Let $x_0 \in \mathbb{R}$ and $f,g: \mathbb{R} \to \mathbb{R}$ such that $f$ is discontinuous at $x_0$ but $g$ is continuous at $x_0$, then $f+g$ at $x_0$ is...

My approach: After some graphical attempts to construct such a pathological case as given above I claim that the described scenario isn't possible. That is $$f \text{ disc. at } x_0 \text{ and } g \text{ cont. at } x_0 \implies f+g \text{ disc. at } x_0 $$

Proof: I was trying to come up with a contradiction, because if they happen to work out I find these kind of proofs to be most satisfying.

Assume $f+g$ is continuous at $x_0$ that is $$\forall \epsilon > 0 ,\exists \delta_1 > 0 : \forall x \in \mathbb{R}, |x-x_0| < \delta_1 \implies |f(x)+g(x)-f(x_0)-g(x_0)| < \epsilon $$

Since this statement must be true for all $\epsilon > 0$ I could choose $\epsilon:=1$. Next I know that also $g$ is continuous at said point $x_0$, thus $$\forall \epsilon > 0, \exists \delta_2 > 0 : \forall x \in \mathbb{R}, |x-x_0| < \delta_2 \implies |g(x)-g(x_0)| < \epsilon $$

and I have that $f$ is NOT continuous at $x_0$ therefore: $$\exists \epsilon^* > 0, \forall \delta>0: \exists x \in \mathbb{R}, |x-x_0| < \delta \wedge |f(x)-f(x_0)| \geq \epsilon $$

My problem is to bring all these three statements nicely together. Mainly I get confused by the quantors. In order to match all three statements I assume I have to choose $\epsilon > 0$ such that the last statement is met, so say $\epsilon:= \epsilon^*>0$, so statement 1 and 2 are still true for said $\epsilon$

I also need to choose $\delta$. Lets say $\delta = \min{(\delta_1,\delta_2)}$ now I should be able to work with all three statements at the same time right? But then I stumble with my argumentation.

Let $|x-x_0|< \delta$, then we obtain $$ |f(x)+g(x)-f(x_0)-g(x_0)| \leq |f(x)-f(x_0)| + |g(x)-g(x_0)| < - \epsilon + \epsilon =0 $$ which means that $|f(x)+g(x)-f(x_0)-g(x_0)| <0$ and I hope that this is a contradiction because $|.| \geq 0$. Please comment (or answer) if I am right or absolutely on the wrong path. I'd also appreciate it if you criticize my formulation.


Well, if $f + g$ were continuous in $x_0$, then $g$ continuous at $x_0$ gives us $-g$ continuous at $x_0$. Hence, we would get $(f+g) - g = f $ continuous at $x_0$. But $f$ was discontinuous there, so...

  • $\begingroup$ That is nice and quick. Mind if I ask, is there nothing true or of value in my elaboration in the post? $\endgroup$
    – Spaced
    Oct 28 '14 at 20:07
  • $\begingroup$ Your last assertion is indeed a contradiction. Looking quickly, seems to me that all that I used in my argument (sum of continuous being continuous, etc), you justified using the definitions. It is good that you noticed the difference and wrote $\epsilon $ and $\epsilon^*$. Most people who are beginning in the subject don't get that they are not the same epsilons. Good studies (: $\endgroup$
    – Ivo Terek
    Oct 28 '14 at 20:13
  • $\begingroup$ Thanks a lot, I appreciate it. Your proof is much more elegant than mine, but I still would like to work with the $\epsilon, \delta$ criteria for a bit, for practice reasons. Hence I was curious if I did choose $\epsilon$ and especially $\delta$ correct. $\endgroup$
    – Spaced
    Oct 28 '14 at 20:16
  • 1
    $\begingroup$ You might find my answer here helpful.. $\endgroup$
    – Ivo Terek
    Oct 28 '14 at 20:23

Suppose that your claim is true. That is, suppose that $f$ is not continuous at $x_0$, but both $g$ and $(f + g)$ are. As sums of continuous functions are continuous, this would give that

$$(f + g) - g = f$$

is continuous at $x_0$. Contradiction.


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.