- #1
SiddharthM
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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?
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?
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