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there are a,b>=0, such that |f(x)|<= ax+b

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hey!!! (Blush)

I have a question..I am given the following exercise:
Let $f:[0,+\infty) \to \mathbb{R}$ uniformly continuous at $[0,+\infty)$.Prove that there are $a,b \geq 0$ such that $|f(x)| \leq ax+b , \forall x \geq 0$.
That's what I did so far:
$f:[0,+\infty) \to \mathbb{R}$ is uniformly continuous at $[0,+\infty)$.
So: $\forall \epsilon>0 \exists \delta>0 $ such that $\forall x,y \in [0,+\infty)$ with $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon$..
But,how can I continue to show that there are $a,b \geq 0$ such that $|f(x)| \leq ax+b , \forall x \geq 0$?? (Thinking)
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
Let $b=|f(0)|$. Suppose that for $\epsilon=1$ the corresponding $\delta$ is $0.1$. This means that from $0$ to $0.1$ the function $|f(x)|$ can grow at most by $1$. By the next $0.1$, $|f(x)|$ can grow at most by another $1$, and so on. Can you find a suitable $a$ in this case? Drawing a sketch may help.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hey!!! (Blush)

I have a question..I am given the following exercise:
Let $f:[0,+\infty) \to \mathbb{R}$ uniformly continuous at $[0,+\infty)$.Prove that there are $a,b \geq 0$ such that $|f(x)| \leq ax+b , \forall x \geq 0$.
That's what I did so far:
$f:[0,+\infty) \to \mathbb{R}$ is uniformly continuous at $[0,+\infty)$.
So: $\forall \epsilon>0 \exists \delta>0 $ such that $\forall x,y \in [0,+\infty)$ with $|x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon$..
But,how can I continue to show that there are $a,b \geq 0$ such that $|f(x)| \leq ax+b , \forall x \geq 0$?? (Thinking)
If f(*) is uniformly continous in $[a, + \infty)$ , then its derivative is bounded in the same inteval, i.e. it exists a constant M > 0 such that for any x in $[a, + \infty)$ is $\displaystyle |f^{\ '} (x)| > M$. In this case You can choose two constant a and b with b > M such that is $\displaystyle |a + b\ x|> |f(x)|$ for any x in $[a, + \infty)$...

Kind regards

$\chi$ $\sigma$
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
If f(*) is uniformly continous in $[a, + \infty)$ , then its derivative is bounded in the same inteval
A function may be uniformly continuous without being differentiable: see StackExchange.
 

Krizalid

Active member
Feb 9, 2012
118
Actually what follows, it's the converse:

Let $f:A\subset\mathbb R\to\mathbb R$ be with bounded derivative then $f$ is Lipschitz on $A,$ hence, uniformly continuous.
 

Krizalid

Active member
Feb 9, 2012
118
Now, more generally:

Let $f:\mathbb R\to\mathbb R$ be a uniform continuous function, then exist $a,b>0$ such that $|f(x)|\le a|x|+b.$

Let $\delta > 0$ be such that $|x-y|\le \delta \implies |f(x)-f(y)| < 1$.

Claim: $|f(x)| \le \frac{2}{\delta}|x| +|f(0)|$ for all $x$.

Proof: WLOG, $x>0,$ so $x\in [n \delta, (n+1) \delta)$ for some $n\in \{0,1,\dots\}$. Write $f(x)$ as a telescoping sum using multiples of $\delta,$ then slap on absolute values. We get \[|f(x)| \le |f(x) -f(n \delta)| +|f(n \delta)-f((n-1) \delta)| + \cdots + |f( \delta)-f(0)|+ |f(0)|\]
So this implies that $|f(x)|\le 1+n + |f(0)|.$ If $n>0,$ the latter expresion can be rewritten as $\frac{(n+1)}{n\delta}n\delta + |f(0)| \le \frac{2}{\delta}|x| + |f(0)|.$ If $n=0,$ the claim holds.

So the case here, it's actually $|f(x)|\le ax+b$ since $x\in[0,\infty).$
 

chisigma

Well-known member
Feb 13, 2012
1,704
I'm not a mathematician and one consequence of that is that I don't consider any type of 'pathological function', the fate of which should be to be closed for life in a 'hospital for incurables' with benefit for all us! :cool:...

