2
$\begingroup$

Let $f$ be a real uniformly continuous function on the bounded set $E$ in $\mathbb R$. Prove that $f$ is bounded on $E$.

What I tried to show was that by contradiction, saying that $f$ is unbounded on $ E$. Then we should be able to find an integer $N$ such that $f(N)>M$ for some integer $M$ which would bound the interval. I know by uniform continuity, that we can choose any epsilon we want. I get stuck about how to show this. Any hints would be wonderful. Thank you

$\endgroup$

4 Answers 4

1
$\begingroup$

For just a hint: Consider $\bar{E}$ which is closed, and use the definition of uniform convergence.

We have $E \subset \mathbb{R}$ is bounded, so $\bar{E} \subset \mathbb{R}$ is closed and bounded, thus compact.
Now, $f$ is uniform continuous on $E$, so for all $\epsilon >0$ there exists a $\delta >0$ such that $d(x,y) < \delta$ implies $d(f(x),f(y)) < \epsilon$.
Now take $\epsilon =37$, or your favorite positive number, then we can find a $\delta$ such that $d(x,y)<\delta$ implies $d(f(x),f(y))<37$.
Now, we can take the covering $\{ B(x,\delta) \mid x \in E\}$ of $\bar{E}$, and $\bar{E}$ is compact, so there exists a finite subcover $\{ B(x_i , \delta ) \mid i \le n \}$ of $\bar{E}$, so also of $E$. Now, we have for $y \in E$, there exists a $x_i\in E$ such that $d(y,x_i)< \delta$, and so $|f(y)| \le |f(x_i)| + 37$.
This yields $\sup_{y \in E} |f(y) | \le \max_{1 \le i \le n} |f(x_i)| + 37$, so $f$ is bounded.

$\endgroup$
3
  • $\begingroup$ how do you know that the closure of E is closed and bounded $\endgroup$ Nov 5, 2015 at 1:04
  • 1
    $\begingroup$ Do you know the definition of the closure of E? $\endgroup$ Nov 5, 2015 at 4:01
  • 1
    $\begingroup$ @JacobRodgers $\bar{E} = \{ x \in \mathbb{R} \mid \forall \epsilon > 0 \exists y \in E \textrm{ s.t. } d(x,y)< \epsilon \}$. Per definition $\bar{E}$ is closed, and $\bar{E}$ is bounded as $E$ is bounded by $K$, so $|x| \le d(x,y) + |y| \le K +\epsilon$ for all $\epsilon >0$, so $\bar{E}$ is also bounded by $K$. $\endgroup$
    – Hetebrij
    Nov 5, 2015 at 10:32
1
$\begingroup$

If $f$ is uniformly continuous on $E$, then there is some $\delta > 0$ such that $|x-y| < \delta$ only if $|f(x) - f(y)| < 1$. Let $a < b$ be the endpoints of $E$; let $n > \frac{b-a}{\delta}$; let \begin{align*} x_{0} &:= a,\\ x_{k} &:= x_{k-1} + \frac{b-a}{n}\quad \text{for all}\quad 1 \leq k \leq n-1,\\ x_{n} &:= b. \end{align*} Then $E \subset \bigcup_{k=1}^{n}[x_{k-1},x_{k}]$; hence for all $x \in E$ there is some $1 \leq k \leq n$ such that $x \in [x_{k-1}, x_{k}]$ and $|f(x) - f(x_{k-1})| < 1$, implying that $$ |f(x)| < 1 + |f(x_{k-1})|; $$ so for all $x \in E$ we have $$ |f(x)| < \sup_{1\leq k \leq n}(1 + |f(x_{k-1})|). $$

$\endgroup$
1
$\begingroup$

We prove that for any $a\in\overline{E}\:$, $\lim\limits_{x\to a}f(x)$ exists.

By definition of uniformly continuous, for any $\epsilon>0$, there is a $\delta>0$ such that for any $x,y\in(a-\delta,a+\delta)$, it is always true that $|f(x)-f(y)|<\epsilon$. By Cauchy criterion, this means that $\lim\limits_{x\to a}f(x)$ exists.

So $f$ can be made continuous on $\overline{E}$. $\overline{E}$ is bounded for $E$ is bounded and any limit point of $E$ is arbitrarily closed to $E$. Since $\overline{E}$ is closed and bounded, it is compact. Since continuous function is bounded on compact set, $f$ is bounded on $\overline{E}$. Thus $f$ is bounded on $E$ for $E\subset\overline{E}$.

$\endgroup$
0
0
$\begingroup$

Proof by contradiction. Suppose $f$ is unbounded.For each $n\in N $ let $x_n\in E$ with $|f(x_n)|+1<|f(x_{n+1}|$.Since $E$ is bounded there is a subsequence $ (x_g(n))_{n\in N}$ which is convergent to a point $y$, where $g:N\to N$ is some strictly increasing function. It does not matter whether or not $y\in E$. What does matter is that every $x_{g(n)}$ belongs to $E$. Now there is no $d>0$ such that $\forall x,x'\in E (|x-x'|<d)\to (|f(x)-f(x')|<1 )$, contradicting the uniform continuity of $f$..... This is because $\forall d>0$ the set $A_d=\{n\in N :|y-x_{g(n)}\not <d/2\}$ is finite so there exist $n(1),n(2)\in N\backslash A_d$ with $n(1)\ne n(2)$.... So $|x_{g(n(1))}-x_{g(n(2))}|<d$ and $|f(x_{g(n(1))}-f(x_{g(n(2))}|>|n(1)-n(2)|\geq 1.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .