Pointwise vs. uniform convergence

[Rectifier]

The problem
I am trying determine wether $f_n$ converges pointwise or/and uniformly when $f(x)=xe^{-x}$ for $x \geq 0$.

Relevant equations
$f_n$ converges pointwise if $\lim_{n \rightarrow \infty} f_n(x) = f(x) \ \ \ \ \$ (1)
$f_n$ converges uniformly if $\lim_{n \rightarrow \infty} || f_n - f || = 0 \ \ \ \ \$ (2)

The attempt
$f_n(x) = f(nx) =nxe^{-nx}$

Pointwise? (1)
$f(x) = \ \lim_{n \rightarrow \infty} f_n(x) = \lim_{n \rightarrow \infty} nxe^{-nx} = \lim_{n \rightarrow \infty} \frac{nx}{e^{nx}} = 0$

Uniformly? (2)
$0 = \ \lim_{n \rightarrow \infty} || f_n - f || = \lim_{n \rightarrow \infty} || nxe^{-nx} - xe^{-x} ||$:I am not sure how to continue from here and wether the last step was correct:

$\lim_{n \rightarrow \infty} || nxe^{-nx} - xe^{-x} || = 0$

 

[member 587159]

Please post the exact wording of the question. Do not post extra info in "Attempt". This makes it very confusing to Homework Helpers.

It appears to me that the pointwise limit of $f_n$ is the 0 function.

Now, you have to show that

$$\forall \epsilon > 0: \exists n_0: \forall n \geq n_0: \forall x: \left|\frac{e^{nx}}{nx}\right| < \epsilon$$

to find uniform convergence.

 

[Rectifier]

The exact wording of this question is in another language. Sorry, if you find my translation and formatting confusing. I have edited the OP and tried to make my point clearer.

 

[member 587159]

The exact wording of this question is in another language. Sorry, if you find my translation and formatting confusing. I have edited the OP and tried to make my point clearer.

Does it ask to determine whether the sequence of functions $(f_n)_n$ converge towards $f$ given by $f(x) = xe^{-x}$?

If so, clearly the answer is no, since it converges to $g = 0$, and limits are unique.

 

[Rectifier]

Does it ask to determine whether the sequence of functions $(f_n)_n$ converge towards $f$ given by $f(x) = xe^{-x}$?

This chapter is about sequence of functions but the notation is different. According to my book its $S_nf(x) = f(nx)$ and sometimes just $f_n(x)$. The question is to determine how (pointwise/uniform) it converges and not whether it does it.

 

[member 587159]

This chapter is about sequence of functions but the notation is different. According to my book its $S_nf(x) = f(nx)$ and sometimes just $f_n(x)$. The question is to determine how (pointwise/uniform) it converges and not whether it does it.

But where does the function $f(x) = xe^{-x}$ come from (see your OP)? Is this included in the problem? Or is your guess for the pointwise limit?

 

[Rectifier]

But where does the function $f(x) = xe^{-x}$ come from (see your OP)? Is this included in the problem? Or is your guess for the pointwise limit?

Its in the problem. Let me try to reword the problem once again:
Examine if the sequence of funtions $f_n(x)$ when $f(x) = xe^{-x}$ converges pointwise for $x \geq 0$. In case the sequence of functions does converge get the limit function and determine if the sequence converges uniformly.

 

[member 587159]

Its in the problem. Let me try to reword the problem once again:
Examine if the funtion $f(x) = xe^{-x}$ converges pointwise for $x \geq 0$. In case the sequence of functions does converge get the limit function and determine if the sequence converges uniformly.

That's my problem! WHAT sequence of functions? The problem does not seem to include that. It is meaningless to consider pointwise convergence of one particular function, unless they consider it as a constant sequence of that function, but then the problem is trivial so I think the problem is flawed.

 

[Rectifier]

Sorry once again. I made an edit just a moment ago. You were quick to reply.

Secuence of funtions is definied as $f_n(x) = f(nx)$ earlier in my book.

 

[member 587159]

Sorry once again. I made an edit just a moment ago. You were quick to reply.

Secuence of funtions is definied as $f_n(x) = f(nx)$ earlier in my book.

