Uniform convergence of f_n(x) = √(x² + 1/n) to |x|

[BrainHurts]

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?

 

[micromass]

$\lim_{ n\rightarrow \infty} \sqrt{x^2 + \dfrac{1}{n}} = \sqrt{x^2 + \lim_{n \rightarrow \infty} \dfrac{1}{n}} = \sqrt{x^2} = |x|$. 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.

 

[BrainHurts]

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?

 

[micromass]

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?

 

[BrainHurts]

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 $|f_n - f(x) | \rightarrow 0$

 

[micromass]

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.

 

[BrainHurts]

Thanks, I think I'm going to have to think about this a little while

 

[dirk_mec1]

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

 

[pasmith]

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 $n \in \mathbb{N}$.

 

[BrainHurts]

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]

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.

 

[BrainHurts]

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

 

[ryanekennedy]

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 $x^2 + \frac{1}{n} > x^2$.

 

[Ray Vickson]

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 $\sup_{x \in I} d_n(x)?$

 

[BrainHurts]

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.