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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.

Spaced
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2 Answers2

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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...

Ivo Terek
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  • That is nice and quick. Mind if I ask, is there nothing true or of value in my elaboration in the post? – Spaced Oct 28 '14 at 20:07
  • 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 (: – Ivo Terek Oct 28 '14 at 20:13
  • 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. – Spaced Oct 28 '14 at 20:16
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    You might find my answer here helpful.. – Ivo Terek Oct 28 '14 at 20:23
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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.

davidlowryduda
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