more continuous functions

Let f:R --> R be a function that satisfies f(x+y)=f(x)+f(y) for all x, y as an element in R.

(a) Suppose that f is continuous at some point c. Prove that f is continuous on R.

(b) Suppose that f is continuous on R and that f(1)=k. Prove that f(x) = kx for all x as an element in R.

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Originally Posted by friday616

Let f:R --> R be a function that satisfies f(x+y)=f(x)+f(y) for all x, y as an element in R.

(a) Suppose that f is continuous at some point c. Prove that f is continuous on R.


f is cont. at x = 0: let {x_n} be any sequence converging to zero ==> the sequence {x_n + c} converges to c ==> by cont. of f at c we get:

f(x_n + c) --> f(c) ; but f(x_n + c) = f(x_n) + f(c), so in fact:

f(x_n) + f(c) --> f(c) ==> by arithmetic of limits we get f(x_n) --> 0.

Since f(0) = f(0 + 0) = f(0) + f(0) ==> f(0) = 0 , and then we proved

f(x_n) --> f(0) = 0 for any seq. converging to zero.

Let now w be any real number and let {y_n} be any seq. s.t. y_n --> w

==> y_n - w --> 0 ==> by cont. of f at zero we get:

f(y_n - w) --> f(0) = 0 and as before f(y_n - w) = f(y_n) - f(w) --> 0

==> f(y_n) --> f(w) and we're done.

(b) Suppose that f is continuous on R and that f(1)=k. Prove that f(x) = kx for all x as an element in R.

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For any natural n, f(n) = f(1 + 1 +...+ 1) = f(1) +...f(1) = nf(1)

Since 0 = f(0) = f(x +(-x)) = f(x) + f(-x) ==> f(x) = -f(-x) and f is an odd

function ==> for any integer m, f(m) = mf(1) using the above.

Now, k = f(1) = f(n*(1/n)) = nf(1/n) ==> f(1/n) = (1/n)f(1) ==> for any

rational number m/n, we get f(m/n) = f(m*(1/n)) = mf(1/n) = (m/n) f(1)

==> for any rational q, f(q) = qf(1) ==> for any real x, if {x_n} is a

rational seq. s.t. x_n --> q , using cont. of f at x we get

f(x) = lim f(x_n) = lim x_n*f(1) = xf(1) and we're done.

Tonio