Quick question about convergence

[STEMucator]

Homework Statement

Let $s_n(x) = \frac{1}{n} e^{-(nx)^2}$. Show there is a function $s(x)$ such that $s_n(x) → s(x)$ uniformly on $ℝ$ and that $s_n'(x) → s'(x)$ for every x, but that the convergence of the derivatives is not uniform in any interval which contains the origin.

Homework Equations

$s_n(x) → s(x)$ as $n→∞$

The Attempt at a Solution

For any real x, $s_n(x) → 0 = s(x)$ as $n→∞$ so we have pointwise convergence.

$\forall ε>0, \exists N(ε) \space | \space n>N \Rightarrow |s_n(x)-s(x)| < ε, \forall x \in ℝ$

$|s_n(x) - s(x)| ≤ 1/n$

So choosing $n > 1/ε$ means we have uniform convergence.

$s_n'(x) = -2xne^{-(nx)^2}$

Now, for all real x $s_n'(x) → 0 = s'(x)$ as $n→∞$.

$|s_n'(x) - s'(x)| = 2|x|ne^{-(nx)^2}$

I'm stuck on showing the convergence of the derivatives is not uniform in any interval containing the origin. Don't really know how to argue this one.

 

[micromass]

Let $(-\varepsilon,\varepsilon)$ be an interval around the origin.

1) Can you calculate

A_n=\sup_{x\in (-\varepsilon,\varepsilon)} |s_n^\prime(x) - s^\prime(x)|

2) Does $A_n\rightarrow 0$?

3) What does (2) say about uniform convergence?

 

[STEMucator]

Let $(-\varepsilon,\varepsilon)$ be an interval around the origin.

1) Can you calculate

A_n=\sup_{x\in (-\varepsilon,\varepsilon)} |s_n^\prime(x) - s^\prime(x)|

2) Does $A_n\rightarrow 0$?

3) What does (2) say about uniform convergence?

A_n=\sup_{x \in (-ε, ε)} |s_n&#039;(x) - s&#039;(x)| = \sup_{x \in (-ε, ε)} 2|x|ne^{-(nx)^2}

That's some new notation on me. I've never seen sup used in such a way.

(2) I would assume not.

(3) Not uniform, but pointwise.

 

[micromass]

A_n=\sup_{x \in (-ε, ε)} |s_n&#039;(x) - s&#039;(x)| = \sup_{x \in (-ε, ε)} 2|x|ne^{-(nx)^2}

That's some new notation on me. I've never seen sup used in such a way.

Let's take a compact interval $[-\varepsilon,\varepsilon]$ instead then. Then the $\sup$ is actually a $\max$ then.

So, when I'm asking for $\sup_{x \in [-\varepsilon,\varepsilon]} 2|x|ne^{-(nx)^2}$, I'm looking for the maximum value of this function.

So use calculus to find where the function $g_n(x)= 2|x|e^{-(nx)^2}$ takes on its maximum and calculate the maximum.

 

[STEMucator]

Let's take a compact interval $[-\varepsilon,\varepsilon]$ instead then. Then the $\sup$ is actually a $\max$ then.

So, when I'm asking for $\sup_{x \in [-\varepsilon,\varepsilon]} 2|x|ne^{-(nx)^2}$, I'm looking for the maximum value of this function.

So use calculus to find where the function $g_n(x)= 2|x|ne^{-(nx)^2}$ takes on its maximum and calculate the maximum.

Alright, that makes sense. I'll even pick the real open interval (-ε,ε) containing 0 since i see it doesn't matter now.

With how the interval and $|s_n(x) - s(x)|$ are defined I have 2 possible functions to examine.

Finding the maximums of these functions with regular calculus seems to be far beyond my understanding.

$|s_n(x) - s(x)| = 2xne^{-(nx)^2}$ for $0 ≤ x < ε$
$|s_n(x) - s(x)| = -2xne^{-(nx)^2}$ for $-ε < x ≤ 0$

How do I find the maximum? It's not like I can take the derivative and find criticals of this monster.

 

[micromass]

It's not like I can take the derivative and find criticals of this monster.

Why not?

 

[STEMucator]

Why not?

It looks terrible, but for the case where x is positive I took the derivative and found $x = ± \frac{1}{ n \sqrt{2}}$ to be the roots of the derivative.

For the case where x is negative, I took the derivative and found the same roots.

 

[micromass]

Ok, so the maximum is reached in $1/(n\sqrt{2})$. The value in that point is

f(1/(n\sqrt{2})) = \frac{2}{\sqrt{2}}e^{-1/2}

So, we see that

\sup_{x\in (-\varepsilon,+\varepsilon)} |s_n^\prime(x)-s^\prime(x)| = \frac{2}{\sqrt{2}} e^{-1/2}

Does this converge to $0$ if n goes to infinity?? (obviously not)

Can you deduce from this that convergence is not uniform?

 

[STEMucator]

Ok, so the maximum is reached in $1/(n\sqrt{2})$. The value in that point is

f(1/(n\sqrt{2})) = \frac{2}{\sqrt{2}}e^{-1/2}

So, we see that

\sup_{x\in (-\varepsilon,+\varepsilon)} |s_n^\prime(x)-s^\prime(x)| = \frac{2}{\sqrt{2}} e^{-1/2}

Does this converge to $0$ if n goes to infinity?? (obviously not)

Can you deduce from this that convergence is not uniform?

Ohhhhhh I see where you were going with this now. So since the supremum over (-ε,ε) of $|s_n(x) - s(x)|$ doesn't go to zero we know that the convergence of the derivatives can't possibly be uniform.

If it were to be uniform, the difference would go to zero if we subtract the sequence from the limiting function ( or vice versa ).