# Uniform Convergence of a Sequence of Functions

## Homework Statement

Define

$f_n : \mathbb{R} \rightarrow \mathbb{R}$ by

$f_n(x) = \left( x^2 + \dfrac{1}{n} \right)^{\frac{1}{2}}$

Show that $f_n(x) \rightarrow |x|$ converges uniformly on compact subsets of $\mathbb{R}$

Show that the convergence is uniform in all of $\mathbb{R}$

## The Attempt at a Solution

Not quite good at these epsilon proofs, not sure if it needs to go that far but by the root law we have that

$\lim_{ n\rightarrow \infty} \sqrt{x^2 + \dfrac{1}{n}} = \sqrt{x^2 + \lim_{n \rightarrow \infty} \dfrac{1}{n}} = \sqrt{x^2} = |x|$. Now this shows that $f_n$ converges uniformly in some subset of $\mathbb{R}$ correct?

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$\lim_{ n\rightarrow \infty} \sqrt{x^2 + \dfrac{1}{n}} = \sqrt{x^2 + \lim_{n \rightarrow \infty} \dfrac{1}{n}} = \sqrt{x^2} = |x|$. Now this shows that $f_n$ converges uniformly in some subset of $\mathbb{R}$ correct?
Well, this only shows that the ##f_n## converge to ##|x|## pointswise. You need some extra argument to show or disprove uniform convergence.

Well I suppose we should then go to the definition right.

Let $\epsilon > 0$, I need to find the right $N(\epsilon)$ such that whenever $n\geq N$, we have that

$\left| \sqrt{x^2 + \dfrac{1}{n}} - |x| \right| < \epsilon$.

So I have a hard time messing with that

$\left| \sqrt{x^2 + \dfrac{1}{n}} - |x| \right|$

any hints?

Well, you need to prove

$$\textrm{sup}_x |f_n(x) - f(x)| \rightarrow 0$$

So I would start by calculating this supremum. Maybe try the usual techniques of calculus?

1 person
Just to be clear, I need to find the

$\lim_{n \rightarrow \infty} \text{sup}_{x \in \mathbb{R}} | f_n(x) - f(x) |$ and show that as $n \rightarrow \infty$ then $|f_n - f(x) | \rightarrow 0$

Just to be clear, I need to find the

$\lim_{n \rightarrow \infty} \text{sup}_{x \in \mathbb{R}} | f_n(x) - f(x) |$
You need to show this limit is ##0## by definition of uniform convergence. Do you see why.

and show that as $n \rightarrow \infty then |f_n - f(x) | \rightarrow 0$
No, that would just be pointswise convergence.

Let me give you a hint: |x| = sqrt(x^2).

pasmith
Homework Helper

## Homework Statement

Define

$f_n : \mathbb{R} \rightarrow \mathbb{R}$ by

$f_n(x) = \left( x^2 + \dfrac{1}{n} \right)^{\frac{1}{2}}$

Show that $f_n(x) \rightarrow |x|$ converges uniformly on compact subsets of $\mathbb{R}$

Show that the convergence is uniform in all of $\mathbb{R}$

## The Attempt at a Solution

Not quite good at these epsilon proofs, not sure if it needs to go that far but by the root law we have that

$\lim_{ n\rightarrow \infty} \sqrt{x^2 + \dfrac{1}{n}} = \sqrt{x^2 + \lim_{n \rightarrow \infty} \dfrac{1}{n}} = \sqrt{x^2} = |x|$. Now this shows that $f_n$ converges uniformly in some subset of $\mathbb{R}$ correct?
Use the hint, and show that convergence is uniform on an arbitrary compact subset $S$ of $\mathbb{R}$.

Note that every $f_n$ is continuous, and $f$ is continuous, so that $g_n = f_n - f$ is continuous, and that $0 < g_{n+1}(x) < g_n(x)$ for every $x \in \mathbb{R}$ and every $n \in \mathbb{N}$.

Use the hint, and show that convergence is uniform on an arbitrary compact subset $S$ of $\mathbb{R}$.

Note that every $f_n$ is continuous, and $f$ is continuous, so that $g_n = f_n - f$ is continuous, and that $0 < g_{n+1}(x) < g_n(x)$ for every $x \in \mathbb{R}$ and every $n \in \mathbb{N}$.
Sorry to bring up old news but I haven't really thought about this problem since, but anyway, pasmith all I'm getting from you is that $g_n$ is one-to-one.

