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There is an assertion that if f and g are both differentiable at x, then so is f + g at x. It is safe if $f'(x)$ and $g'(x)$ are both finite, I wonder if it still holds for infinite derivative. To be precise, suppose $f,g: \mathbb R\to\mathbb R, x\in \mathbb R, f'(x)=\lim\limits_{t\to x}\frac{f(t)-f(x)}{t-x}=+\infty, g'(x)=\lim\limits_{t\to x}\frac{g(t)-g(x)}{t-x}=-\infty$, f and g are both continuous at x, do we always have $f+g$ is still differentiable at x, that is, does the limit $\lim\limits_{t\to x}\frac{(f+g)(t)-(f+g)(x)}{t-x}$ exist in extended real number system? and, if it is, what's the value of $(f+g)'(x)$? is it $+\infty$ or $-\infty$ or a finite real number? if it isn't, could you please come up with an concrete counterexample? Thanks!

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If the limit is infinite, then $f$ is not differentiable at $x$. –  Arturo Magidin Apr 1 '11 at 16:18
    
I used the term "f is differentiable at x" just for convienience. I mean the limit exists in extended real number system. –  zzzhhh Apr 1 '11 at 16:23
    
@Arturo: I think it was intended using an "extended" definition of differentiability, i.e. if the limit of the difference quotient is $\infty$, then the derivative is $\infty$ and the function is differentiable in the "extended" sense. –  joriki Apr 1 '11 at 16:25
    
@zzzhhh: Right; the thing about "convenience" is that when one uses a term for convenience in a way that makes an expression false, then it is very in convenient. –  Arturo Magidin Apr 1 '11 at 16:27
    
@Arturo: Each of the "shortcut" expressions was followed by "To be precise, ..." or "that is, ...", followed by a well-defined "conventional" statement. I thought it was clear enough. –  joriki Apr 1 '11 at 16:29

2 Answers 2

up vote 8 down vote accepted

First, if the limit $$\lim_{t\to x}\frac{f(t)-f(x)}{t-x}$$ is $\infty$ or $-\infty$, then we usually say that $f$ is not differentiable at $x$. (We don't talk about "infinite derivative").

That said: you still have the very basic limit laws in the extended reals $$\lim_{t\to x}\frac{(f+g)(t) \pm (f+g)(x)}{x-t} = \lim_{t\to x}\frac{f(t)-f(x)}{t-x}\pm\lim_{t\to x}\frac{g(t)-g(x)}{t-x}$$ if both sides make sense in the extended reals. If you are taking $f+g$ and the limit for $f'$ and for $g'$ are $\infty$, then so is the limit for $f+g$; if the limit for $f$ is $\infty$, the limit for $g$ is $-\infty$, then the limit for $f-g$ is $\infty$, etc.

But you have problems if the operation in the extended reals is undefined, like $\infty-\infty$, which is precisely the example you are looking at. In that case, the limit for $f+g$ may be $\infty$, may be $-\infty$, may be any particular number you care to specify, or may fail to exist altogether.

In other words, this amounts to asking whether $$\lim_{x\to a}(f(x)+g(x)) = \lim_{x\to a}f(x) + \lim_{x\to a}g(x)$$ holds in the extended reals (provided the two limits on the right are defined). The answer is that it holds provided the expression on the right is defined in the extended reals: if it is of the form $\infty+\infty$, $-\infty-\infty$, $\infty+M$, $-\infty+M$, or $L+M$, with $L$ and $M$ (finite) real numbers. But if the expression on the right is $\infty-\infty$ or $-\infty+\infty$, then the equality does not make sense, because the operation on the right hand side is undefined.

For specific examples, take $f(x) = x^{1/3}+kx$ and $g(x)=-x^{1/3}$. Then $$ \begin{align*} \lim_{t\to 0}\frac{g(t)-g(0)}{t} &= \lim_{t\to 0}\frac{-t^{1/3}}{t}\\ &= -\lim_{t\to x}\frac{1}{t^{2/3}} = -\infty,\\ \lim_{t\to 0}\frac{f(t)-f(0)}{t} &= \lim_{t\to 0}\frac{t^{1/3}+kt}{t}\\ &=\lim_{t\to 0}t^{-2/3}+k = \infty,\\ \lim_{t\to 0}\frac{(f+g)(t) - (f+g)(0)}{t} &= \lim_{t\to 0}\frac{t^{1/3}+kt - t^{1/3}}{t}\\ &= \lim_{t\to 0}k = k. \end{align*} $$ So here we have an example where $(f+g)'(0)$ is finite and anything you want. To get an example where $\lim\limits_{t\to 0}\frac{(f+g)(t)-(f+g)(0)}{t} = \infty$, take $f(x) = g(x) = x^{1/3}$.

