Additive Functions!

**frenchguy87** · February 27th 2010, 05:12 PM

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

**NonCommAlg** · February 27th 2010, 05:28 PM

Originally Posted by frenchguy87

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

given $a \in \mathbb{R},$ take the limit of both sieds of the identity $f(x)+f(x_0)=f(x-a+x_0)+f(a)$ as $x \to a.$

**Drexel28** · February 27th 2010, 10:00 PM

Originally Posted by frenchguy87

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

A little more formal.

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

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

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

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

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

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 $f:\mathbb{R}\mapsto\mathbb{R}$ which is continuous at a single point and $f(x+y)=f(x)+f(y)$ is the function $f(1)x$ .

**HallsofIvy** · February 28th 2010, 01:18 AM

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 $\lim_{x\to b}f(x)= \lim_{y\to 0}f(b+ y)$ where y= x- b. Then $\lim_{y\to 0}f(b+ y)= f(b)+ \lim_{y\to 0} f(y)$ . Since f is continuous at x= b, $\lim_{x\to b} f(x)= f(b)$ so we have $\lim_{x\to b}f(x)= f(b)= f(b)+ \lim_{y\to 0}f(y)$ 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)= $\lim_{y\to 0}f(a+ y)$ where y= x- a. But f(a+y)= f(a)+ f(y) so $\lim_{y\to 0}f(a+y)= f(a)+ \lim_{y\to 0}f(y)$ and since f is continous at 0, $\lim_{y\to 0}f(x)= f(0)= 0$ . That is, if f is continous at x= 0, then $\lim_{x\to a} f(x)= f(a)$ so f is continuous at x= a.