Proving that $f(x+y) = f(x) + f(y)$ and $f$ being continuous at one point implies that $f$ is continuous on ${\bf R}$ [duplicate]

Suppose $f(x+y)=f(x) + f(y)$ for all $x,y \in \mathbb R$ and $f$ is continuous at a point $a \in \mathbb R$. Prove that $f$ is continuous at every $b \in \mathbb R$.

I know that in order to prove continuity we can use the definition that states that $\lim_{x\to b}f(x+y)=f(b)$ then the function is continuous however I do not know how to show that the limit will be $f(b)$ for the function. Thanks in Advance.

marked as duplicate by Jack, Cave Johnson, Xander Henderson, user99914, JMPOct 9 '17 at 4:09

• Pretty sure this is a duplicate – Jam Oct 7 '17 at 20:02

It follows that for any $h\in \mathbb{R}$, $f(a+h)-f(a)=f(b)-f(b-h)$, because $f(b)+f(a)=f(a+b)=f(a+h)+f(b-h)$. Then we have that $f(a+h)\to f(a)$ as $h\to 0$ if and only if $f(b-h)\to f(b)$ $h\to 0$.

• where did $f(a+h)$ come from? – Skrrrrrtttt Oct 7 '17 at 20:04
• @StrahinjaSukiban $h$ is an any prescribe real number – Ice sea Oct 7 '17 at 20:05

HINT:

We have $f(x+h)-f(x)=f(h)$ for all $x$ and all $h$ (Just let $y=h$.).

Then, this holds in particular for $x=a$, the point for which we are given that $f$ is continuous.

Use this to show that $f$ is continuous at $0$ by letting $h\to 0$ and noting $f(0)=0$.

Then, show this implies that $f$ is continuous everywhere since $f(b+h)-f(b)=f(a+h)-f(a)=f(h)$.

• where is the h coming from and what about the y in $f(x+y)$ – Skrrrrrtttt Oct 7 '17 at 20:01
• Let $y=h$. That's all we need to do. – Mark Viola Oct 7 '17 at 20:04
• cant you simply say that since $f(x+y)$ is defined for all $x,y \in \mathbb R$ that $f(b+y) -f(y) = f(b)$ which shows that $f(x+y)$ is continuous for any $b \in \mathbb R$ ? – Skrrrrrtttt Oct 7 '17 at 20:18
• How does that conclusion follow from $f(b+y)-f(y)=f(b)$ alone? Where did you use the fact that $f$ is continuous at $a$? – Mark Viola Oct 7 '17 at 20:22
• can we show this proof without showing continuity at 0 – Skrrrrrtttt Oct 7 '17 at 20:24

Take any $b$, then exist $c$ such that $b=c+a$. So for each $\epsilon >0$ exist $\delta >0$ such that if $|h|<\delta$ we have:

$$|f(b+h)-f(b)| = f(a+h)+f(c)-f(a)-f(c)| = |f(a+h)-f(a)|<\epsilon$$