Let $f_n(x)=\displaystyle \frac {x^n}{1+x^2}$, for $n\in \Bbb N$. Then which of the following statements is true?
(A)$\;\;f_n$ converges uniformly on $[0,1]$.
(B)$\;\;f_n$ converges uniformly on $[-1,1-\epsilon]$ for $\epsilon \in (0,1)$.
(C)$\;\; \sum f_n$ converges uniformly on $[0,1)$.
(D)$\;\; \sum f_n$ converges uniformly on $[-1+\epsilon,1-\epsilon]$ for $\epsilon \in (0,1)$.
My Attempt:
Let us denote the limit function by $f(x)$.
For (A):
$f(x) = \begin{cases} 0, & x\in[0,1) \\ 1, & x=1 \end{cases}$
Convergence cannot be uniform as the limit function is discontinuous.
For (B):
At $x=-1$, $f(x) $ does not exist.
So the sequence is not even pointwise convergent in the domain.
For (C):
We make use of the following result, 'If $\sum f_n$ converges uniformly $\implies$ $f_n \to 0$ uniformly'
Consider, $x_k=\displaystyle \left( \frac {1}{2} \right)^{1/k}$ and $n_k=k$, then $$f_{n_k}(x_k)=\displaystyle \frac { \left(\left (\frac {1}{2}\right) ^{1/k}\right)^k}{1+\left(\frac {1}{2}\right) ^{2/k}}\to \frac{1}{4}$$
Thus, $f_n$ does not converge to $0$ uniformly and hence $\sum f_n$ cannot be uniformly convergent.
For (D):
We have, $$\displaystyle \frac {x^n}{1+x^2} \le x^n \le (1-\epsilon)^n$$
By Weierstrass M-Test, we conclude $\sum f_n$ converges uniformly in its domain.
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I want to know whether all my arguments are correct.
The way the intervals have been defined using $\epsilon$ is what confused me a bit.