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Suppose that the domain of $f(x)$ is $\mathbb{R}$ and is continuous at $0$. Then if $f(x_1+x_2)=f(x_1)+f(x_2)$ for all real values $x_1$ and $x_2$, show that $f(x)$ is continuous for all reals.

We know that, $$\lim_{x\to0}f(x)=f(0)$$

From that I wanted to do something like,

Let $x_1+x_2=0$ and show that $$\lim_{x\to0}f(x_1+x_2)=f(0)$$ Am I even going about this the right way? To show that $f(x)$ is continuous on $\mathbb{R}$, I also thought to maybe relate it to the Intermediate Value Theorem. I am stuck, and I also can not find a forsure answer on whether given a domain of all real values, does it mean that $f$ is continuous on $\mathbb{R}$? I am Lost! I can not use derivatives, or integration as I have not learned either. Only using the basic limit theorems of continuity or the the intermediate value theorem.

UPDATE: I found an $\epsilon-\delta$ proof! I posted it below as the answer!

Nick Pavini
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5 Answers5

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Consider the sequence $x_n \to x $ for any $x \in \mathbb{R}$. Then the sequence $(x_n-x) \to 0$. Now we use linearity: $$ |f(x_n)-f(x)|=|f((x_n-x)-0)|=|f(x_n-x)-f(0)| \to 0 $$ The intermediate value theorem only works if already know that your function is continuous. Consult your textbook/class if you have a basic lack of understanding, they can handle and explain your problems better and thats what they are there for!

F. Conrad
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  • Thats not the case, this is a bonus challenge question. – Nick Pavini Feb 20 '17 at 20:13
  • Even though I do not like your assumption that I am lacking understanding, I appreciate your answer. Thank you – Nick Pavini Feb 20 '17 at 22:07
  • I answered your other question earlier and then you said "I am Lost!" in your post. Overall, there is no shame in asking and consulting the help that is given you outside of MSE. I am sorry if you felt offended by the way my post was written. – F. Conrad Feb 20 '17 at 22:37
  • Ya they were so much different and I'm doing hw early lol so I am kind of ahead no problems tho! :) – Nick Pavini Feb 20 '17 at 22:43
  • If you can check out the delta epsilon proof i posted! :) – Nick Pavini Feb 22 '17 at 03:45
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My making $x_1=x_2 =0$ we get $f(0)=0$. Then $0=f(0)=f(x + (-x))=f(x) + f(-x)$, therefore $f(-x)=-f(x), \forall x$.

Now let $x$ arbitrary and $x_n \rightarrow x$. Then $f(x_n) - f(x) = f(x_n) + f(-x)=f(x_n -x)$. Because $x_n - x \rightarrow 0$ and $f$ continous at $0$ it follows $lim_{n \rightarrow \infty} f(x_n -x) = f(0) = 0$ therefore $lim_{n \rightarrow \infty} f(x_n) - f(x) = 0$

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A one line answer:

$$\lim_{h \to 0}f(x + h) = \lim_{h \to 0}f(x) + f(h) = f(x) + f(0) = f(x + 0) = f(x)$$

I looked at other answers and I could not understand the need for all the extra stuff done there. The problem is a relatively simple one which requires you to use just the definition of continuity and nothing more. There is no need to calculate $f(0)$ in particular.

Paramanand Singh
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  • Can you elaborate a little bit please on how this proves continuity for all real values? – Nick Pavini Feb 21 '17 at 03:50
  • @NickPavini: In my equation $x$ is any general real number and the equation proves that $f$ is continuous at $x$. Thus $f$ is continuous at all real numbers. – Paramanand Singh Feb 21 '17 at 03:52
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    Hmm ok I like it. I am going to read some more on some definitions of continuity to try to make sure I have a complete understanding and can reproduce with reason instead of just copy lol. Thank you! :) – Nick Pavini Feb 21 '17 at 03:54
  • @NickPavini: I think you are trying to go overboard. Understand that your problem is a very simple one (like say of 1 mark rather than 5 marks). The definition of continuity at a single point $a$ is $\lim_{x \to a}f(x) = f(a)$ or $\lim_{h \to 0}f(a + h) = f(a)$. That's all we need. – Paramanand Singh Feb 21 '17 at 03:56
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    @NickPavini: But I like your attitute of "reproduce with reason instead of just copy". This will help you a lot! +1 for your comment and your question. – Paramanand Singh Feb 21 '17 at 03:58
  • Im sure I am! Haha I just felt as though i had to get back to $f(0)$ for some reason? Because that is the only point of continuity i know. When left with just $f(x)+f(0)$ in your example, how do I know $f(0)=0$? Sorry if im being difficult here – Nick Pavini Feb 21 '17 at 04:02
  • @NickPavini: Now i understand why everyone wants to know about $f(0)$. we actually don't need to know that because $f(x) + f(0) = f(x + 0) = f(x)$. I have used $x_1 = x, x_{2} = 0$ in the equation $f(x_1 + x_2) = f(x_1) + f(x_2)$. I also updated my answer to add this logic. – Paramanand Singh Feb 21 '17 at 04:06
  • Let us continue this discussion in chat. – Paramanand Singh Feb 21 '17 at 04:08
  • Check out my new proof I posted as an answer! I am very excited about it and it is very clear!:) – Nick Pavini Feb 22 '17 at 02:54
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$f(x)=f(x+0)=f(x)+f(0) \implies f(0)=0$ and

