Proving $f(x) = x^5$ is not uniformly continuous

Question

Is this line of reasoning legitimate?

Let $f: \mathbb{R} \rightarrow \mathbb{R}, f(x) = x^5$. $f$ is not uniformly continuous.

Suppose it were. Fix $\epsilon>0$ and the associated $\delta$, and choose $x,~ y>0, ~x>y$, st $|x-y|< \delta$.

Then, $|x-y|<\delta \Rightarrow |x^5 - y^5|<\epsilon ~\forall x, y$

Fix $|x-y| = C$. Then,

$$|x^5 - y^5| = |x-y||x^4 + x^3y + x^2y^2 + xy^3 + y^4|= $$

$$C|x^4 + x^3y + x^2y^2 + xy^3 + y^4|<\epsilon$$.

This is valid for all $x, y$ sufficiently close, but because the left hand side increases monotonically, it can be made arbitrarily large while keeping $|x-y|<\delta$, so this is a contradiction. Therefore f is not uniformly continuous.

I'm pretty sure this is logically sound kinda and I think my intuition (that because f is increasing increasingly fast without bound, no set of fixed-sized intervals in the domain map to a set of bounded intervals in the image) but I think I can get away with a lot less. Any advice?

Answer 1

An easy way of arguing is as follows:

Suppose it were, then for every $\epsilon$, there exists a $\delta$ for which $|x - y| < \delta \implies |x^5 - y^5| < \epsilon$

However, consider $\epsilon = 1$; if such $\delta$ existed and $y = x + \frac{\delta}{2}$, we would find that

$$|x^5 - (x + \frac{\delta}{2})^5| < 1$$

for every real $x$; however, this a clear contradiction, since we can choose $x$ large.

Answer 2

This argument works for any function with an unbounded derivative. Since $f'(x) >M$ for any $M>0$, $f(x+h)-f(x)>Mh$ which contradicts uniform continuity for large enough $M$.

Answer 3

Take , $\displaystyle x=\frac{\delta}{2}+\frac{1}{(\delta)^{1/4}}$ and $\displaystyle y=\frac{1}{(\delta)^{1/4}}$. Then , $\displaystyle |x-y|=\frac{\delta}{2}<\delta$.

But , $|f(x)-f(y)|>5/2$.

Note : Any function $f:\mathbb R \to \mathbb R$ defined by $f(x)=x^n$ is NOT uniformly continuous for $n>1$.