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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!

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    See here: http://math.stackexchange.com/questions/423492/overview-of-basic-facts-about-cauchy-functional-equation – 2017-02-20
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    I need to relate to the basic limit theorems @carmichael561 – 2017-02-20
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    As a hint what is the value of $f(0)$ – 2017-02-20
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    @kingW3: Do we really need to know the value of $f(0)$? – 2017-02-21
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    @ParamanandSingh Well we don't *really* need to know though it makes it easier,also replacing $f(x)+f(0)=f(x+0)$ is just another way of phrasing $f(0)=0$ though maybe its a bit more intuitive. – 2017-02-21

5 Answers 5

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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 $\lim\limits_{x\to x_0} f(x) = \lim\limits_{x\to 0} f(x_0-x) = f(x_0)+\lim\limits_{x\to 0}-f(x)=f(x_0)$.

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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? – 2017-02-21
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    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 – 2017-02-21
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    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! – 2017-02-21
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    Is it based on the domain being all reals? – 2017-02-21
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    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 $x\to 0$. – 2017-02-22
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    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. – 2017-02-22
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    Don't worry about sequences. They are a powerful tool but not needed here. – 2017-02-22
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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!

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    Thats not the case, this is a bonus challenge question. – 2017-02-20
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    Even though I do not like your assumption that I am lacking understanding, I appreciate your answer. Thank you – 2017-02-20
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    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. – 2017-02-20
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    Ya they were so much different and I'm doing hw early lol so I am kind of ahead no problems tho! :) – 2017-02-20
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    If you can check out the delta epsilon proof i posted! :) – 2017-02-22
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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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    Ok that's kinda where my brain was going thank you! – 2017-02-20
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    What about $\lim_{x\to -\infty}f(x_n)-f(x)$? – 2017-02-20
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    I think there is no need to find out $f(0)$ and $f(-x)$. See my answer. – 2017-02-21
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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.

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    Can you elaborate a little bit please on how this proves continuity for all real values? – 2017-02-21
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    @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. – 2017-02-21
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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! :) – 2017-02-21
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    @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. – 2017-02-21
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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. – 2017-02-21
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    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 – 2017-02-21
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    @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. – 2017-02-21
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53970/discussion-between-paramanand-singh-and-nick-pavini). – 2017-02-21
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    Check out my new proof I posted as an answer! I am very excited about it and it is very clear!:) – 2017-02-22
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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$(-\infty,\infty)!$

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    Good try and you succeeded! – 2017-02-22
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    @ParamanandSingh thank you! This is exactly what i needed to see! Now im going to try this with the problem you gave me:) – 2017-02-22
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    Looks good! I would edit the last equation line to avoid having that misleading equal sign. Otherwise it seems you understand what is going on. Good job! – 2017-02-22
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    Where I say it is $=f(a)$? @user394946 But i thought I proved that with my definition? Ill change it if need be lol just wondering:) – 2017-02-22
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    That one is fine. I am talking about the line above. The way it is written can lead one into thinking $\varepsilon = |f(x)-f(a)|$. You can simply remove the first "$<\varepsilon$" and leave the equal sign – 2017-02-22
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    Oh i see!@user394946 – 2017-02-22