# Uniform Convergence verification for Sequence of functions - NBHM

Following is a list of problems from an exam for admission into Ph.D program. I have just compiled all previous questions on uniform convergence of sequence of functions and i tried to work out . I would be thankful if some one can check the solutions and please suggest if there are any better ways to do and if the solution is wrong please let me know what could be done for them.

• $f_n(x)=\sin^n x$ on $[0,\frac{\pi}{2}]$

• $f_n(x) = \frac{x^n}{n}+1$ on $[0,1)$

• $f_n(x) = \dfrac{1}{1+(x-n)^2}$ on $(-∞,0)$

• $f_n(x) = \dfrac{1}{1+(x-n)^2}$ on $(0, ∞)$

• $f_n(x) = nxe^{−nx}$ on $(0,∞)$

• $f_n(x) = x^n$ on $[0, 1]$

• $f_n(x) = \dfrac{\sin(nx)}{\sqrt{n}}$ on $\mathbb{R}$

• $f_n(x)=\dfrac{nx}{1 + nx}$ on $(0,1)$

• $f_n(x) = \dfrac{x^n}{1 + x^n}$ ; on $[0, 2]$

• $f_n(x)=n^2x^2e^{-nx}$ over $(0,\infty)$

• $f_n(x)=(\cos (\pi n!x))^{2n}$ on $[0,1]$

• $f_n(x)=n^2x(1-x^2)^n$ on $[0,1]$

I would like to explain what all i have tried and if there is any better way to do please let Me know.

• For $f_n(x)=\sin^n x$ on $[0,\frac{\pi}{2}]$ we do as follows :

for $x=\frac{\pi}{2}$ we have $f_n(x)=1$ and for $x\in [0,\frac{\pi}{2})$ we have $f_n(x)=(a)^n$ where $0\leq a <1$ so, $f_n(x) \rightarrow 0$ for each $x\in [0,\frac{\pi}{2})$ so limit function is :

$$\lim_{n\rightarrow \infty}f_n(x) = \begin{cases} 0, & x \in [0,\frac{\pi}{2}) \\ 1, & x=\frac{\pi}{2}. \end{cases}$$

so the limit function is not continuous so convergence is not uniform.

• For $f_n(x) = \frac{x^n}{n}+1$ on $[0,1)$ limit function is $f(x)=1$ .

Now $|f_n(x)-f(x)|=\frac{x^n}{n}$ let $d_n =\sup \{ |f_n(x)-f(x)| : x\in [0,1) \}==\sup \{ \frac{x^n}{n} : x\in [0,1) \}$ I could see that $d_n\rightarrow 0$ intuitively but could not produce a complete proof.please help me with that.keeping that gap aside, as $d_n\rightarrow 0$ we see that convergence is uniform

• For $f_n(x) = \dfrac{1}{1+(x-n)^2}$ on $(-∞,0)$ we see that the limit function is $f\equiv 0$ I do not see any problem that stops this from being uniform convergence (please see below solution for what i mean when i say problem) at the same time i can not say that it is uniform convergent for sure.

• For $f_n(x) = \dfrac{1}{1+(x-n)^2}$ on $(0, ∞)$ we have limit function to be $f(x)\equiv 0$ but then, $|f_n(x)-f(x)|=|f_n(x)|$ equals $1$ at $x=n$ so the condition $$|f_n(x)-f(x)|<\epsilon \textbf { for all } x\in (0,\infty)$$ does not hold so the convergence is not uniform.

• For $f_n(x) = nxe^{−nx}$ on $(0,∞)$ we have $f_n(x)=\frac{nx}{e^{nx}}$.

But $e^{nx}$ diverges faster that $nx$ so limit function is $f\equiv 0$.

But then at $x=\frac{1}{n}$ we have $f_n(x)=\frac{1}{e}$ so I can not say that $|f_n(x)-f(x)|< \epsilon$ for all $x$ and for all $n\geq N$ for some $N$ if my epsilon is less than $\frac{1}{e}$ thus the convergence is Not uniform.

• For $f_n(x) = x^n$ on $[0, 1]$ we see that limit function is $$\lim_{n\rightarrow \infty}f_n(x) = \begin{cases} 0, & x \in [0,1) \\ 1, & x=1. \end{cases}$$ so the limit function is not continuous so convergence is not uniform.

• For $f_n(x) = \dfrac{\sin(nx)}{\sqrt{n}}$ on $\mathbb{R}$ we see that the limit function is $f\equiv 0$. $|f_n(x)-f(x)|=|f_n(x)|=|\dfrac{\sin(nx)}{\sqrt{n}}|\leq\frac{1}{\sqrt{n}}$.

Now, for given $\epsilon>0$ if i choose $N$ such that $\frac{1}{\sqrt{N}}<\epsilon$ then we have $|f_n(x)-f(x)|\leq\frac{1}{\sqrt{n}}<\epsilon$ for all $n\geq N$ irrespective of choice of $x$. thus the convergence is uniform.

• For $f_n(x)=\dfrac{nx}{1 + nx}$ on $(0,1)$ we have $f_n(x)=\dfrac{1}{\frac{1}{nx}+1}$ the limit function is $1$.

Now $|f_n(x)-f(x)|=|\dfrac{nx}{1 + nx}-1|=|\frac{nx-1-nx}{1+nx}|=\frac{1}{1+nx}$

Now $0<x\Rightarrow 0<nx \Rightarrow 1<1+nx \Rightarrow \frac{1}{1+nx} < 1$ But then I do not know how to proceed further...

