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Let $f_n(x)=\frac{1}{nx+1}$ and $g_n(x)=\frac{x}{nx+1}$ for $x\in (0,1)$. Show that $f_n$ doesn't converge uniformly on $(0,1)$ but that $g_n$ does. I know that it is not something difficult, but I still would like to know if my proofs are correct, please.

First of all, we remark that $\lim_{n\to\infty}f_n(x)=1$ and $\lim_{n\to\infty}g_n(x)=x$. We prove now that $f_n$ doesn't converge uniformly to $1$:

$\bullet$ First approach: By definition, to show that $f_n$ doesn't converge uniformly to $1$, we have to satisfy the following: $ \exists \epsilon>0 \ \forall N \ \exists n\ge N \ \exists x\in (0,1)$: $|\frac{1}{nx+1}-1|>\epsilon$. If we take $x=1/n$ and $\epsilon=1/3$, we satisfy the definition. Therefore, $f_n(x)$ doesn't converge uniformly to $1$ on $(0,1)$

$\bullet$ Second approach: Suppose by absurd that $f_n$ converges uniformly to $1$. Then, we have the following: $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x\in(0,1)$:

$|\frac{1}{nx+1}-1|<\epsilon$

But, $|\frac{1}{nx+1}-1|=|\frac{-nx}{nx+1}|\le|\frac{nx}{nx}|=1$. Taking $\epsilon=2$ we got the contradiction. So, $f_n$ doesn't converge uniformly to $1$.

Now, we show that $g_n$ converges uniofrmly to $x$ on $(0,1)$. By definition we have to satisfy the following: $\forall \epsilon>0 \ \exists N \ \forall n\ge N \ \forall x\in(0,1)$:

$|\frac{x}{nx+1}-x|<\epsilon$.

But,

$|\frac{x}{nx+1}-x|\le|\frac{x}{nx+1}|\le|\frac{x}{nx}|\le|\frac{1}{n}|$.

If we pick $N$ such that $\frac{1}{N}< \epsilon$ we obtain that $f_n$ converges uniforomly to $x$ on $(0,1)$

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    $\begingroup$ Both of the claims are false. Both of those functions converge pointwise to zero. $\endgroup$ Feb 22, 2021 at 19:07
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    $\begingroup$ Don't they both converge to $0$? In this case, your proof that $f_n$ doesn't converge to $1$ uniformly is obvious since it doesn't even converge to $1$ pointwise. For $g_n$, this line isn't true $|\frac{x}{nx+1}-x|\le|\frac{x}{nx+1}|\le|\frac{x}{nx}|\le|\frac{1}{n}|$ $\endgroup$
    – ProfOak
    Feb 22, 2021 at 19:07

2 Answers 2

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Both your pointwise limits for $f_n$ and $g_n$ are wrong.


Hint.

For every $x\in (0,1)$, $$ \lim_{n\to\infty}\frac{1}{nx+1}=0,\quad \lim_{n\to\infty}\frac{x}{nx+1}=x\lim_{n\to\infty}\frac{1}{nx+1}=0 $$

So both $f_n$ and $g_n$ converges to $0$ pointwise on $(0,1)$.

To show that $f_n$ does not converge uniformly to zero, consider the sequence $x_n=\frac1n$ and estimate $$ |f_n(x_n)-f(x_n)| $$ It is easy to see that this quantity does not go to zero.

For $g_n$, consider the estimate $$ |\frac{x}{nx+1}|=|\frac{1}{n+\frac{1}{x}}|\le\frac1n $$

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hint

You made a mistake in your $\lim_{n\to+\infty}f_n(x)=1$.

In fact, $$\lim_{n\to +\infty}f_n(x)=0$$

and $$M_n=\sup_{0<x<1}|f_n(x)-0|=1$$

thus, the convergence is not uniform at $(0,1)$. Or

$$M_n\ge f_n(\frac 1n)=\frac 12$$

By the same, let $$G_n(x)=|g_n(x)-0|$$ then $$G_n'(x)=\frac{1}{(nx+1)^2}$$ and $$\sup_{0<x<1}G_n(x)\le g_n(1)=\frac{1}{n+1}\to 0$$

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  • $\begingroup$ Forget, I'm stupid... I confused with $x^n$. Thank you for an answer $\endgroup$
    – Daniil
    Feb 22, 2021 at 19:15

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