$f(0)=f(x-x)=f(x)+f(-x)=0 \implies f(-x)=-f(x)$ $\forall x$

$\epsilon-\delta$ Proof:

Let $\epsilon>0$ and choose $\delta>0$ such that, $$|t-0|<\delta \implies |f(t)-f(0)|<\epsilon$$ i.e $$|t|<\delta \implies |f(t)|<\epsilon$$ Let $a$ be any real number.

Assume $t=x-a$. Then, $$|x-a|<\delta \implies |f(x-a)|=|f(x)-f(a)|<\epsilon$$ Therefore, $\lim_{x\to a}f(x)=f(a)$ $\forall a\in \mathbb{R}$. i.e. $f$ is continuous on$(-\infty,\infty)!$

Nick Pavini
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First we have, for all $x$, $f(x)=f(0+x)=f(0)+f(x)$, which shows that $f(0)=0$.

Now, for all $x$, $f(0)=f(x-x)=f(x)+f(-x)$, so $f(-x)=-f(x)$.

Finally, for any $x_0$ in $\mathbb{R}$, we have $\lim\limits_{x\to x_0} f(x) = \lim\limits_{x\to 0} f(x_0-x) = f(x_0)+\lim\limits_{x\to 0}-f(x)=f(x_0)$.

user394946
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  • I know this is a late response but, when you say $$\lim_{x\to x_0}f(x)=\lim_{x\to0}f(x_0-x)$$ Is that a propert of limits? Can you please explain a little bit more to me? – Nick Pavini Feb 21 '17 at 15:45
  • Do you agree that when $x\to 0$ we have $x_0-x\to x_0$? So we are still passing to $f$ something that tends to $x_0$. Note that $x_0+x$ does the same--the sign only matters if you need to approach $x_0$ from a specific side. This is what Paramanand Singh used, I am just showing where it comes from. And indeed looking at $x_0-x\to x_0$ would have been more straight forward. Please don't hesitate if still unclear, I can reformulate – user394946 Feb 21 '17 at 18:43
  • I agree with your statements, for some reason I am having trouble just trying to see exactly how it happens.:/ could you reformulate? I would appreciate it! – Nick Pavini Feb 21 '17 at 19:06
  • Is it based on the domain being all reals? – Nick Pavini Feb 21 '17 at 19:17
  • A slightly different way of looking at this: in the notation $f(x)$, we can replace $x$ with anything, it is just a notation. We can say $f(\Box)$ for example, as long as box is real we're fine. Now we want to know what happens when $\Box$ gets very close to a certain point $x_0$. In other words we are studying the limit of $f$ when $\Box$ tends to $x_0$. If we take $\Box$ to be $x$ then we are looking at $f$ when $x\to x_0$. If we take $\Box$ to be $x+x_0$, then we are looking at $f$ when $(x+x_0)\to x_0$. Now another way of writing $(x+x_0)\to x_0$, because $x_0$ doesn't move, is $x\to 0$. – user394946 Feb 22 '17 at 03:06
  • Tell me how you like my proof that I posted as an answer! I understand it way more in epsilon delta format.:) The sequences are hard because I havent learned those. – Nick Pavini Feb 22 '17 at 03:07
  • Don't worry about sequences. They are a powerful tool but not needed here. – user394946 Feb 22 '17 at 03:37