Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a uniformly continuous function values in $f: \mathbb{R}{\to}\mathbb{R}$ such that its first derivative is not bounded and is defined on a non-compact set?
And if $f: X_{1}{\to}\mathbb{R}$? (let $(X_{1},d{1})$ metric space ).
And if $f: X_{1}{\to}X_{2}$? (let $(X_{2},d{2})$ metric space ).

I know that the first derivative is bounded f:I-->R (let I interval) iff f is a lipschitz function; If X1 is a compact thanks to the Heine-Cantor theorem f is uniformly continuous.

share|improve this question
    
not limited = not bounded? –  Hagen von Eitzen Apr 5 '13 at 16:56
    
yes, I m sorry .-) –  Agenog Apr 5 '13 at 17:04
add comment

2 Answers 2

up vote 5 down vote accepted

The function $f(x)=\frac{\sin(e^x)}{1+x^2}$ has derivative $\frac{e^x\cos(e^x)(1+x^2)-2x\sin(e^x)}{(1+x^2)^2}$, which is clearly unbounded. Yet, $f$ is uniformly continuous as for $f(x)-f(y)$ to be $\ge \epsilon$, we clearly need at least one of $|f(x)|, |f(y)|$ to be $\ge \frac\epsilon2$, hence $|x|$ or $|y|$ must be $\le\sqrt{ \frac2\epsilon-1}$, so we may essentially consider $f$ on a compact interval.

EDIT: To elaborate, I'll give an explicit proof of uniform continuity - the above short argument seems to have been ambiguous.

Let $\epsilon>0$ be given. We are looking for $\delta>0$ such that for all $x,y$ with $|x-y|<\delta$ we have $|f(x)-f(y)|<\epsilon$. Let $\epsilon_1=\min\{\epsilon,1\}$ and $L=1+\sqrt{\frac2{\epsilon_1}-1}$. Then $L\ge2$. The restriction $g\colon [-L,L]\to \mathbb R$, $x\mapsto f(x)$ of $f$ to the compact interval $[-L,L]$ is continuous and hence uniformly continuous. Therefore, there exists $\delta_0>0$ such that for all $x,y\in[-L,L]$ with $|x-y|<\delta_0$, we have $|g(x)-g(y)|<\epsilon$. Let $\delta=\min\{\delta_0,1\}$. Now if $x,y\in\mathbb R$ with $|x-y|<\delta$, I claim that $|f(x)-f(y)|<\epsilon$. Indeed, if both $x,y$ are $\in[-L,L]$, then $|f(x)-f(y)|=|g(x)-g(y)|<\epsilon$ because $|x-y|<\delta\le\delta_0$. If on the other hand one of $x,y$ has absolute value $>L$, then the other has absolute value $>L-\delta\ge L-1$, hence both $|x|,|y|$ are $>L-1=\sqrt{\frac2{\epsilon_1}-1}$. This implies $x^2+1>\frac2{\epsilon_1}$, hence $|f(x)|<\frac{\epsilon_1}2$ because the sine is bounded by $1$. Since similarly $|f(y)|<\frac{\epsilon_1}2$, we find $|f(x)-f(y)|<|f(x)|+|f(y)|<\epsilon_1\le\epsilon$. Thus in all possible cases $|f(x)-f(y)|<\epsilon$, as was to be shown.$_\square$

Finally, an explicit argument why $\frac{e^x\cos(e^x)(1+x^2)-2x\sin(e^x)}{(1+x^2)^2}$ is unbounded. The inequality $e^t\ge1+t$ for all $t\in\mathbb R$ should be well-known. It implies $e^t=(e^{t/3})^3\ge(1+\frac t3)^3$ for all $t\ge-3$. Therefore, if $x\ge-3$ then
$$ \frac{e^x}{1+x^2}\ge \frac{1+x+\frac13x^2+\frac1{27}x^3}{1+x^2}=\frac{x}27+\frac13+\frac{26x+18}{27(x^2+1)}.$$ The last summand is bounded,hence the total expression is $$ \frac{e^x}{1+x^2}\ge\frac{x}{27}+C$$ for some constant $C$. Given $M\in\mathbb R$ (where without loss of generality, $M>C$), select an integer $k>\frac{e^{27(M-C)}}{2\pi}>0$ and let $x=\ln(2\pi k)\ge\ln(2\pi)>0$. Then $$f'(x)=\frac{e^x\cos(e^x)(1+x^2)-2x\sin(e^x)}{(1+x^2)^2} = \frac{e^x\cos(2k\pi)(1+x^2)-2x\sin(2k\pi)}{(1+x^2)^2} =\frac{e^x}{1+x^2} \ge \frac x{27}+C.$$ But $x=\ln(2\pi k)>27(M-C)$ then implies $$ f'(x)>M.$$ Similarly, by letting $x=\ln((2k+1)\pi)$ we find $x$ with $f'(x)<-M$. Thus $f'$ is neither bounded from above nor from below.$_\square$

share|improve this answer
    
So f(x) is not uniformly continuous (f:R->R), only if f is define on a compact interval. f(x) not satisfies the requests –  Agenog Apr 5 '13 at 17:21
    
@Giacomo Maybe you misunderstood: $f$ is uniformly continuous because everything outside a compact interval (depending on $\epsilon$) is small. –  Hagen von Eitzen Apr 6 '13 at 13:00
    
yes, now everything of your proof is clear. Thank you very much :-) –  Agenog Apr 6 '13 at 14:14
add comment

There is a simpler function than $g(x)= \frac{sin(e^{x})}{1+x^2}$ uniformly continuous such that its first derivative is not bounded and is defined on a non-compact set.

The function $f:R^{+}\to R^{+}$ $f(x):=\sqrt{x}$ is a 0.5-holder function and then $f(x)$ is uniformly continuous. $f^{'}(x)= \frac{1}{2\sqrt{x}}$ which is unbounded in $\mathbb{R^{+}}$. Obviously the set $\mathbb{R^{+}}$ is non-compact with euclidean metric because is unbounded.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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