Limit and Integration of ##f_n (x)##

[songoku]

Homework Statement

Please see below

Relevant Equations

Limit
Integration

My attempt:
(a)
I don't think I completely understand the question. By "evaluate $\lim_{n\to \infty f_n (x)}$", does the question ask in numerical value or in terms of $x$?

As $x$ approaches 1 or -1, the value of $f_n (x)$ approaches zero. As $x$ approaches zero, the value of $f_n (x)$ approaches $\frac{n+1}{2}$ so if $n \to \infty$, then $f_n (0) \to \infty$.

There would be a certain value of $x \in [-1,1]$ where $\lim_{n\to \infty} f_n (x)=\infty$ so the limit does not exist.

Does it make any sense?(b)
$$\lim_{n\to \infty} \int_{-1}^{1} f_n (x) dx$$
$$=\lim_{n\to \infty} \int_{-1}^{1} \frac{n+1}{2} (1-|x|)^n dx$$
$$=\lim_{n\to \infty} \frac{n+1}{2} \int_{-1}^{1} (1-x)^n dx$$
$$=\lim_{n\to \infty} \frac{n+1}{2} \left[-\frac{1}{n+1} (1-x)^{n+1}\right]^{1}_{-1} dx$$
$$=\lim_{n\to \infty} (2)^n$$

The limit does not converge so it does not exist. Is this correct?

Thanks

Edit: wait, I realize my mistake for (b). I will revise it in post#2

 

[songoku]

(b)
$$\lim_{n \to \infty} \int_{-1}^{1} f_n (x) dx$$
$$=\lim_{n\to \infty} \int_{-1}^{1} \frac{n+1}{2} (1-|x|)^n dx$$
$$=\lim_{n\to \infty} \frac{n+1}{2}\left(\int_{-1}^{0} (1+x)^n dx + \int_{0}^{1} (1-x)^n dx\right)$$
$$=\lim_{n\to \infty} \frac{n+1}{2} \left(\frac{1}{n+1}[(1+x)^{n+1}]_{-1}^{0} - \frac{1}{n+1} [(1-x)^{n+1}]_{0}^{1}\right)$$
$$=1$$

 

[WWGD]

Have you seen some of the results/theorems regarding convergence of Integrals, like Monotone, Dominated Convergence, etc?

 

[PeroK]

(a)
I don't think I completely understand the question. By "evaluate $\lim_{n\to \infty f_n (x)}$", does the question ask in numerical value or in terms of $x$?

This is called the pointwise limit or pointwise convergence. For each $x$ you have a sequence $f_n(x)$ and you are asked to calculate the limit of this sequence.

As $x$ approaches 1 or -1, the value of $f_n (x)$ approaches zero. As $x$ approaches zero, the value of $f_n (x)$ approaches $\frac{n+1}{2}$ so if $n \to \infty$, then $f_n (0) \to \infty$.

There would be a certain value of $x \in [-1,1]$ where $\lim_{n\to \infty} f_n (x)=\infty$ so the limit does not exist.

Does it make any sense?

This is not quite right. What is $\lim_{n \to \infty} f_n(0)$? And, for $x \ne 0$, what is $\lim_{n \to \infty} f_n(x)$?

(b)
$$\lim_{n \to \infty} \int_{-1}^{1} f_n (x) dx$$
$$=\lim_{n\to \infty} \int_{-1}^{1} \frac{n+1}{2} (1-|x|)^n dx$$
$$=\lim_{n\to \infty} \frac{n+1}{2}\left(\int_{-1}^{0} (1+x)^n dx + \int_{0}^{1} (1-x)^n dx\right)$$
$$=\lim_{n\to \infty} \frac{n+1}{2} \left(\frac{1}{n+1}[(1+x)^{n+1}]_{-1}^{0} - \frac{1}{n+1} [(1-x)^{n+1}]_{0}^{1}\right)$$
$$=1$$

That's right, although you could have saved some work by noting that the function is even (symmetrical about the y-axis).

If you are wondering about the purpose of this question, you have a sequence of functions whose limit looks like the Dirac Delta function.

 

[songoku]

Have you seen some of the results/theorems regarding convergence of Integrals, like Monotone, Dominated Convergence, etc?

I have not

This is called the pointwise limit or pointwise convergence. For each $x$ you have a sequence $f_n(x)$ and you are asked to calculate the limit of this sequence.

This is not quite right. What is $\lim_{n \to \infty} f_n(0)$? And, for $x \ne 0$, what is $\lim_{n \to \infty} f_n(x)$?

$$\lim_{n \to \infty} f_n(0)=\lim_{n \to \infty} \frac{n+1}{2} (1)^n=\lim_{n \to \infty} \frac{n+1}{2} \to \text{diverge}$$

For $x \ne 0, \lim_{n \to \infty} f_n(x)=0$

So:
$$\lim_{n\to \infty} f_n (x) = f(x) =
\begin{cases}
\text{does not exist }&\text{if } x = 0 \\
0 & \text{if } x \neq 0, x \in [-1,1]
\end{cases}
$$

Is that correct? Thanks

 

[fresh_42]

I have not$$\lim_{n \to \infty} f_n(0)=\lim_{n \to \infty} \frac{n+1}{2} (1)^n=\lim_{n \to \infty} \frac{n+1}{2} \to \text{diverge}$$

For $x \ne 0, \lim_{n \to \infty} f_n(x)=0$

So:
$$\lim_{n\to \infty} f_n (x) = f(x) =
\begin{cases}
\text{does not exist }&\text{if } x = 0 \\
0 & \text{if } x \neq 0, x \in [-1,1]
\end{cases}
$$

Is that correct? Thanks

Yes. If $x\neq 0$ then $1-|x| = r\in [0,1)$ and $r^n$ goes faster to zero than $n$ goes to infinity.

The meaning of this exercise is that $1= \lim \int \neq \int \lim =0.$

Pointwise convergence does in general not allow for exchange limits and integrals.

The theorem that grants the exchange requires $|f_n|<h$ and $\int h <\infty .$

 

[songoku]

Thank you very much for all the help and explanation WWGD, PeroK, fresh_42

 

[fresh_42]

Thank you very much for all the help and explanation WWGD, PeroK, fresh_42

P.S.: The phenomenon is called the vanishing mass at infinity. The "buckle" can vanish to the left or right, e.g. if you consider functions like $g_n =\chi([0,1])-\chi([n,n+1])$ where $\chi$ is the indicator function ($=1$ on the interval, and $=0$ elsewhere), or as in the case of the $f_n$ above to the top by getting larger and larger and slimmer and slimmer at the same time.

 

[PeroK]

So:
$$\lim_{n\to \infty} f_n (x) = f(x) =
\begin{cases}
\text{does not exist }&\text{if } x = 0 \\
0 & \text{if } x \neq 0, x \in [-1,1]
\end{cases}
$$

Is that correct? Thanks

I would say:
$$\lim_{n\to \infty} f_n (x) =
\begin{cases}
+\infty&\text{if } x = 0 \\
0 & \text{if } x \neq 0, x \in [-1,1]
\end{cases}
$$