Hoping to be absolved for that, since there are uniformly continous functions 'nowhere differentiable', lets try to find a more general solution of thye proposed question. It is not limitative to suppose f(*) a positive non decreasing uniformly continous function in $[0, + \infty)$ and we indicate with $\Phi(*)$ the set of such functions. The following propositions are easy to be demonstrated...

a) if $f(*) \in \Phi(*)$ and a>0 then $a\ f(*) \in \Phi(*)$...

b) if $f(*) \in \Phi(*)$ and $g(*) \in \Phi(*)$ then $f(*) + g(*) \in \Phi(*)$...

A little more complex but feasible is to demonstrate the following theorem...

c) the function $f(x) = x^{a}$ is uniformly continous in $[0, + \infty)$ if and only if $0 \le a \le 1$

Consequence of a), b) and c) is that, given a>0 and b>0, $g(x) = a x + b \in \Phi(*)$ and because value of a and b for which $g(*) = f(*) + h(*)$ being $h(*) \in \Phi(*)$ exist, the OP has been demonstrated. If necessary the demonstration of c) will be supplied...

Kind regards

$\chi$ $\sigma$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
It is not limitative to suppose f(*) a positive non decreasing uniformly continous function in $[0, + \infty)$ and we indicate with $\Phi(*)$ the set of such functions. The following propositions are easy to be demonstrated...

a) if $f(*) \in \Phi(*)$ and a>0 then $a\ f(*) \in \Phi(*)$...

b) if $f(*) \in \Phi(*)$ and $g(*) \in \Phi(*)$ then $f(*) + g(*) \in \Phi(*)$...

A little more complex but feasible is to demonstrate the following theorem...

c) the function $f(x) = x^{a}$ is uniformly continous in $[0, + \infty)$ if and only if $0 \le a \le 1$
Unfortunately the function $f(x) = x^\alpha$ is not uniformly continuous on $[0,\infty)$ for $\alpha<1$. In fact, $f'(x) = \alpha x^{\alpha-1}$, which becomes unbounded as $x\to0$. A function with an unbounded derivative can never be uniformly continuous. So the only value of $\alpha$ for which the function is uniformly continuous on $[0,\infty)$ is $\alpha=1.$

Examples of 'non-pathological' functions in the set $\Phi(*)$ include $(x+1)^\alpha$ (for $0\leqslant \alpha\leqslant 1$), $\arctan x$ and $\tanh x$.

*Edit* See retraction below!! I was confusing uniform continuity with Lipschitz continuity when I wrote the above nonsense.
 
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chisigma

Well-known member
Feb 13, 2012
1,704
Unfortunately the function $f(x) = x^\alpha$ is not uniformly continuous on $[0,\infty)$ for $\alpha<1$. In fact, $f'(x) = \alpha x^{\alpha-1}$, which becomes unbounded as $x\to0$. A function with an unbounded derivative can never be uniformly continuous. So the only value of $\alpha$ for which the function is uniformly continuous on $[0,\infty)$ is $\alpha=1.$

Examples of 'non-pathological' functions in the set $\Phi(*)$ include $(x+1)^\alpha$ (for $0\leqslant \alpha\leqslant 1$), $\arctan x$ and $\tanh x$.
I have to confess to be a little embarassed at this point, because I remember that a 'classical' example of uniformly continous function on $[0, + \infty)$ is $f(x)= \sqrt{x}$. Among the [numerous] posts on Web about this argument there is the following...

Show that sqrt(x) is uniformly continuous on [0,infinity) - Math Help Forum

Kind regards

$\chi$ $\sigma$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,703
I have to confess to be a little embarassed at this point, because I remember that a 'classical' example of uniformly continous function on $[0, + \infty)$ is $f(x)= \sqrt{x}$. Among the [numerous] posts on Web about this argument there is the following...

Show that sqrt(x) is uniformly continuous on [0,infinity) - Math Help Forum

Kind regards

$\chi$ $\sigma$
Now it's my turn to be embarrassed, because of course $x^\alpha$ is uniformly continuous on $[0,\infty)$ when $0\leqslant \alpha \leqslant 1.$ I must be getting old. (Worried)
 

chisigma

Well-known member
Feb 13, 2012
1,704
Now it's my turn to be embarrassed, because of course $x^\alpha$ is uniformly continuous on $[0,\infty)$ when $0\leqslant \alpha \leqslant 1.$ I must be getting old. (Worried)
... not only You! (Worried)...

Kind regards

$\chi$ $\sigma$