# Show that $f(x) = x^2$ is not uniformly continuous on $[0,\infty)$

Ok, I know the same question has already been asked here, and I am not looking for an answer even though my proof looks kind of the same. But, I need to know whether or not I am on the right track. Also, the choice of $y$ on the other proof doesn't many much sense to me. So, here it goes:

Show that $f(x) = x^2$ is not uniformly continuous on $[0,\infty)$

This is what I did:

Suppose, it is. Then, fix $\epsilon = 1 > 0.$ Let, $x < \delta \in [0, \infty)$ and $y = 2x \in [0,\infty)$. Then, according to the definition,

$$\forall \epsilon > 0, \quad\exists \delta > 0\quad\text{such that} \quad\forall x,y \in [0,\infty),\qquad\mid x-y\mid < \delta\quad \implies\quad\mid f(x) - f(y)\mid < \epsilon$$

If we replace $y = 2x$, then $$\mid x -2x\mid = \mid -x\mid = x < \delta.$$ So, that holds. Now, $$\mid f(x) - f(y)\mid = \mid x^2 - 4x^2\mid = \mid -3x^2\mid = 3x^2 > \epsilon = 1,$$ which is a contradiction depending on the choice of $x$.

Is it correct? Can I do that? Thanks.

• No this proof is not correct. Did you study thoroughly the answer to the question on the other page? What did you fail to get about it?
– Did
Jun 26, 2015 at 20:18
• Absolute continuity is stronger than uniform continuity. Jun 26, 2015 at 20:18
• @Jellyfish You can't choose $x$ as big as you want because you need to satisfy $|x|<\delta$ in order to be able to choose $y=2x$. Indeed, you could choose $y=x+\delta/2$.
– Surb
Jun 26, 2015 at 20:18
• @JackD'Aurizio ??? What's your definition of "absolutely continuous"? The term has a standard definition, and absolutely continuous functions certainly need not be sublinear. Same for uniformly continuous functions - I don't see how absolute continuity has anything to do with the question. Maybe you were thinking of $Lip_1$ functions? Jun 26, 2015 at 20:19
• @JackD'Aurizio Do you think this can help an OP asking for a proof that $x\mapsto x^2$ is not uniformly continuous?
– Did
Jun 26, 2015 at 20:19

Your proof is wrong but the idea of contradiction is good. The problem in setting $y=2x$, is that the condition $|x-y|<\delta$ forces you to choose $|x|<\delta$ and so you can't guarantee a contradiction with $3x^2 >\epsilon$ ($\delta$ might be extremely small).

Suppose by contradiction that for $\epsilon =1$ there exists $\delta >0$ such that $|f(x)-f(y)|<\epsilon$ for every $|x-y|<\delta$.

Note that $\delta$ here is given and you don't know what it is, it can be anything very small or very big (but usually very small).

You want to find $x,y>0$ such that $|x-y|<\delta$ and $|f(x)-f(y)|>\epsilon$, this would imply a contradiction.

Your idea to choose $y$ so that $x$ comes out when you evaluate $|f(x)-f(y)|$ (for getting the contradiction) is good. Since you need $|x-y|<\delta$ you have to choose $y\in ]x-\delta,x+\delta[$.

For example, we can take $y=x+\delta/2$ to get $$|f(x)-f(y)|=|x^2-(x+\delta/2)^2| =|x\delta +\delta^2/4|=x\delta + \delta^2/4\qquad \qquad \forall x\geq 0$$

Finally, choose $x$ big enough to get the contradiction. That is, we want $x\geq 0$ such that $$|f(x)-f(y)|=x\delta+\delta^2/4\geq \epsilon = 1 \implies x\geq \dfrac{1-\delta^2/4}{\delta}.$$ It follows that any $x$ such that $x\geq \dfrac{1-\delta^2/4}{\delta}$ will lead to a contradiction.

You can try to find such bound for $y=x+\alpha$ with $|\alpha|<\delta$. This is a good exercise to check if you really understood the proof.

• you can choose $y=x+\alpha$ for any $0<\alpha<\delta$ if you prefer, e.g. $y=x+\delta/123798$. The important thing is to be sure that $|x-y|<\delta$.
– Surb
Jun 26, 2015 at 20:51
• Thank you so much for your detail answer. Jun 26, 2015 at 21:38
• @Jellyfish I edited my answer. Is it ok now? (don't be sorry, it's a pleasure to answer someone who is really interested in understanding the answer :) )
– Surb
Jun 29, 2015 at 21:00
• Sorry! one more thing. I think there is a computational mistake where you computed $\mid x^2 - (x+\delta/2)^2\mid$ I think it should be $\mid x^2 - (x^2 + 2*x*\delta/2 + \delta^2 / 4)\mid = \mid -x\delta-\delta^2/4\mid = x\delta + \delta^2/4$. But, then again, it's more about understanding the idea that counts. Thank you so much. Jul 5, 2015 at 0:43
• @Jellyfish good job :), thanks for pointing out this miscalculation. I edited in consequence. Should be correct now :). Thank you for your feedback.
– Surb
Jul 5, 2015 at 13:59

No, your proof method is invalid. Uniform continuity means, effectively, that if I give you any value $\epsilon > 0$, no matter how small, there is some number $\delta$ such that the neighborhood of size $\delta$ around every point $x$ is mapped to a neighborhood of size $\epsilon$ or smaller around $f(x)$. Importantly, this $\delta$ has to be the same for all $x$. What you've shown, on the other hand, is that the neighborhoods of size $\delta$ around the specific point $x = \delta$ (in other words, the intervals $(0, 2 \delta)$) will be mapped to a neighborhood around $f(\delta)$ whose size is greater than 1 if we choose $\delta$ to be sufficiently large. These are not the same proposition.

The easiest way to actually prove that a function is not uniformly continuous is to show that $\delta$ can't exist. In other words, you need to show that for a given $\epsilon$ and $\delta$, you can always find an $x$ value such that the $\delta$-neighborhood around $x$ will be mapped to a neighborhood of $f(x)$ whose size is greater than $\epsilon$.