: n
: 2. Let f in C[0,1] and let f_n(x) = f(x ), n = 1,2,.... Assume {f_n} is
: equicontinuous on [0,1], show that f ≡ constant.
Clearly, {fn} is uniformly bounded and [0,1] is a closed bdd. interval,
then we can use the Arzela-Ascoli thm.
From the theorem, we can find a subsequence of {fn}, {fn_k} uniformly
converges to a function g.
Since f_n(x) tends to f(0) if 0≦x <1 and f(1) if x = 1,
so g(x) = f(1) if x = 1; otherwise, g(x) = f(0).
Consider g is continuous since g is a limit function of a sequnece of
continous functions which is uniformly convergent. so f(0) = f(1) = g(x)
for all x in [0,1].
Finally, let x,y be in [0,1] and f(0)=f(1) = g(x)=: A
Given an ε>0, we can find N>0, such that for all n ≧ N,
|f(t^n) - A| < ε/2 for all t in [0,1].
Let t1 = x^(1/N) and t2 = y^(1/N), so (t1)^N = x and (t2)^N = y.
Hence, |f(x) - f(y)| = |f((t1)^N)) - f((t2)^N))| < ε.
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