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!
A question about infinite derivative
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0If the limit is infinite, then $f$ is not differentiable at $x$. – 2011-04-01
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0I used the term "f is differentiable at x" just for convienience. I mean the limit exists in extended real number system. – 2011-04-01
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0@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. – 2011-04-01
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0@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. – 2011-04-01
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0@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. – 2011-04-01
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0@joriki: Yes, it was clear enough; but it's still not a correct expression to use, hence the correction. It's not much of a shortcut if you must follow it every time with the long-way-around expression... (-; – 2011-04-01
2 Answers
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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0So 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? – 2011-04-01
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0@zzzhhh: See the edit to my answer. – 2011-04-01
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0@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. – 2011-04-01
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0@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. :-) – 2011-04-01
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0@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 $g(x)=-x^{1/9}$? – 2011-04-01
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0@joriki: Yes, your second example works. There was an errant $-$ when I took $f(x)=g(x)=x^{1/3}$; I *meant* to get a limit equal to $\infty$. The other was a lapse; thanks. – 2011-04-01
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0There is a new question I'm facing :-) : Both answers have solved my question, Arturo's counterexample shows that the limit can be anything in the extended real number system, while joriki gave an counterexample that this limit does not exist at all. I hope to adopt both answers but this website requires that I have to choose only one. Now I don't know how to choose; this is a serious problem. Maybe I have to throw a coin ... – 2011-04-01
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0@zzzhhh: My final example also shows you can have the limit not exist at all, though in my case the reason is because the limits from either side are different; I don't think I can get an example where the functions are "nice" everywhere **except** at $x$, be continuous at $x$, and for the limit to simply not exist from either side, though. – 2011-04-01
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0Thank you very much professor. This question itself is solved, but I now have a bigger one: Many theorems in calculus books are expressed in a shortcut form, e.g. "f is differentiable" as in my question or "... exists", some of them may not hold (like this question) if we extend the meaning to the extended real number system. (...to be continued) – 2011-04-01
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0Now I have a shade in my heart: those theorems I used safely once, may not hold now. Should I have to check them to see if it is still true if infinite derivative or infinite value are adopted before attempting to use them? I feel that I am living in a insecure world, this make me feel terrible. Do you have the same feeling? how do you tackle this problem? – 2011-04-01
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0To give an example of my above worry, there is a famous assertion that if f is differentiable at x, f is continuous at x. But a jump is a discontinuity having infinite derivative. If I encounters a problem in the exam asking me to prove a statement that are expressed in this shortcut and I need to use this assertion, what shall I do? – 2011-04-01
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0@zzzhhh: In the extended reals, *some* operations are undefined: $\infty-\infty$, $0\times\infty$, $\frac{0}{0}$, $\frac{\infty}{0}$, etc. When those "limit laws" you want to use result in expressions that *are* defined in the extended reals, you are okay to use them. But if they result in undefined expressions, then what you have are "indeterminate limits", and you cannot just use them. In summary: you have to think about what you are doing when you are doing something. – 2011-04-01
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0@zzzhhh: Again: that's why we **don't** call a function "differentiable" when the limit of the difference quotient is infinite. It is **not** differentiable in that situation. So, once again, I point out that your "short-cut" expression is a **bad** expression, because it leads to *false statements.* If you are in an exam that uses such bad nomenclature, then you'll have to consider separate cases, and avoid that professor in future semesters. – 2011-04-01
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0I see. Thank you very much. You are really a very nice professor, both outer and inner. ^_^ – 2011-04-01
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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0Thank you but, I prescribed that both f and g are continuous at x, so this counterexample may not meet the requirement. – 2011-04-01
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0Wow, $x\chi_{\mathbb Q}(x)$ is continuous but has NO derivative at 0, It's cool! – 2011-04-01
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0I'm glad you like it -- but I realized later that a more mundane example would have been $h(x)=x^{1/3}$ and $a(x)=\lvert x \rvert$. – 2011-04-01