# Real analysis Help

1. Nov 29, 2013

### tiger2030

1. The problem statement, all variables and given/known data

Let f : (a, b) → R be a continuous function on (a, b) such that |f'(x)| <= 1 for all x that are elements of (a,b). Prove that f is uniformly continuous function on (a,b).

2. Relevant equations

3. The attempt at a solution

Proof:For the sequence {xn}, where the limit(n→∞) xn = xo, then limit(n→∞) |((f(xn)-f(xo))/(xn-xo))| <= 1. This can be rewritten as limit(n→∞) |(f(xn)-f(xo)) <= limit(n→∞) |xn-xo|. For uniformly continuous, for all ε>0 and x,y that are elements of (a,b), there exists a δ such that |x-y|<δ implies |f(x)-f(y)|<ε. Let x=Xn and y=Xo. Then if ε=δ, then limit(n→∞) |(f(xn)-f(xo)) <= limit(n→∞) |xn-xo|<δ=ε and therefore f is uniformly continuous on (a,b)

Any help on where to improve this or if there are any steps that are incorrect would be greatly appreciated.

2. Nov 29, 2013

### LCKurtz

My suggestion would be to abandon that approach and think about using the mean value theorem.

3. Nov 29, 2013

### pasmith

So $f$ is actually a differentiable function (which is sufficient, but not necessary, for $f$ to be continuous).

This just says that $0 \leq 0$, which is true but not useful.

You can say that for all $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that if $n \geq N$ then
$$-1 - \epsilon \leq f'(x_0) - \epsilon < \frac{f(x_n) - f(x_0)}{x_n - x_0} < f'(x_0) + \epsilon \leq 1 + \epsilon$$
so that
$$|f(x_n) - f(x_0)| < (1 + \epsilon)|x_n - x_0|$$
which doesn't get you there; to take $\epsilon \to 0$ you must also take $N \to \infty$ which again gives you $0 \leq 0$.

It is best to use the mean value theorem: Let $x \in (a,b)$ and $y \in (a,b)$ be arbitrary such that $|x - y| < \epsilon$.
If $x < y$ then $f$ is continuous on $[x,y]$ and differentiable on $(x,y)$. Thus by the MVT there exists $c \in (x,y)$ such that
$$f'(c) = \frac{f(y) - f(x)}{y - x}.$$

Last edited: Nov 29, 2013
4. Nov 29, 2013

### tiger2030

Ok so I see that I can get f(y)-f(x)=f'(c)(y-x) and then I know from the assumptions that |f'(c)|≤1. From a previous exercise I know that any contraction is uniformly continuous on the domain. (The exercise says: f is said to be a contraction if there exists a c such that 0<c<1 such that for all x,y that are elements of the domain, we have |f(x)-f(y)|≤c|x-y|). Basically, if I can get from f(y)-f(x)=f'(c)(y-x) to |f(x)-f(y)|≤c|x-y| I will be able to say that it is uniformly continuous. Is it enough to use triangle inequality to say these are equivalent?

5. Nov 29, 2013

### LCKurtz

You don't need that contraction mapping theorem. You have$$|f(y) - f(x)|= |f'(c)||y-x| \le 1\cdot |y-x|$$So given $\epsilon > 0$ pick $\delta$...

6. Nov 29, 2013

### tiger2030

So if we pick δ=ε, then |f(y)−f(x)|=|f′(c)||y−x|≤1⋅|y−x|≤δ=ε and therefore |x-y|≤δ implies |f(y)−f(x)|≤ε, making it uniformly continuous