# Proving $f(x)=\frac{x^2}{1+x}$ is uniformly continuous on $[0,\infty)$

I have a homework assignment to complete and I am having trouble proving that $f(x)=\frac{x^2}{1+x}$ is uniformly continuous on $[0,\infty)$.

For some reason I can't find the way to solve this one...

EDIT: I made a mistake. I meant to write $$f(x)=\frac{x^2}{1+x}$$ rather than $f(x) = \frac{x}{1+x}$.

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$f'(x)=1/(1+x)^2$ is bounded... –  yoyo Dec 15 '11 at 19:38
Use two facts: $f$ has a limit when $x\to\infty$ and an continuous function on a compact is uniformly continuous. –  Davide Giraudo Dec 15 '11 at 19:38
@DavideGiraudo ah.. Ok thanks I think I see a way to do it now... –  Jason Dec 15 '11 at 19:40

We can write, since $x,y\geq 0$ $$|f(x)-f(y)|=\left|1-\frac 1{1+x}-\left(1-\frac 1{1+y}\right)\right|=\frac{|1+y-(1-x)|}{(1+x)(1+y)}=\frac{|x-y|}{(1+x)(1+y)}\leq |x-y|$$ so $f$ is $1$-Lipschitz, and uniformly continuous (take $\delta:=\varepsilon$ in the definition).

For $f(x)=\frac{x^2}{1+x}$, write $$f(x)=x\frac{x+1-1}{1+x}=x-\frac x{1+x}=x-\frac{x+1-1}{1+x}=x-1+\frac 1{1+x},$$ so $$|f(x)-f(y)|=\left|x-1+\frac 1{1+x}-\left(y-1+\frac 1{1+y}\right)\right|\leq 2|x-y|$$ (so this time you have to take $\delta:=\frac{\varepsilon}2$).

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Thanks for the answere :) Can you explain the left equality - I don't see it :S –  Jason Dec 15 '11 at 20:04
I only use the fact that $(1+x)(1+y)\geq 1$. –  Davide Giraudo Dec 15 '11 at 20:05
Yea the right part of the equation is understood I don't get the first step from the left.. I don't get how you get a 1 there instead of $x$ and $y$ –  Jason Dec 15 '11 at 20:23
I wrote $\frac x{x+1}=\frac{x+1-1}{x+1}=1-\frac 1{x+1}$. –  Davide Giraudo Dec 15 '11 at 20:25
Ah!!! darn it - I wrote the equation wrong! I meant to write $f(x)=\frac{x^2}{x+1}$ –  Jason Dec 15 '11 at 20:26

Any function which has bounded derivative on the underlying set is uniformly continuous on the underlying set. More generally, any function which is Hölder continuous on the underlying set is uniformly continuous on the underlying set. In your example, where $$f(x) = \frac{x}{1+x} = 1 - \frac1{1+x}$$ we have that $$f'(x) = \frac1{(1+x)^2}$$ On the set $[0, \infty)$, we have that $$\lvert f'(x)\rvert \leq 1.$$ Hence, $\displaystyle f(x) = \frac{x}{1+x}$ is uniformly continuous on $[0,\infty)$.

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This is before learning derivatives so we must prove it in a different way - but thanks for the answer –  Jason Dec 15 '11 at 19:46

Let $f(x):=\frac{x^2}{1+x}$ for $x\geq 0$.

Function $f$ is continuous in $[0,\infty[$ and its graph has an oblique asymptote when $x$ tends to $\infty$ (the asymptote being the straight line of equation $y=x-1$). Finally you can conclude using the following lemma:

Let $a\in \mathbb{R}$ and $f:[a,\infty[\to \mathbb{R}$ a continuous function.

If the graph of $f$ has a horizontal or an oblique asymptote when $x$ tends to $\infty$, then $f$ is uniformly continuous in $[a,\infty[$.

The lemma cited above is a corollary of the following theorem:

Let $a\in \mathbb{R}$ and $f,g:[a,\infty[\to \mathbb{R}$ be continuous functions such that: $$\tag{A} \lim_{x\to \infty} f(x)-g(x)=0\; .$$ Then $f$ is uniformly continuous in $[a,\infty[$ if and only if $g$ is uniformly continuous in $[a,\infty[$.

The proof is very simple.

Since the statement is symmetric w.r.t. $f$ and $g$, it suffices to prove only one implication; therefore we assume, e.g., that $g$ is u.c. in $[a,\infty[$ and prove that also $f$ is u.c. in $[a,\infty[$. Moreover, w.l.o.g., we assume also $a=0$.

Let $\varepsilon >0$ be fixed. By (A) we can find $r\geq 0$ such that: $$\tag{I} \forall x\geq r,\quad |f(x)-g(x)|\leq \frac{\varepsilon}{8}\; ;$$ since $g$ is u.c. in $[0,\infty[$ we can find $\delta_1>0$ such that: $$\tag{II} \forall x,y\geq 0,\quad |x-y|\leq \delta_1\ \Rightarrow\ |g(x)-g(y)|\leq \frac{\varepsilon}{4}\; ;$$ finally, by Heine-Cantor theorem, we can find $\delta_2>0$ such that: $$\tag{III} \forall 0\leq x,y\leq r,\quad |x-y|\leq \delta_2\ \Rightarrow\ |f(x)-f(y)|\leq \frac{\varepsilon}{2}\; .$$ Now, set $\delta:=\min \{\delta_1, \delta_2\}$ and take two arbitrary $x,y\geq 0$ s.t. $|x-y|\leq \delta$: then, due to the symmetry w.r.t. $x,y$, three cases arise:

1. $r\leq x,y$: in this case we find: $$|f(x)-f(y)|\leq |f(x)-g(x)|+|g(x)-g(y)|+|g(y)-f(y)|\; ,$$ hence $|f(x)-f(y)|\leq \frac{\varepsilon}{8}+\frac{\varepsilon}{4}+\frac{\varepsilon}{8}=\frac{\varepsilon}{2}\leq \varepsilon$ because of (I) & (II)

2. $0\leq x,y\leq r$: in such a case $|f(x)-f(y)|\leq \frac{\varepsilon}{2}\leq \varepsilon$ by (III);

3. $0\leq x\leq r < y$: in this case we have: $$|f(x)-f(y)|\leq |f(x)-f(r)| +|f(r)-f(y)|\; ,$$ thus $|f(x)-f(y)|\leq \frac{\varepsilon}{2} +\frac{\varepsilon}{2}=\varepsilon$ for (III) & case 1.

Therefore $|x-y|\leq \delta\ \Rightarrow\ |f(x)-f(y)|\leq \varepsilon$ for all $x,y\geq 0$ and $f$ is u.c. in $[0,\infty[$. $\square$

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Welcome back!${}{}$ –  The Chaz 2.0 Dec 16 '11 at 4:56
@TheChaz: Thank you! –  Pacciu Dec 17 '11 at 0:05

On $[0,\infty)$ the derivative of this function is non-negative but always less than $1$, so $|f(u)-f(v)| < |u-v|$ for all $u,v$ (you can use the mean value theorem to justify what comes after the word "so" above). Consequently you can just take $\delta=\varepsilon$.

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As others have noticed, $$f(x)=x-1+\frac 1{1+x}$$ Now this is a sum of the identity function, which certainly is uniformly continuous, with a bounded decreasing function (one that even goes to $0$ as $x\to\infty$), which also is uniformly continuous. Hence $f$ is uniformly continuous.

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