I'm trying to use continuity to show that $f(x)=x^3$ is uniformly continuous on $[0,1]$ but not $[0,\infty)$.
I've tried setting up an epsilon-delta proof, but I'm struggling a little:
By definition of uniform continuity, we know that $\forall \epsilon >0, \exists \delta >0$ such that
$|x-y|<\delta \Rightarrow |f(x) – f(y)| < \epsilon$.
And so, forcing
$\delta = \min\{1, \frac{\epsilon}{p_x}\}$ where $p_x = (x^2+xy+y^2)$
And so, we havve that
$|(x)^3 – (y)^3|=|(x-y)(x^2+xy+y^2)| < |\delta (x^2 + xy+y^2)| < \epsilon$
I'm not sure if this is the correct way to go about proving it, or if I landed myself into a circular argument. Furthermore, intuitively I'm guessing we only have uniform continuity on $[0,1]$ but not [0,$\infty)$ because our $p_x$ would get too large?
Best Answer
Note that in the definition of uniform continuity, given $\varepsilon > 0$, you need to provide a $\delta > 0$ that works for all $x,y \in [0,1]$. In particular, you cannot have a $\delta$ that depends on $x$. This is in contrast to proving that $f$ is continuous at a specific $x$ where the $\delta$ you provide can depend on $x$.
In your case, we have
$$ |x^3 - y^3| = |(x - y)(x^2 + xy + y^2)| \leq |x - y||x^2 + xy + y^2|. $$
Now, if $x,y \in [0,1]$, we have $|x^2 + xy + y^2| \leq 3$ so we can deduce that
$$ |x^3 - y^3| \leq 3|x - y|. $$
Hence, given $\varepsilon > 0$, we can take $\delta = \frac{\varepsilon}{3}$ and then if $|x - y| < \delta$ then $|x^3 - y^3| \leq 3|x - y| < 3\delta = \varepsilon$.
Note that the same argument would work if you wanted to prove uniform continuity on $[0,L]$ with the constant $3$ replaced by a different constant $C_L$ which bounds $|x^2 + xy + y^2|$ on $[0,L]$ (for example, $C_L$ can be $3L^2$).
Expressed in this way, we see that your basic intuition is correct. As $L$ gets larger, our constant $C_L$ also gets larger and hence our $\delta = \frac{\varepsilon}{C_L}$ gets smaller and smaller. In the limit, this should lead us to suspect (but this is not a formal proof!) that uniform continuity will fail.