Proving that a function satisfying $|f(x)-f(y)|

Question

Let $f$ be a function defined on the interval $[a,b]$ such that $0<a<b$:

$\forall x,y\in[a,b],x\neq y,|f(x)-f(y)|<k|x^{3}-y^{3}|$ where $k>0$ .

Prove that the function $f$ is continuous.

My attempt

I let $\epsilon>0$ and I let $\eta=\frac{\epsilon}{k}$ so by definition of uniform continuity:

$|x^{3}-y^{3}|<\eta\Rightarrow|f(x)-f(y)|<k|x^{3}-y^{3}|<k\eta=\epsilon$

So therefore $f$ is uniformly continuous. Can anyone check my work please? Thank you in advance.

Yea but in my cause I have to show that $f$ is uniformly continuous no just continuous at a certain point. I also believe this is a Lipschitz function. — Module, Dec 14 '13 at 14:41
When you factor $|x^3-y^3|$, you left with a factor you need to bound, which can be done by noting that $x,y \in [a,b]$. — Mhenni Benghorbal, Dec 14 '13 at 14:51
@MhenniBenghorbal Yes thank you very much for your help take a look at my comment to your answer. I believe that's what I have to do right? — Module, Dec 14 '13 at 14:59

Sugata Adhya · Answer 1 · 2013-12-14T14:52:10.297

1

Since $x,y\ge a$

$$|f(x)-f(y)|<k|x^{3}-y^{3}|=k|x^2+xy+y^2||x-y|\le 4a^2k|x-y|$$

and hence $f$ is Lipschitz continuous.

edited Dec 14 '13 at 14:52

answered Dec 14 '13 at 14:46

Sugata Adhya

Alright thank you very much so all I have to chose is $\eta=\frac{\epsilon}{4a^{2}k}$ so I get $4a^{2}k|x-y|<4a^{2}k(\eta)=\epsilon$ – Module Dec 14 '13 at 14:55
Yep...$$$$$$$$$$ – user113578 Dec 14 '13 at 15:01

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