Prove that a function $f:R \to R$ which is unbounded on the interval $(a, b)$ cannot be uniformly continuous on this interval.
$$$$Consider the closed interval $[a, b]$, Now as the function is unbounded on $(a, b)$, so it is unbounded on $[a, b]$. Now divide $[a, b]$ into two subintervals of equal width, then $f(x)$ is unbounded on atleast any one of them say $[a_1, b_1]$. Now again divide this subinterval into subintervals of equal width, then again $f(x)$ is unbounded on any one of them say $[a_2, b_2]$. Now if we continuously divide these intervals, then by nested interval Theorem there exists a $c$ which lies in every interval and hence $f(x)$ is unbounded in every neighbourhood of $c$ and hence for any given $N$, for every $\delta>0$ we can find a $x_{\delta} \neq c$ such that $$|x_{\delta}-c|<\delta$$ and $$f(x_{\delta})>N$$. Now chose a $\epsilon_0>0$ then for every $\delta>0$ we can find a $x_{\delta} \neq c$ such that $$|x_{\delta}-c|<\delta$$ and $$f(x_{\delta})>f(c)+\epsilon_0$$ and hence there exists a $\epsilon =\epsilon_0$ such that for every $\delta>0$ we can find a $x=x_{\delta}$ and $y=c$ such that $$|f(x)-f(y)|>\epsilon_0$$ and hence $f(x)$ is not uniformly continuous on $(a, b)$
$$$$Is My Proof Correct??
Function Unbounded On Bounded Intervals Is Not Uniformly Continuous
calculus
Best Answer
It can be $c=a$ or $c=b$, so it must be fixed like this:
$f$ is unbounded on $[a, b]$, so we can take decending chain of closed intervals $[a, b]\supset [a_1, b_1] \supset \cdots $ which satisfies the following condition: (1)$f$ is unbounded on $[a_i, b_i]$. (2)$\lim_{n\to \infty}(a_i-b_i)=0$.
Then $\{c\}=\bigcap_{i\in \mathbb{N}}[a_i, b_i]$, for some $c\in [a, b]$. It is clear that $$\forall r>0 \forall \delta>0\exists x\in(a,b):|x-c|<\delta \land|f(x)|>r. $$ Take $\varepsilon =1$. For every $\delta>0$, you can take $x\in (a,b)\cap(c-\delta/2, c+\delta/2) $, since this set is not empty. Then take $y\in (a,b)\cap(c-\delta/2, c+\delta/2)$ such that $|f(y)|>|f(x)|+1$. It follows from the triangle equality that
$$|x-y|\leq|x-c|+|c-y|<\delta/2+\delta/2=\delta$$ and $$|f(x)-f(y)|\geq|f(x)|-|f(y)|>1=\varepsilon$$ Therefore $f$ is not uniform continuous.