Additive function on R with one point of continuity?

Question

Suppose f:R-->R satisfies f(x+y)=f(x)+f(y) for every x,y in R. If f is continuous at some point p, prove that there is some constant a in R s.t. f(x)=ax for all x. that is, an additive function that is continuous at even one point is linear-and hence continuous on all of R.

*Remark: In the original statement I had to fix a s.t. f(ax)=ax, but that is trivial because f(0)=0. For a nonzero a, the condition in the book could not have been made generally. Say for f(x) = 2x.

Answer 1

By induction (details skipped), it's easy to show that f(nx) = n f(x) for every positive integer n and every real x.

According to the given functional equation, f(0 + 0) = f(0) = f(0) + f(0) = 2f(0), which implies f(0) = 0. If we let y = -x, then the functional equation now implies that f(x +(-x)) = f(0) = 0 = f(x) + f(-x), so that f(-x) = - f(x). If n is any negative integer, then f(nx) = f(|n| (-x)).Since |n| >=0, it follows from what we've already seen that f(nx) = |n| f(-x) = -|n| f(x) = n f(x), so that this equation holds for every real x and every integer n.

If we let a = f(1), then for every integer n, f(n) = f(n * 1) = n f(1) = a*n. If n<>0 is any integer, then, f(1) = f(n * 1/n) = n f(1/n) = a, so that f(1/n) = a/n. And now, if x = m/n is any rational, then f(x) = f(m * 1/n) = m f(1/n) = m * a/n = m/n * a = ax. And so, we see the equation f(x) = ax holds for every rational x.

Now, suppose f is continuous at some real k. Since every element of R is an accumulation point of R, this is equivalent to lim ( h -> 0) f(k + h) = f(k). So, lim (h ->0) f(k) + f(h) = f(k) + lim (h -> 0) f(h) = f(k) => lim (h -> 0) f(h) = 0 = f(0) => f is continuous at 0. For every real x, it then follows that lim (h ->0) f(x + h) = lim (h -> 0) f(a) + f(x) = f(a) + lim (h -> 0) f(x) = f(a) + f(0) = f(a). So, f is continuous on all of R.

Let g:R->R be given by g(x) = ax. Then, g is continuous on all of R. And, from what we've seen, f and g agree on Q, that is f(x) = g(x) for every rational x. Since Q is dense in R and f and g are continuous on R, it then follows that f=g all over R. So, f(x) = a(x) for every real x, with a = f(1).

Not sure if you're aware of that conclusion about Q. There's a theorem (valid not only in R but also in other metric spaces and even in other topological spaces) thats says: If D is dense in R and f and g are real valued functions continuous on R that agree on D, then f = g. To see this, pick x in R and let (x_n) be a sequence in D that converges to x. Since D is dense in R, this sequence exists. By the continuity of f, it follows f(x_n) -> f(x). Since f and g agrees on D, f(x_n) and g(x_n) are the same sequence. By continuity of g, it follows g(x_n) = f(x_n) -> g(x). So, f(x_n) converges to f(x) and to g(x). Since the limit of a convergent sequence in any metric space is unique, f(x) = g(x) for every real x. That is f = g.

Answer 2

If n is a positive integer, then f(nx) = nf(x) by induction. Since f(0) = 0, then f(-x) = -f(x) and we get f(nx) = nf(x) also for any negative integer n. Next, for any rational r = m/n, set mx = ny. Then:

mf(x) = f(mx) = f(ny) = nf(y)

so we get f(rx) = rf(x). Now use the continuity at a point p. Take a real x, and a sequence (p_n) converging to p. Then

f(x+p_n) = f(x) + f(p_n)

Since f(p_n) converges to f(p), it follows that f(x+p_n) converges to f(x) + f(p) = f(x+p), thus f is continuous everywhere. Let now be "a" a real number, and a_n a sequence of rationals converging to a. Since, for any x, (a_n)*x converges to ax, by continuity we get that f((a_n)*x) converges to f(ax). But f((a_n)*x)=(a_n)*f(x) and the latter converges to af(x). Thus f(ax) = af(x) for any x.

Answer 3

Discontinuity potential that there are regulations on the variety and area so there are open circles on numbers that are no longer coated and closed circles on numbers the place the line of the graph starts off. that could be a step graph too.