Let $f,g:\Bbb R\to\Bbb R$ be real variable continuous functions and define $h:=f+g$; suppose $h$ non-constant.

If $h$ is differentiable at some $x_0\in\Bbb R$, does the same holds for both $f$ and $g$? Or there exist a couple of functions $f,g$ not differentiable at $x_0$ such that their sum does?

Maybe is trivial but I can't find counterexamples.

In what case is this true?

  • 2
    $\begingroup$ What if you take $g$ to be $-f$? $\endgroup$ – Lord Shark the Unknown Oct 4 '17 at 19:25
  • 7
    $\begingroup$ $g = -f$ where $f$ is whatever mess you want. $\endgroup$ – user296602 Oct 4 '17 at 19:25
  • $\begingroup$ Ok, that is trivial; any other counterexample? $\endgroup$ – Joe Oct 4 '17 at 19:27
  • 5
    $\begingroup$ Let $h$ be a differentiable function, $f$ be literally any non-differentiable function, and $g := h-f$. $\endgroup$ – Dustan Levenstein Oct 4 '17 at 19:28

As others have pointed out, if you allow $f$ and $g$ to be any continuous functions, then knowing that $f+g$ is differentiable will tell you nothing about the differentiability of $f$ and $g$.

If you know that $f + g$ is differentiable and you assume that $f$ is also differentiable while making no assumptions at all on $g$, then $g$ will also be differentiable. (because $g= (f+g) - f$ and differences of diffentiable functions are differentiable.)


If you know what a group and a subgroup are, think that the set of differentiable functions is a proper subgroup of the group of the continuous functions.

Similar examples:

If $ab>0$ are $a$ and $b$ positive?
If $a+b$ is a multiple of $3$, then both $a$ and $b$ are?
If $x,y$ are real and $x+y$ are rational, then are $x$ and $y$ rational?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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