Prove uniform continuity of $\frac1{1 + e^x}$

Question

I tried applying the standard definition of uniform convergence

$$ \forall \varepsilon >0 ~\exists \delta > 0 ~\forall x,x_0 \in \mathbb{R}: |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon $$

Here's my attempt

Let $\varepsilon > 0 $ be fixed. $$ |f(x) - f(x_0)| = \left|\frac{1}{1+e^x} - \frac{1}{1 + e^{x_0}}\right| = \left| \frac{e^x - e^{x_0}}{(1 + e^x)(1 + e^{x_0})} \right| \leq \left|\frac{e^x - e^{x_0}}{e^x + e^{x_0}}\right| $$ How can I choose $\delta$ and obtain a factor $|x - x_0|$ ? I've tried using the series definition of $e^x$ but could progress from there, too.

So this is my work using all your help

Since by Lagrange's Mean Value Theorem $$ \exists c \in (x_0,x): |f(x)−f(x_0)|=|(x − x_0)~f′(c)| $$ and $$ |f′(c)|= \left|\frac{e^x}{(e^x + 1)^2}\right| \leq 1 ~∀c \in \mathbb{R} \supset (x_0, x) $$ by transitivity we have $$ \frac{|f(x)−f(x_0)|}{|x − x_0|} = |f'(c)| \leq 1 \iff |f(x)−f(x_0)| \leq |x − x_0| ~\forall x,x_0 \in \mathbb{R} $$ Now we can either argue that $f(x)$ is Lipschitz continuous with $L = 1$ and therefore uniformly continuous OR use the definition of uniform continuity:

Let $\varepsilon = \delta > 0$ be fixed. Then it follows for all $x,x_0 \in \mathbb{R}$ that $$ |x - x_0| < \delta \implies |f(x) - f(x_0)| \leq |x - x_0| < \delta = \varepsilon $$

Answer 1

Probably the easiest way is to look at the derivative of $f$ and use Lagrange's mean value theorem.

That theorem tells you that $|f(x_0)-f(x)| = |(x_0-x)f'(c)| = |x_0-x|\cdot |f'(c)|$ for some $c\in(x, x_0)$

If you can find some upper bound for $f'(c)$, you are done.

Answer 2

You can use the mean value theorem:

$$|f'(x)| = \left| \frac{e^x}{(1+e^x)^2} \right| \leq 1$$ so the function is $1$-Lipschitz and in particular uniformly continuous.

Answer 3

Continue the string of inequalities by factoring out the smaller of $e^x$ and $e^{x_0}$ to obtain

$$\left|e^x-e^{x_0}\over e^x+e^{x_0}\right|=\left|e^{|x-x_0|}-1\over e^{|x-x_0|}+1 \right|\lt\left|e^\delta-1\over e^0+1 \right|\lt e^\delta-1.$$

Uniform continuity (a single $\delta$ works for all choices of $x$ and $x_0$) now follows from simple continuity of $e^u$ at $u=0$.