Real Analysis Help: Prove f Uniformly Continuous on (a,b)

[tiger2030]

Homework Statement

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).

Homework Equations

The Attempt at a Solution

Proof:For the sequence {x_n}, where the limit(n→∞) x_n = x_o, then limit(n→∞) |((f(x_n)-f(x_o))/(x_n-x_o))| <= 1. This can be rewritten as limit(n→∞) |(f(x_n)-f(x_o)) <= limit(n→∞) |x_n-x_o|. 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(x_n)-f(x_o)) <= limit(n→∞) |x_n-x_o|<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.

 

[LCKurtz]

Homework Statement

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).

Homework Equations

The Attempt at a Solution

Proof:For the sequence {x_n}, where the limit(n→∞) x_n = x_o, then limit(n→∞) |((f(x_n)-f(x_o))/(x_n-x_o))| <= 1. This can be rewritten as limit(n→∞) |(f(x_n)-f(x_o)) <= limit(n→∞) |x_n-x_o|. 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(x_n)-f(x_o)) <= limit(n→∞) |x_n-x_o|<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.

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

 

[pasmith]

Homework Statement

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).

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

Prove that f is uniformly continuous function on (a,b).

Homework Equations

The Attempt at a Solution

Proof:For the sequence {x_n}, where the limit(n→∞) x_n = x_o, then limit(n→∞) |((f(x_n)-f(x_o))/(x_n-x_o))| <= 1. This can be rewritten as limit(n→∞) |(f(x_n)-f(x_o)) <= limit(n→∞) |x_n-x_o|.

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
$$f'(c) = \frac{f(y) - f(x)}{y - x}.$$

 

[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?

 

[LCKurtz]

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?

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 $\delta$...

 

[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