I have to prove that $f(x)=\log(1+x^2)$ is Uniform continuous in $[0,\infty)$ (with $\epsilon ,\delta$ formulas…)
I wrote the definition: (what I have to prove):
$\forall \epsilon>0 \quad \exists \delta>0 \quad s.t. \quad \forall x,y\in [0,\infty) : \quad |x-y|<\delta \quad \Rightarrow \quad |f(x)-f(y)|<\epsilon $
So I tried developing $|f(x)-f(y)| = |\log(1+x^2)-\log(1+y^2)| = |\log(\frac{1+x^2}{1+y^2})| $…
now this is where I try to make it bigger and simplify the expression so i can choose the right $\delta$ which depends on the $\epsilon$, and then say that if the simplified bigger expression is still smaller than $\epsilon$ then of course the original $|f(x)-f(y)|$ is smaller than $\epsilon$
but what can I do with this expression? any algebraic tricks?
Best Answer
By the mean value theorem, $$f(x)-f(y) = f'(c)(x-y) = \frac{2c}{1+c^2}(x-y),$$ for some $c$ between $x$ and $y$.
Show that $$ \left|\frac{2c}{1+c^2}\right| \le 1,$$ for example using the AM-GM inequality.