# Proving that an additive function $f$ is continuous if it is continuous at a single point

Suppose that $f$ is continuous at $x_0$ and $f$ satisfies $f(x)+f(y)=f(x+y)$. Then how can we prove that $f$ is continuous at $x$ for all $x$? I seems to have problem doing anything with it. Thanks in advance.

• $f(x)-f(b)=f(x-b+a)-f(a)$. Commented Dec 24, 2011 at 4:26
• @Sabyasachi: I have edited your question. Check if this is what you would like to ask.
– Paul
Commented Dec 24, 2011 at 4:28
• Thank you.That's what I meant. Commented Dec 24, 2011 at 4:40
• In fact, you can do more. A solution $f$ to the Cauchy equation $f(x+y)=f(x)+f(y)$ is not of the form $f(x)=mx$ if and only if its graph $$\Gamma=\{(x,f(x)):x\in\mathbb{R}\}$$ is dense in $\mathbb{R}^{2}$. Commented Dec 24, 2011 at 15:16

Fix $a\in \mathbb{R}.$

Then

\begin{align*}\displaystyle\lim_{x \rightarrow a} f(x) &= \displaystyle\lim_{x \rightarrow x_0} f(x - x_0 + a)\\ &= \displaystyle\lim_{x \rightarrow x_0} [f(x) - f(x_0) + f(a)]\\& = (\displaystyle\lim_{x \rightarrow x_0} f(x)) - f(x_0) + f(a)\\ & = f(x_0) -f(x_0) + f(a)\\ & = f(a). \end{align*}

It follows $f$ is continuous at $a.$

• Doesn't this work in any topological group? Commented Dec 25, 2011 at 16:18

Let's examine your situation. You have that $\lim_{x\rightarrow a} f(x) = f(a)$ for some $a\in \mathbb{R}$ and that for any $x,y\in \mathbb{R}, f(x)+f(y)=f(x+y)$. You want to prove that for any $c\in \mathbb{R}, \lim_{x\rightarrow c} f(x) = f(c)$. The key step here is to realize that $$\lim_{x\rightarrow c} f(x) = \lim_{x\rightarrow a} f(x-a+c)$$ because $|(x-a+c)-c| = |x-a|$, so in plain english $x$ is close to $a$ if and only if $x-a+c$ is close to $c$. We can then complete the proof as follows: $$\lim_{x\rightarrow a} f(x-a+c) = \lim_{x\rightarrow a} (f(x) + f(c-a)) = f(a) + f(c-a) = f(c)$$

Suppose $x=x_0, f(x_0) + f(y) = f(x_0 + y)$ taking limit as $y$ tends to $0$, we get $\lim_{y\rightarrow 0} (f(x_0) + f(y)) = \lim_{y\rightarrow 0} f(x_0 + y)$ From the continuity at $x_0$, we know that RHS of the above equation is $f(x_0)$ which means that $\lim_{y\rightarrow 0} f(y) =0$

Next bit it simple. For any $x,y$ in the domain, $f(x+y)=f(x)+f(y)$ continuity can be established by checking whether $\lim_{y\rightarrow 0} f(x+y) =f(x)$ which is true since $f(x+y)=f(x)+f(y)$ and $\lim_{y\rightarrow 0}f(y)=0$. Hence $f(x)$ is continuous everywhere in the domain.

I'm new here so I don't know how to input equations using latex. Pls bear with it.

• To format equations in LaTeX, you simply place dollar signs around the expression. For example, \$f(y)\$ gives $f(y)$. Of course, there a more things you can do beyond this to get nicer format. I will edit your answer to put everything into LaTeX, and you can right-click on the expressions and select "view source" to see how to format them in LaTeX. Commented Dec 24, 2011 at 4:41

Preliminaries: $$f(0)=f(0+0)=f(0)+f(0)\implies 2f(0)=f(0)\implies$$ $$f(0)=0$$ $$f(0)=f(x-x)=f(x+(-x))=f(x)+f(-x)\implies f(x)+f(-x)=0\implies$$ $$f(-x)=-f(x)$$ $$\displaystyle{\lim_{x\to x_0}(f(x-x))}=\displaystyle{\lim_{x\to x_0}(f(x)-f(x))}=\displaystyle{\lim_{x\to x_0}(f(x))-\lim_{x\to x_0}(f(x))}=f(x_0)-f(x_0)=0$$ $$\mbox{By theorem: If }f~\mbox{and }g~\mbox{are continuous, then }f-g~\mbox{is continuous}.$$ $$\therefore f~\mbox{is continuous at }0$$

Proof: Choose any $$c\in \mathbb{R}$$ and let $$h=x-c$$. (as $$x$$ approaches $$c$$, $$h$$ approaches $$0$$) $$x\to c\iff h\to 0$$ \begin{align*} \lim_{x\to c}(f(x))&=\lim_{h\to 0}(f(h+c))~~~~~~~~~~~~~~(\mbox{Because }x=h+c)\\ &=\lim_{h\to 0}(f(h)+f(c))\\ &=\lim_{h\to 0}(f(h))+\lim_{h\to 0}(f(c))\\ &=f(0)+f(c)~~~~~~~~~~~~~~~~~~(f~\mbox{continuous at 0 and }f(c)~\mbox{constant})\\ &=0+f(c)~~~~~~~~~~~~~~~~~~~~~~~(f(0)=0)\\ &=f(c) \end{align*} Because $$\displaystyle{\lim_{x\to c}(f(x))=f(c)}$$ ($$f$$ continuous at c) and $$c$$ is arbitrary, then $$f$$ is continuous for all real numbers.