To get an example where $\lim\limits_{t\to 0}\frac{(f+g)(t)-(f+g)(0)}{t}=-\infty$, try $f(x) = x^{1/3}$ and $g(x) = -x^{1/9}$. Then $$\lim\limits_{t\to 0}\frac{t^{1/3}-t^{1/9}}{t} = \lim_{t\to 0}\left(\frac{1}{t^{2/3}} - \frac{1}{t^{8/9}}\right)$$ and since $t^{8/9}$ goes to $0$ faster than $t^{2/3}$, the second term will dominate and give you $-\infty$ in the limit.

For an example where the limit does not exist and the functions are continuous everywhere, take $f(x) = x^{1/3}$, and $$g(x) = \left\{\begin{array}{ll} -2x^{1/3} &\mbox{if $x\lt 0$,}\\ -x^{1/3}&\mbox{if $x\geq 0$.} \end{array}\right.$$ Then $g$ is continuous everywhere, and has $$\begin{align*} \lim_{t\to 0^+}\frac{g(t)-g(0)}{t} &= \lim_{t\to 0^+}\frac{-t^{1/3}}{t} = -\infty,\\ \lim_{t\to 0^-}\frac{g(t)-g(0)}{t} &=\lim_{t\to 0^-}\frac{-2t^{1/3}}{t} = -\infty. \end{align*}$$ On the other hand, we have $$(f+g)(x) = \left\{\begin{array}{ll} 0&\mbox{if $x\geq 0$,}\\ -x^{1/3} &\mbox{if $x\lt 0$} \end{array}\right.$$ so that $$\begin{align*} \lim_{t\to 0^+}\frac{(f+g)(t) - (f+g)(0)}{t} &= 0\\ \lim_{t\to 0^-}\frac{(f+g)(t) - (f+g)(0)}{t} &= \lim_{t\to 0^-}\frac{-t^{1/3}}{t} = -\infty \end{align*}$$ so the limit does not exist.

Changing $g$ so that it is $-\frac{1}{2}t^{1/3}$ on the nonnegative reals will give you that the difference quotient for $f+g$ has infinite limit when we approach $0$ from the right, and limit $-\infty$ when we approach from the left.

Note that these functions are continuous everywhere, and have derivatives defined everywhere outside of $0$.

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So you think the answer is "no", that is, the limit of the difference quotient for f+g does not always exist, right? Then we need to find a counterexample. Since the derivative does not has first-kind discontinuity, we need to use an oscillating function, right? But I only know sin(1/x)-type functions, what's its indefinite integral? –  zzzhhh Apr 1 '11 at 16:49
    
@zzzhhh: See the edit to my answer. –  joriki Apr 1 '11 at 16:50
    
@zzzhhh: I just posted example where you get any limit you want (with continuous functions), including $\infty$. I'll post examples where you get $-\infty$, and try to find one where you don't have a limit. –  Arturo Magidin Apr 1 '11 at 16:50
    
@Arturo: I believe in my second example (which is quite similar to your example) there's no limit. (Of course I should have used $x^{1/3}$ like you did. :-) –  joriki Apr 1 '11 at 16:52
    
@Arturo: I think there's a mix up in your later examples. $f(x)=g(x)=x^{1/3}$ isn't an example of getting anything you want, since the limits of both are $+\infty$, so the limit of the sum has to be $+\infty$. And for $x^{1/3}$ and $x^{1/9}$, you're adding them in the first limit, which would give $+\infty$, not $-\infty$, but then you subtract them in the second limit. I guess you mean $f(x)=x^{1/3}$ and $g(x)=-x^{1/9}$? –  joriki Apr 1 '11 at 17:08

Consider $f(x)=h(x)+a(x)$ and $g(x)=-h(x)$, where $h(x)$ is some jump function with infinite derivative (say, $-1$ for $x<0$, $1$ for $x>0$ and $0$ for $x=0$) and $a(x)$ is some non-differentiable function with variation less than the jump in $a$ (say, $1/2$ times the characteristic function of the rationals). Then $f$ and $g$ fulfill your premises, but $f+g$ is not differentiable, not even continuous.

Edit in response to the comment:

Sorry, I overlooked that requirement. Then would $h(x)=\frac{x}{\lvert x \rvert}\sqrt{\lvert x \rvert}$ and $a(x) = x \chi_\mathbb{Q}(x)$ do?

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Thank you but, I prescribed that both f and g are continuous at x, so this counterexample may not meet the requirement. –  zzzhhh Apr 1 '11 at 16:35
    
Wow, $x\chi_{\mathbb Q}(x)$ is continuous but has NO derivative at 0, It's cool! –  zzzhhh Apr 1 '11 at 17:14
    
I'm glad you like it -- but I realized later that a more mundane example would have been $h(x)=x^{1/3}$ and $a(x)=\lvert x \rvert$. –  joriki Apr 1 '11 at 17:19

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