• For $f_n(x) = \dfrac{x^n}{1 + x^n}$ ; on $[0, 2]$ we see that $f_n(x)=0$ for all $x\in [0,1)$ at $x=1$ we have $f_n(x)=\frac{1}{2}$ and $f_n(x)=1$ for all $x\in[1,2]$ so limit function is : $$\lim_{n\rightarrow \infty}f_n(x) = \begin{cases} 0, & x \in [0,1) \\ \frac{1}{2}, & x=1,\\ 1&x\in[1,2] \end{cases}$$ so the limit function is not continuous so convergence is not uniform.

• For $f_n(x)=n^2x^2e^{-nx}$ over $(0,\infty)$ we have $f_n(x)=\frac{n^2x^2}{e^{nx}}$.

But $e^{nx}$ diverges faster that $n^2x^2$ so limit function is $f\equiv 0$.

But then at $x=\frac{1}{n}$ we have $f_n(x)=\frac{1}{e}$ so I can not say that $|f_n(x)-f(x)|< \epsilon$ for all $x$ and for all $n\geq N$ for some $N$ if my epsilon is less than $\frac{1}{e}$ thus the convergence is Not uniform.

• For $f_n(x)=(\cos (\pi n!x))^{2n}$ on $[0,1]$ at $x=0$ we have $f_n(x)=1$ and for $x\in(0,1]$ we have $f_n(x)\rightarrow 0$ $$\lim_{n\rightarrow \infty}f_n(x) = \begin{cases} 1, & x=0 \\ 0, & x\in(0,1]. \end{cases}$$ so the limit function is not continuous so convergence is not uniform.

• $For f_n(x)=n^2x(1-x^2)^n$ on $[0,1]$ limit function is $f\equiv 0$ Now if i consider $|f_n(x)-f(x)|=n^2x(1-x^2)^n$ by considering derivative i have seen that maximum value is obtained at $x^2=\frac{1}{2n+1}$ so for uniform convergence we should make sure $f_n(x)$ for $x^2=\frac{1}{2n+1}$ goes to $0$ but then$f_n(x)=\dfrac{n^2}{}\sqrt{2n+1}(\frac{2n}{2n+1})^n$ but the limit does not even exist.

So, the convergence is not uniform.

Thank you for sparing your valuable time in checking my solutions. Once somebody confirm that every thing is fine and no modifications needed i would delete this question so please post only comments.

Thank you :)

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It makes little difference in your analysis but for the $f_n(x)=n^2x^2e^{-nx}$ problem, I think that the max occurs at $x = 2/n$ –  MSEoris Aug 4 at 15:24
@MSEoris : I am sorry, I did not get your point... –  Praphulla Koushik Aug 4 at 15:58
We are interested in showing that the sup($f_n$) converges to 0. Thus we take the derivative of $f_n$ with respect to $x$, and see that the max occurs at $x_{max} = 2/n$. Then $f_n(x_{max})$ has a slightly different value than the one you proposed. –  MSEoris Aug 5 at 15:32

• For $f_n(x)=\sin^n(x)$ I think your solution is good.

• For $f_n(x)=\frac{x^n}{n}+1$, if we define $d_n$ as above, note that $0\leq d_n\leq 1/n$ (because $0\leq x<1$ or simpler: because the application $x\mapsto x^n/n$ is an increasing function of $x$). This implies that $d_n\to 0$

• For $f_n(x)=\frac{1}{1+(x-n)^2}, x<0$, note that the restriction over $x$ implies that $n^2<1+(n-x)^2$ and therefore, that $f_n(x)<n^{-2}$. Using the same argument used in the last point, we have that the convergence is uniform.

• For $f_n(x)=\frac{1}{1+(x-n)^2}, x>0$ it's fine.

• For $f_n(x)= nxe^{-nx}$ it's fine.

• For $f_n(x)=x^n, x\in [0,1]$ it's fine.

• For $f_n(x)= \frac{\sin(nx)}{\sqrt{n}}$ on $\mathbb{R}$ it's fine.

• For $f_n(x)=\frac{nx}{1+nx}, x\in (0,1)$ the convergence is not uniform since for every $N$ there exists an $n\geq N$ and an $0<x<1$ such that $\vert f_n(x)-f(x)\vert \geq 1/4$ : just take $x=1/n$ and note that $\vert f_n(1/n)-f(1/n)\vert=1/2$.

• For $f_n(x)=\frac{x^n}{1+ x^n}, x\in [0,2]$ it's fine.

• For $f_n(x)=n^2x^2e^{-nx}$ over $(0,\infty)$ it's fine.

• For $f_n(x)=(\cos(\pi n!x))^{2n}, x\in [0, 1]$ your conclusion is correct, but the limit function is not the one that you said: note that if $x$ is a rational number, for $n$ large enough we will have $n!x\in\mathbb{Z}$ and then, $(\cos(\pi n!x))^{2n}=1$. Then, the limit is $$f(x)=\begin{cases} 1 & \mathrm{if} \quad x \quad \mathrm{rational}\\ 0 & \mathrm{if} \quad x \quad \mathrm{not \, rational} \end{cases}$$ Since $f$ is clearly not continuous, the convergence is not uniform.

• For $f_n(x)=n^2x(1-x^2)^n$ I think it is fine too

I hope this will be useful for you

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for $f_n(x)=\frac{nx}{1+nx}$ and $f_n(x)=\frac{x^n}{n}+1$ I had similar ideas But the most beautiful thing in your answer is the case of $f_n(x)=\cos (\pi n! x)^{2n}$.. I really love that! Thank you so much :) .... you may want to edit $f_n(x)=(\cos(\pi n!x))^2n$ to $f_n(x)=(\cos(\pi n!x))^{2n}$ –  Praphulla Koushik Jan 14 at 8:43
Thank you, I have edited it. –  Brandon Jan 14 at 8:51