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Function space with sup metric

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Poirot

Banned
Feb 15, 2012
250
Let F=C([0,1],[0,1]) be the set of continuous functions from [0,1] to itself, endowed with the sup metric d. You may take it that the space is complete.

Let {$f_{n}$} be given by $f_{n}(x)=x^n$. Prove the sequence is not cauchy

My 'proof': Since F is complete, we can prove it doesn't converge. The sequence converges pointwise to f given by f(1)=1 and f(x)=0 otherwise. Since uniform convergence implies pointwise convergence, if this sequence is goin to converge, it will do so to f.
two cases to consider.

x=1; Then $sup{|f_{n}(x)-f(x)|}=sup({0})=0$ so certainly whatever epsion is chosen, we can choose N=1 and we are fine.

x in [0,1); then $sup{|{f_n}(x)-f(x)|}=sup({x^n})=1$ so choosing epsilon =0.5 shows that $f_n{x}$ doesn't converge to f(x) for x not 1.
Hence sequence is not cauchy.
Not sure if this is right. If it is, is there a better way.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
I think the proof is fine, but it can be simplified. You showed that if $f_n$ converge uniformly, it has to be to $f$. Then you showed using the definition of uniform convergence that $f_n$ do not uniformly converge to $f$. Concerning the second step, since the function space is complete, if $f_n$ converge uniformly to $f$, $f$ has to be continuous, which it is not, so $f_n$ do not uniformly converge to $f$.