Proof that f is Constant on the Nonnegative Reals

[SiddharthM]

say a function f is continuous on the reals and there exists a sequence of functions so that f_n (t)=f(nt) for n=1,2,3... also the sequence of functions is equicontinuous on [0,1].

what does show about f? I show f is constant on the nonnegative reals

Let epsilon (from here onwards we shall call it e)>0, then because f_n is equicontinuous there exists a delta>0 s.t. d(x,y)<delta implies |f_n(x)-f_n(y)|<e for all n as long as x,y in [0,1]. Now fix any two positive numbers x,y. since x/n and y/n go to zero there exists a N so that |x/n-y/n|<delta for all n>N and that they are both in [0,1]. Then this means that |f_n(x/n)-f_n(y/n)|<e for all n>N. This means |f(x)-f(y)|<e. Since epsilon is arbitrary we see that f(x)=f(y).

I feel like there should be more to this or maybe that I'm wrong altogether.

What examples satisfy the hypothesis?

 

[rudinreader]

f must be constant on [0,\infty). The way I proved it was by looking at f(0) = f_n(0). Fix any c in [0,\infty), and e > 0. Find d > 0 such that |x-y|<d implies |f_n(x)-f_n(y)| < e for all n, x,y in [0,1]. Then choose n such that |c/n| < min{d,1}. Then |f(0)-f(c)| = |f_n(0)-f_n(c/n)| < e. It follows that for any c in [0,\infty), and any e > 0, |f(0)-f(c)| < e. Hence f is constant on [0,\infty).

(edit: changed the proof to restrict c to [0,\infty)).

 

[rudinreader]

Actually, I didn't notice the restriction "equicontinuous on [0,1]"... when I did this problem I must have glanced over that.. I assumed f_n was equicontinuous on R.

So all I really proved was that f is constant on [0,\infty).

 

[rudinreader]

correction

And after a second look, the condition that f is constant on [0,\infty) is necessary and sufficient to satisfy the hypothesis.

That is, you can show that as long as f is constant on [0,\infty), regardless of what f does on (-\infty,0) (even not continuous), then f_n(x)=f(nx) will be equicontinuous on [0,1]. This is actually easy to see: f_n(x)=f(nx)=C for all x in [0,1], hence it's obviously equicontinuous there.