kn(x) = -n on (-∞, -n], x on [-n, n], and n on [n, ∞). Prove that the (real valued) function f(x) is continuous iff all kn( f(x) ) are continuous.
Solution
This is apparently almost trivial.
The kn are continuous, so if f is continuous, then so are all knf.
If f is not continuous, then there must be a point x0 and an infinite sequence of points xm in the interval (x0 - 1, x0 + 1) but not equal to x0, such that xm tend to x0, but f(xm) do not tend to f(x0). Take n sufficiently large that (x0 - 1, x0 + 1) lies inside (-n, n). Then knf(xi) = f(xi), so knf is not continuous at x0.
© John Scholes
jscholes@kalva.demon.co.uk
25 Aug 2001