$g_n(x) = f_n(x) - f(x) = \left( x^2 - \dfrac{1}{n} \right)^{\frac{1}{2}} - |x| = \left(x^2 - \dfrac{1}{n} \right)^{\frac{1}{2}} - \sqrt{x}$

So

$\lim_{n \rightarrow \infty} \text{sup}_{x \in \mathbb{R}} \left|f_n(x) - f(x) \right| = \lim_{n \rightarrow \infty} \text{sup}_{x \in \mathbb{R}} \left| \left(x^2 - \dfrac{1}{n} \right)^{\frac{1}{2}} - \sqrt{x} \right|$

I feel like I need to use the squeeze theorem or something. Am I supposed to use the following argument along the lines of

$0 < |g_{n+1}| < |g_n|$ and as $n \rightarrow \infty, |g_n| \rightarrow 0$

jbunniii
Homework Helper
Gold Member
Sorry to bring up old news but I haven't really thought about this problem since, but anyway, pasmith all I'm getting from you is that $g_n$ is one-to-one.

$g_n(x) = f_n(x) - f(x) = \left( x^2 - \dfrac{1}{n} \right)^{\frac{1}{2}} - |x| = \left(x^2 - \dfrac{1}{n} \right)^{\frac{1}{2}} - \sqrt{x}$
I assume you intended that last ##\sqrt{x}## to be ##\sqrt{x^2}##, and also I assume you meant
$$\left(x^2 + \frac{1}{n}\right)^{1/2}$$
as in the problem statement.

Try using the fact that
$$\sqrt{a} - \sqrt{b} = \frac{a - b}{\sqrt{a} + \sqrt{b}}$$
assuming ##a## and ##b## are positive.

mmmmmmm OK I'll do that! but yes that's what I meant

Currently working on the same problem for my Intro Real Analysis II course at Rutgers University.

We know ## x^2 > 0 ## and ## \frac{1}{n} > 0 ## for all ##x \in \mathbb{R}## and for all ##n \in \mathbb{N}##, hence using the hint above we obtain $$\left | \sqrt{x^2 + \frac{1}{n}} - \sqrt{x^2} \right | = \left | \frac{1}{n (\sqrt{x^2 + \frac{1}{n}} + \sqrt{x^2})} \right | .$$

I think we need to split the reasoning for "##f_n## converges uniformly to ##f## on ##\mathbb{R}##" into two parts. Consider ##|x| \geq 1##, which implies ##x^2 \geq 1## and certainly ##x^2 + \frac{1}{n} \geq x^2## for natural number ##n##, hence ##\sqrt{x^2 + \frac{1}{n}} \geq \sqrt{1} = 1##. Moreover, ##\sqrt{x^2 + \frac{1}{n}} + \sqrt{x^2} \geq \sqrt{x^2 + \frac{1}{n}} \geq 1##. Certainly ##n \geq N## gives ## \frac{1}{n} \leq \frac {1}{N}##, so combining these facts we have that $$\frac{1}{n (\sqrt{x^2 + \frac{1}{n}} + \sqrt{x^2})} \leq \frac{1}{N}$$ for ##|x| > 1##. So setting ##\frac{1}{N} < \epsilon## gives $$\forall \epsilon > 0, \rm{choose}\ N \in \mathbb{N}\ \rm{such\ that}\ N > \frac{1}{\epsilon} \implies \\ \forall |x| > 1 \ \forall n \geq N\ \rm{we\ have}\ \left | \frac{1}{n (\sqrt{x^2 + \frac{1}{n}} + \sqrt{x^2})} \right | \leq \frac{1}{n} \leq \frac{1}{N} < \epsilon.$$

To argue for ##|x|<1##, I am at a loss. Thoughts?

Attempt:
We know ##x^2 < 1##; however ##x^2 + \frac{1}{n} > x^2##.

Last edited:
Ray Vickson
Homework Helper
Dearly Missed
Well I suppose we should then go to the definition right.

Let $\epsilon > 0$, I need to find the right $N(\epsilon)$ such that whenever $n\geq N$, we have that

$\left| \sqrt{x^2 + \dfrac{1}{n}} - |x| \right| < \epsilon$.

So I have a hard time messing with that

$\left| \sqrt{x^2 + \dfrac{1}{n}} - |x| \right|$

any hints?
Look at the graph of ##d_n(x) = \sqrt{x^2 + \frac{1}{n}} - |x| > 0##. For any interval ##I## that contains ##0##, what is the value of ##\sup_{x \in I} d_n(x)?##

I don't remember, but in looking at your argument, why don't you include [\itex]|x|\geq 0[\itex]. Secondly, if you're going to do it for x<0, use the definition of |\cdot|, do the same thing, multiply both top and bottom by the conjugate and I think you have what you got. I hope this helps.