Question #122939
Show that {fn}, where fn : [0, 1] → R is the map fn(x) = x
n, is not uniformly convergent.
1
Expert's answer
2020-06-22T18:07:34-0400

Given that , (fn)(f_n) be a sequence of function.

Where fn:[0,1]Rf_n:[0,1]\rightarrow\R defined by fn(x)=xnf_n(x)=x^n .

Clearly, if x=1x=1 then the sequence (fn(1))(f_n(1)) converges to 1 .

If 0x<10\leq x <1 then the sequence (fn(x))(f_n(x)) converges to 0 as we known that limnxn=0lim_{n\to \infty} x^n=0 if 0x<10\leq x<1 .

Let f(x)={1if x=10if 0x<1f(x)=\begin{cases} 1 & \text{if} \ x=1 \\ 0 & \text{if} \ 0\leq x <1 \end{cases}

Thus the sequence of function (fn(x))(f_n(x)) convergent to ff on the set [0,1][0,1] .

If nk=k and xk=(12)1kn_k=k \ and \ x_k=(\frac{1}{2})^{ \frac{1}{k}} then

fnk(xk)f(xk)=120=12| f_{n_k}(x_k)-f(x_k)|=|\frac{1}{2}-0|=\frac{1}{2} .


Therefore the sequence (fk)(f_k) doesn't converge uniformly on [0,1] to f[0,1] \ to \ f .



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