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 $$ |x - x_0| < \delta \implies |f(x) - f(x_0)| \leq |x - x_0| < \delta = \varepsilon $$


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.


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 $u=0$.

  • 1
    $\begingroup$ This is a nice example, especially for a course where uniform continuity comes before derivatives. $\endgroup$ – Lee Mosher Apr 9 '18 at 13:12

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.