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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.

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  • $\begingroup$ For D, you need an absoulte value sign, else all looks good as a first glance. $\endgroup$ – Mark Viola Jun 4 '17 at 17:35
  • $\begingroup$ For $A$ you should also mention that $f_n(x)$ is continuous, so that if $f_n$ converges uniformly, it must do so to a continuous function. $\endgroup$ – Alex R. Jun 4 '17 at 17:36
  • $\begingroup$ @MarkViola Yup, the absolute value sign's missing. $\endgroup$ – Naive Jun 4 '17 at 17:41
  • $\begingroup$ @AlexR. Ahh, I should include that too. $\endgroup$ – Naive Jun 4 '17 at 17:42

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