfectHacker In my Real Analysis book I found a very interesting problem. Let f be a function on R such that: |f(x-y)| <= (x-y)^2 for all x,y in R. Show that, f is a constant function on R. Is f a function of one variable? For example, might it be something like f(a), where in this case we are letting a = x - y?
Originally Posted by ecMathGeek Is f a function of one variable? For example, might it be something like f(a), where in this case we are letting a = x - y? Yes. If it were a function in two variables I would have said a function on R^2.
f(x)=x^2 satisfies this and is not Konstant.
Originally Posted by ThePerfectHacker Yes. If it were a function in two variables I would have said a function on R^2. I figured that was the case, but I wanted to be sure there was no typo in the problem. Originally Posted by Rebesques f(x)=x^2 satisfies this and is not Konstant. I was thinking the same thing. But I think we are misunderstanding the logic of the problem.
Originally Posted by Rebesques f(x)=x^2 satisfies this and is not Konstant. Sorry I meant to write: |f(x)-f(y)|<=(x-y)^2
Yes that would make a difference. See the file.
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