Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $f(x) = g(x) + h(x)$ where $g(x)$ is known to be non-differentiable. When $h(x) \ne -g(x)$ is some other function (differentiable or not), can $f(x)$ ever be differentiable? We can assume $f, g, h$ are real valued functions.

edit: I should have specified that $h(x)$ is not some function that cancels out $g(x)$ . Bill Dubuque's answer is intriguing.

share|cite|improve this question
What does "some function that cancels out $g(x)$" mean? – Chris Eagle May 2 '11 at 21:52
The difficulty is that there is no rigorous notion of what it means to say "$h(x)$ is not some function that cancels out $g(x)$". Any $h(x)$ is a function that "cancels out $g(x)$", because we have that $h(x)=f(x)-g(x)$. – Zev Chonoles May 2 '11 at 21:55

Easy counterexamples have been mentioned. What is true, however, is the fact that sum of a differentiable function and a nondifferentiable function is nondifferentiable. This is true because differentiable functions are closed under subtraction, i.e. they comprise a subgroup of all functions. Hence the claimed property is simply a special case of the following complementary form of the subgroup property from my prior post.

THEOREM $\ $ A nonempty subset $\rm\:S\:$ of abelian group $\rm\:G\:$ comprises a subgroup $\rm\iff\ S\ + \ \bar S\ =\ \bar S\ $ where $\rm\: \bar S\:$ is the complement of $\rm\:S\:$ in $\rm\:G$

Instances of this are ubiquitous in concrete number systems, e.g.

     algebraic * nonalgebraic  =  nonalgebraic  if  nonzero 
      rational * irrrational   =   irrational   if  nonzero 
          real *   nonreal     =    nonreal     if  nonzero 

         even  +     odd       =      odd          additive example
       integer + noninteger    =   noninteger
share|cite|improve this answer
For the theorem to work the space of functions must form an abelian group, and the space of differentiable functions forms a subgroup of this abelian group, right? Are these standard ideas? – ijems May 2 '11 at 21:51
Yes, both your suppositions are correct. – Bill Dubuque May 2 '11 at 23:35
This is a nice answer. Are you sure you mean "sum" in the beginning? – Glen Wheeler May 5 '11 at 16:38
This is nice. In the theorem, "comprises" is ambiguous, since it can mean either "forms" or "contains"; I think "forms" would be clearer. – joriki May 28 '11 at 23:47

Let $h(x)=-g(x)+k(x)$ where $k\neq0$ is any differentiable function other than the constant 0 function. Then $f(x)=g(x)-g(x)+k(x)=k(x)$ is differentiable.

share|cite|improve this answer

Yes, of course. For example, if $h(x)=1-g(x)$, then $f(x)=1$ is differentiable.

share|cite|improve this answer

And to answer the other side of the question (is the function of a continuous function and a discontinuous function always discontinuous?), suppose that $f(x) = g(x)+h(x)$ is differentiable as before, with $g(x)$ non-differentiable and $h(x)$ differentiable. Then $g(x) = f(x) - h(x)$ is the difference of differentiable functions, and it's a pretty quick (and standard) exercise in deltas and epsilons to show that this must be differentiable itself.

share|cite|improve this answer

Your Answer


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.