If $f$ is uniformly continuous on $(a, b)$, then $f$ is bounded on $(a, b)$.

Question

So I know that since $f$ is uniformly continuous on $(a,b)$, then for every $\varepsilon > 0$, there exists $\delta > 0$ such that for all $x$ and $y$ in $(a,b)$, if $\,\lvert x - y\rvert < \delta$, then $\,\lvert\,f(x) - f(y)\rvert < \varepsilon$.

I also know that I need to show that there are some numbers $M$, $N$, so that $M ≤ f(x) ≤ N$, for all $x$ and $y$ in $(a,b)$.

It makes logical sense to me because it means that for every x and y that you pick, the function values can only jump by epsilon, so there would be no way for the function suddenly approach infinity at the endpoints. However, I don't know really how to make a rigorous argument. I was thinking that maybe I could do something by contradiction, like saying that $f(y)>N$, but $f(x)<N$, and then somehow showing that this jump is now greater than $\varepsilon$, but I'm not sure that would really work since we only have an arbitrary function with arbitrary bounds.

Answer 1

Indirect proof: Suppose $f$ is not bounded above; choose $x_0$; then there exists $x_1$ such that $f(x_0)+1< f(x_1)$ (otherwise $f(x_0)+1$ would be a bound); now you may continue: there exists $x_2$ such that $f(x_2)> f(x_1)+1$, and so on. By construction, for $i\ne j$, $\vert f(x_i)-f(x_j)\vert >1$. By assumption there exists $\delta>0$ such that $\vert x-y\vert<\delta$ implies $\vert f(x)-f(y)\vert < 1$. Now you just have to argue that among the elements of the sequence $(x_i)$ constructed at the beginning there are $x_i,x_j$, with $i\ne j$ such that $\vert x_i-x_j\vert<\delta$ (why?) which leads to a contradiction.

Answer 2

Hint: Choose any $\varepsilon>0$ and $\delta>0$ such that $|x-y|<\delta\rightarrow |f(x)-f(y)|<\varepsilon$. Now, choose some $x\in (a,b)$. It is clear from definition that, for any $y\in (x-\delta,x+\delta)$ the value $|f(y)-f(x)|<\varepsilon$. By repeating this process on $y$, we can get that any value $z\in (y-\delta,y+\delta)$ has $|f(z)-f(x)|<2\varepsilon$ and hence, any value in $z\in (x-2\delta,x+2\delta)$ has $|f(z)-f(x)|<2\varepsilon$.

Proceed onto intervals of the form $(x-k\delta,x+k\delta)$ until you cover $(a,b)$.

Answer 3

Let $x_n\to 1$, then $\{x_n\}_{n\in\mathbb N}$ is a Cauchy sequence, and so is $\{f(x_n)\}_{n\in\mathbb N}$, due to uniform continuity. This implies that $f$ extends continuously to $x=1$ and similarly to $x=0$. Hence, $f$ extends to a continuous function on a compact set $[0,1]$. Therefore, $f$ is bounded.

Note. In general, if $D$ is dense in $X$, and $f:D\to\mathbb R$ is uniformly continuous, then $f$ extends uniquely to a continuous function on $X$.

Answer 4

For $\epsilon=1$ there exists $\delta>0$ such that $|x-y|<\delta\implies |f(x)-f(y)|<1.$ Consider $n\in \mathbb{N}$ such that $1/n<\delta.$ Thus, $|x-y|\le 1/n\implies |f(x)-f(y)|<1.$

Now, any $x\in [a,b]$ belong to an interval of the form $\left[a+\frac{k(b-a)}{n},a+\frac{(k+1)(b-a)}{n}\right]$ for some $k\in\{0,\cdots,n\}.$ Thus:

$$|f(x)-f(a)|=\left|f(x)-f\left(a+\frac{k(b-a)}{n}\right)+f\left(a+\frac{k(b-a)}{n}\right)-f\left(a+\frac{(k-1)(b-a)}{n}\right)+\cdots +f\left(a+\frac{2(b-a)}{n}\right)-f\left(a+\frac{b-a}{n}\right)+f\left(a+\frac{b-a}{n}\right)-f(a)\right| $$

$$\le\left|f(x)-f\left(a+\frac{k(b-a)}{n}\right)\right| +\left|f\left(a+\frac{k(b-a)}{n}\right)-f\left(a+\frac{(k-1)(b-a)}{n}\right)\right|+\cdots +\left|f\left(a+\frac{2(b-a)}{n}\right)-f\left(a+\frac{b-a}{n}\right)\right|+\left|f\left(a+\frac{b-a}{n}\right)-f(a)\right|\le k\le n. $$

Thus, $|f(x)-f(a)|\le n$ from where $|f(x)|\le |f(a)|+n,\forall x\in [a,b].$

Answer 5

Let $\epsilon = 1$ in the definition of uniform continuity.

There is $\delta$ such as $$ |x-y| < \delta \implies |f(x) - f(y)|<1 $$

Now let $N\in \Bbb N$, $N>\frac{b-a}{\delta}$. Then for every $x\in (a,b)$ one can find $k$ points $x_k$ with $|x_{m+1} - x_m| < \delta$, $x_1 = x$, $x_k = \frac{a+b}2$ and $k\le N$.

Using the triangle inequality, $$ |f(x)|\le \left|f\left(\frac{a+b}2\right)\right| + \left|f\left(\frac{a+b}2 \right)- f(x) \right| \le \left|f\left(\frac{a+b}2\right)\right| + \sum_{m=1}^{k-1} \left| f\left(x_{m+1}\right)- f(x_m) \right| \\ \le \left|f\left(\frac{a+b}2\right)\right| + k \le \left|f\left(\frac{a+b}2\right)\right| + N $$ which is a universal bound.