# prove that $f(x)=\log(1+x^2)$ is Uniform continuous with $\epsilon ,\delta$ …

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?

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Do you know that every function which is differentiable and has bounded derivative is uniformly continuous? –  Fabian Jan 6 '13 at 21:58
No, unfortunately we haven't proved that yet. we have to try to prove it without it –  Dor Shalom Jan 6 '13 at 22:06

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.

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How did you get $\frac{2c}{1+c^2}$ ? –  GinKin Dec 28 '13 at 10:21
@GinKin: $$f'(x)=\frac{2x}{1+x^2}$$ –  mrf Dec 28 '13 at 10:44
Thanks I should have known that. About the AM-GM inequality, is this the right use of it ? $$\frac{2x}{1+x^2}\le\frac{2x}{x^2}\Rightarrow\frac{2x}{x^2}\le1\Rightarrow \frac{x+x}{2}\le\sqrt{x^2}\Rightarrow x=x$$ –  GinKin Dec 28 '13 at 11:00
You may use it, but directly $\frac{1+c^2}{2} \geq \sqrt{c^2}$. –  Galc127 Dec 28 '13 at 16:40
@mrf, don't we need to show that the function is differentiable in $(0,\infty)$? –  Galc127 Dec 28 '13 at 16:41

Hint: You can use that $|\log (1+ z)| \leq z$ valid for $z\geq 1$. Thus, for $x>y$ $$\left|\log \left(\frac{1+x^2}{1+y^2}\right)\right| \leq \frac{x^2-y^2}{1+y^2} \leq \frac{x+y}{1+y^2} (x-y)= \frac{2y +(x-y)}{1+y^2}(x-y) .$$

Moreover, for $x<y$ $$\left|\log \left(\frac{1+x^2}{1+y^2}\right) \right| = \left|\log \left(\frac{1+y^2}{1+x^2}\right) \right| \leq \frac{y^2-x^2}{1+x^2} \leq \frac{x+y}{1+x^2} (y-x) =\frac{2x+(y-x)}{1+x^2} (y-x).$$

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I dont understand this first transition:$$\left|\log \left(\frac{1+x^2}{1+y^2}\right)\right| \leq \frac{x^2-y^2}{1+y^2}$$ can you please elaborate? –  Dor Shalom Jan 6 '13 at 22:34
You apply the inequality $\log(1+z) \leq z$ with $z=(x^2-y^2)/(1+y^2)$ –  Fabian Jan 6 '13 at 23:25
Could you please explain also how you got rid of the 2y or 2x ? –  user1685224 Jun 4 '13 at 15:28
@user1685224: as in mrf's answer, you have to show that $|2c/(1+c^2)| \leq 1$. –  Fabian Jun 4 '13 at 19:48
but that means using the derivative isn't it? I thought we're not allowed to use it in this question. –  user1685224 Jun 4 '13 at 20:35