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.
Best Answer
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.