Okay, let us get this thing straight. I will formulate the problem as I understand it:

Let $f: \mathbb{R}^+ \to \mathbb{R}$ be given by $f(x) = xe^{-x}$. Define a sequence of functions $(f_n: \mathbb{R}^+ \to \mathbb{R})_n$ by letting $f_n(x) = f(nx) = nxe^{-nx}$. Determine if $f_n \to f$ pointswise and/or uniformly.

Is this the correct problem?

 

[fresh_42]

The exact wording of this question is in another language. Sorry, if you find my translation and formatting confusing. I have edited the OP and tried to make my point clearer.

This doesn't explain the fact that the $f_n$ are not defined. It would make more sense to me to ask, whether $f(x)$ is uniformly continuous or not.

 

[nuuskur]

I am confused. We'd attempt to show $f_n(x)\xrightarrow[n\to\infty]{} 0, x\geq 0$ (this does hold), why would we even further contemplate whether $f_n$ converges to anything else (pointwise or uniformly) other than $g(x) = 0\neq xe^{-x}, x\geq 0$?

 

[member 587159]

I am confused. We'd attempt to show $f_n(x)\xrightarrow[n\to\infty]{} 0, x\geq 0$ (this does hold), why would we even further contemplate whether $f_n$ converges to anything else (pointwise or uniformly) other than $g(x) = 0\neq xe^{-x}, x\geq 0$?

Indeed, this was what I was addressing in #4.

 

[Rectifier]

Okay, let us get this thing straight. I will formulate the problem as I understand it:

Let $f: \mathbb{R}^+ \to \mathbb{R}$ be given by $f(x) = xe^{-x}$. Define a sequence of functions $(f_n: \mathbb{R}^+ \to \mathbb{R})_n$ by letting $f_n(x) = f(nx) = nxe^{-nx}$. Determine if $f_n \to f$ pointswise and/or uniformly.

Is this the correct problem?

Yes, I think so.

 

[member 587159]

Yes, I think so.

Okay. Then the problem is actually very easy. See post #4 and #12.

 

[Rectifier]

If so, clearly the answer is no, since it converges to $g = 0$, and limits are unique.

We'd attempt to show $f_n(x)\xrightarrow[n\to\infty]{} 0, x\geq 0$ (this does hold), why would we even further contemplate whether $f_n$ converges to anything else (pointwise or uniformly) other than $g(x) = 0\neq xe^{-x}, x\geq 0$?

What is $g$ here? Is that f(x) in (1) ?

$f_n$ converges pointwise if $\lim_{n \rightarrow \infty} f_n(x) = f(x) \ \ \ \ \$ (1)
$f_n$ converges uniformly if $\lim_{n \rightarrow \infty} || f_n - f || = 0 \ \ \ \ \$ (2)

 

[member 587159]

What is $g$ here? Is that f(x) in (1) ?

$g = 0$ is shorthand notation for $g: \mathbb{R}^+ \to \mathbb{R}: x \mapsto 0$

It is easy to see that this is the pointwise limit, since for any $x$ in the domain

$\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} nxe^{-nx} = 0 = g(x)$

 

[Rectifier]

And now I am somehow supposed to show that the function series does not cenverge uniformly by showing that
$\lim_{n \rightarrow \infty} || f_n - g || = \sup_{x \geq 0} | f_n - g | \neq 0 \ \ \ \ \$
So in other words, should I just insert $g = 0$ and $f_n = nxe^{-nx}$ since that's what we got earlier?

 

[member 587159]

And now I am somehow supposed to show that the function series does not cenverge uniformly by showing that
$\lim_{n \rightarrow \infty} || f_n - g || = \sup_{x \geq 0} | f_n - g | \neq 0 \ \ \ \ \$
So in other words, should I just insert $g = 0$ and $f_n = nxe^{-nx}$ since that's what we got earlier?

You are already done.

Indeed, it is true:

$f_n \to f$ uniform $\implies f_n \to f$ pointwise

and by applying contraposition on this, we find that $f$ can't be a uniform limit of $(f_n)_n$, because it isn't a pointwise limit.