# Then which of the following statement(s) is(are) true? about uniform convergence

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

• 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 $x^n(1-x^n) = \frac{1}{4}$. – 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 $$f_n\text{ converges uniformly to }f \Leftrightarrow \|f_n-f\|_\infty\to 0\\ \text{ and } g_n\text{ converges uniformly to }g \Leftrightarrow \|g_n-g\|_\infty\to 0.$$
• 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 – jasmine Oct 7 '18 at 10:37
• 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. – Mundron Schmidt Oct 7 '18 at 10:37