examples which satistfy f(x+y)=f(x) + f(y)

[HappyN]

The question states:
Give two different examples of f:R->R such that f is continuous and satisfies f(x+y)=f(x)+f(y) for every x,y e R. Find all continuous functions f:R->R having this property. Justify your answer with a proof.

I came up with one example:
f(x)=ax
then f(x+y)=a(x+y)=ax+ay=f(x)+f(y)

however, I cant seem to think of another example, any hints?

[lurflurf]

ax is the only answer, so maybe that makes it count double.

[HallsofIvy]

Your example is "general". Get two functions by taking a to be two different values! As lurflurf says, any continuous function satisfying f(x+ y)= f(x)+ f(y) must be of the form f(x)= ax so different example just have different values of a.

To prove that is straightforward but tedius. Here's an outline of how I would do it.

1) Prove, by induction, that f(nx)= nf(x) for all real numbers, x, and n any positive integer.

2) Use f((0+ n)x)= f(0x)+ f(nx) to show that f(0x)= 0= 0f(x).

3) Use f((n+(-n))x)= f(0)= 0 to show that f(-nx)= -f(nx) and so -nf(x) for any positive integer n.

3) Use f(nx)= nf(x) to show that f(n(1/n)y)= nf((1/n)y)= f(y) so f((1/n)y)= (1/n)f(y).

4) Use f(nx)= nf(x) to show that f((m/n)y)= mf((1/n)y)= (m/n)f(y) for any rational number m/n and any real number y.

5) Use continuity to show that f(rx)= rf(x) for any real numbers r and x and, taking x= 1, that f(r)= rf(1)= ar where a= f(1). (If r is any real number, there exist a sequence of rational numbers, \{r_n\}, converging to r. By continuity, f(rx)= f((\lim r_n)x)= \lim f(r_nx)= (\lim r_n) f(x).)

That means, of course, that the graph of a continuous function satisfying f(x+y)= f(x)+ f(y) is a straight line through the origin. There do, however, exist non-continuous functions satisfying that equation- and the graph of such a function is dense in the plane!

[micromass]

That means, of course, that the graph of a continuous function satisfying f(x+y)= f(x)+ f(y) is a straight line through the origin. There do, however, exist non-continuous functions satisfying that equation- and the graph of such a function is dense in the plane!

Only if you accept the axiom of choice! :biggrin:

[lavinia]

The question states:
Give two different examples of f:R->R such that f is continuous and satisfies f(x+y)=f(x)+f(y) for every x,y e R. Find all continuous functions f:R->R having this property. Justify your answer with a proof.

I came up with one example:
f(x)=ax
then f(x+y)=a(x+y)=ax+ay=f(x)+f(y)

however, I cant seem to think of another example, any hints?

try doing it by taking a Newton quotient.

f(x+h) -f(x)/h = f(h)/h so the Newton quotient is constant for all x.

So there is a derivative and it must be constant. Since f(0) = 2f(0) the function must be ax for some a.

[HallsofIvy]

try doing it by taking a Newton quotient.

f(x+h) -f(x)/h = f(h)/h so the Newton quotient is constant for all x.

So there is a derivative and it must be constant. Since f(0) = 2f(0) the function must be ax for some a.
HOW do you assert "so there is a derivative"? There is a derivative if and only if \lim_{h\to 0} f(h)/h exists but how do you show that?

[lavinia]

HOW do you assert "so there is a derivative"? There is a derivative if and only if \lim_{h\to 0} f(h)/h exists but how do you show that?

i wasn't asserting anything - just suggesting a line of argument.

I guess at some point you have to use continuity. look at expressions like f(h/n)/h/n = f(h)/h so the ratio is constant for rationals. Then use continuity.
The thing is that without continuity the assertion is false.

[HappyN]

Thanks HallsofIvy, I managed to make it work from your outline :)
Wasn't sure about the Newton quotient though...