By definition, a function is additive if $\displaystyle f(x+y) = f(x) + f(y)$ for all $\displaystyle x$, $\displaystyle y \in \mathbb{R}$.
How would you show that if $\displaystyle f$ is continuous at some point $\displaystyle x_0$, then it is continuous at every point of $\displaystyle \mathbb{R}$?

2. Originally Posted by frenchguy87
By definition, a function is additive if $\displaystyle f(x+y) = f(x) + f(y)$ for all $\displaystyle x$, $\displaystyle y \in \mathbb{R}$.
How would you show that if $\displaystyle f$ is continuous at some point $\displaystyle x_0$, then it is continuous at every point of $\displaystyle \mathbb{R}$?
given $\displaystyle a \in \mathbb{R},$ take the limit of both sieds of the identity $\displaystyle f(x)+f(x_0)=f(x-a+x_0)+f(a)$ as $\displaystyle x \to a.$

3. Originally Posted by frenchguy87
By definition, a function is additive if $\displaystyle f(x+y) = f(x) + f(y)$ for all $\displaystyle x$, $\displaystyle y \in \mathbb{R}$.
How would you show that if $\displaystyle f$ is continuous at some point $\displaystyle x_0$, then it is continuous at every point of $\displaystyle \mathbb{R}$?
A little more formal.

Notice two things first, namely $\displaystyle f(0)=f(0+0)=f(0)+f(0)\implies f(0)=0$ and so $\displaystyle 0=f(0)=f(x-x)=f(x)+f(-x)\implies f(-x)=-f(x)$. Thus,

$\displaystyle f(x)-f(y)=f(x)+f(-y)=f(x-y)$

A little less "limity". Suppose that there existed some $\displaystyle x_0\in\mathbb{R}$ such that $\displaystyle f$ is continuous at that point. Let $\displaystyle x_1\in\mathbb{R}$ be arbitrary. Given any $\displaystyle \varepsilon>0$ we have that

$\displaystyle \left|f(x_1)-f(x)\right|=\left|f\left(x_1-x\right)\right|=\left|-f(x_1-x)\right|=\left|f\left(-(x_1-x)\right)\right|$$\displaystyle =\left|f\left(x_0-(x_1-x)\right)-f(x_0)\right|$. Notice that since $\displaystyle f$ is continuous at $\displaystyle x_0$ we know that for the previously mentioned $\displaystyle \varepsilon$ there exists

some $\displaystyle \delta>0$ such that $\displaystyle \left|x-x_0\right|<\delta\implies |f(x)-f(x_0)|<\varepsilon$. So, let $\displaystyle z=x_0-(x_1-x)$, then notice that choosing $\displaystyle \left|x_1-x\right|<\delta\implies \left|z-x_0\right|<\delta\implies \left|f(z)-f(x_0)\right|=\left|f(x_1)-f(x)\right|<\varepsilon$.

The conclusion follows. The point of all of these is to shift the point of continuity to an arbitrary point.

Note that from this it readily follows that the only function $\displaystyle f:\mathbb{R}\mapsto\mathbb{R}$ which is continuous at a single point and $\displaystyle f(x+y)=f(x)+f(y)$ is the function $\displaystyle f(1)x$.

4. First you need to prove that f(0)= 0. That is true because f(a+0)= f(a)+ f(0)= f(a).

If f is continuous at any x= b, then it is continuous at x= 0.

That is true because $\displaystyle \lim_{x\to b}f(x)= \lim_{y\to 0}f(b+ y)$ where y= x- b. Then $\displaystyle \lim_{y\to 0}f(b+ y)= f(b)+ \lim_{y\to 0} f(y)$. Since f is continuous at x= b, $\displaystyle \lim_{x\to b} f(x)= f(b)$ so we have $\displaystyle \lim_{x\to b}f(x)= f(b)= f(b)+ \lim_{y\to 0}f(y)$ and so [tex]\lim_{y\to 0}f(y)= 0= f(0) and f is continuous at 0.

If f is continuous at x= 0 then it is continuous at every point, a.

Almost the same proof: [tex]lim_{x\to a} f(x)= $\displaystyle \lim_{y\to 0}f(a+ y)$ where y= x- a. But f(a+y)= f(a)+ f(y) so $\displaystyle \lim_{y\to 0}f(a+y)= f(a)+ \lim_{y\to 0}f(y)$ and since f is continous at 0, $\displaystyle \lim_{y\to 0}f(x)= f(0)= 0$. That is, if f is continous at x= 0, then $\displaystyle \lim_{x\to a} f(x)= f(a)$ so f is continuous at x= a.

5. Originally Posted by Drexel28
$\displaystyle f(0)=f(0+0)=f(0)+f(0)\implies f(0)=0$
I don't get this implication, how do you conclude that $\displaystyle f(0)=0$?

6. You have $\displaystyle f(0)=f(0)+f(0)=2f(0)$

So $\displaystyle 2f(0)-f(0)=0$ and finally $\displaystyle f(0)=0$...