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

Question

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.

Answer 1

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.

Answer 2

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 $\epsilon$.