Let $f_n, g_n : (0, 1) → \mathbb{R}$ be the sequences of functions defined

by $f_n :=x^n$ and $g_n(x) = x^n(1-x^n)$

for $x \in (0,1)$ and $n = 1,2,.......$

Then which of the following statement(s) is(are) true?

$(a)$ Both $(f_n)$ and $(g_n)$ converge uniformly in $(0, 1).$

$(b)$ $(f_n)$ converges uniformly in $(0, 1)$ but $(g_n)$ does not converge uniformly in $(0, 1).$

$(c)$ $(g_n)$ converges uniformly in $(0, 1)$ but $(f_n)$ does not converge uniformly in $(0, 1).$

$(d)$ Both $(f_n)$ and $(g_n)$ do not converge uniformly in $(0, 1)$

My attempt : for $f_n(x) = \begin{cases} 1 \text{ if}\ x ∈ (0,1) \\ 0 \ \text{if x =0}. \end{cases} $

so option $1)$ will not true because $f_n(x) =x^n$ doesnot converge uniformly

option $2)$ will also not true

im doubt/confusion in option $3)$ and option $4)$

Any hints/solution will be appreciated

thanks u

  • 1
    $\begingroup$ So you just are asking if $g_n$ converges uniformly in $(0,1)$. Since $g_n \to 0$ pointwise, you're just asking if $g_n$ converges to $0$ uniformly. This is the same as asking if $\lim_{n \to \infty} \sup_{x \in (0,1)} |x^n(1-x^n)| = 0$, but this is false. For $n \ge 1$, if $x = 2^{-1/n}$, then $x^n(1-x^n) = \frac{1}{4}$. $\endgroup$ – mathworker21 Oct 7 '18 at 10:20

I don't understand your attempt $f_n(x)=\begin{cases}1 &\text{ if }x\in (0,1)\\0 &\text{ if }x=0\end{cases}$. This equation is definitely wrong...

If you read carefully, then you see that you just have to decide which of $f_n$ and $g_n$ congerves uniformly.

First, you need to find the pointwise limit of $f_n$ and $g_n$. For each $x\in (0,1)$ you define $$ f(x):=\lim_{n\to\infty}f_n(x)\text{ and }g(x):=\lim_{n\to\infty}g_n(x). $$ This should be not so hard.

Next, you compute the difference of $f_n$ to $f$ and $g_n$ to $g$ with respect to the supremum norm: $$ \|f_n-f\|_\infty=\sup_{x\in(0,1)}|f_n(x)-f(x)|\text{ and }\|g_n-g\|_\infty=\sup_{x\in(0,1)}|g_n(x)-g(x)|. $$ Finally, you get $$ f_n\text{ converges uniformly to }f \Leftrightarrow \|f_n-f\|_\infty\to 0\\ \text{ and } g_n\text{ converges uniformly to }g \Leftrightarrow \|g_n-g\|_\infty\to 0. $$

  • $\begingroup$ thanks u @Mundron im little bit confusing $\|f_n-f\|_\infty=\sup_{x\in(0,1)}|f_n(x)-f(x)|=\sup_{x\in(0,1)}|x^n- 0|= \sup_{x\in(0,1)}|x^n| = 1 \neq 0$ but u have written $\rightarrow$ $ 0$ pliz elaborate this $\endgroup$ – jasmine Oct 7 '18 at 10:37
  • $\begingroup$ That is exactly the point. $f_n$ converges uniformly to $f$ if and only if $\|f_n-f\|_\infty$ goes to zero. This is the definition of uniformly convergence. Since you get correctly $\|f_n-f\|_\infty=1\not\to 0$, you can conclude that $f_n$ doesn't converge uniformly. $\endgroup$ – Mundron Schmidt Oct 7 '18 at 10:37
  • $\begingroup$ okss i got its option D is the corret answer $\endgroup$ – jasmine Oct 7 '18 at 10:39

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