## If $f_n \rightarrow 0$ with $f_n ‘ \rightarrow g$, then is $g=0$ in some sense?

Suppose $$f_n :[a,b] \rightarrow \mathbb{R}$$ are differentiable functions (need not be $$C^1$$) with $$f_n \rightarrow 0$$, $$f_n ‘ \rightarrow g$$ pointwise. Can we say that $$g=0$$ in some sense? (Say, a.e.)

In particular, is it possible for $$g$$ to equal $$1$$ everywhere?

cf) Interchanging pointwise limit and derivative of a sequence of C1 functions This question deals with the $$C^1$$ case.

## Determine the generating function for the number $f_n$ of solutions of integers non-negative of the inequality

Determine the generating function for the number $$f_n$$ of solutions of integers non-negative of the inequality: $$3x_1+4x_2+x_3+7x_4\leq n$$ Approximate value of $$f_{10000}$$ ?

So I don’t have any idea of how to begin with this. The only thing that came to my mind that the solutions of the inequality would be the same of this system: $$3x_1+4x_2+x_3+7x_4+x_5\leq n$$ or the sum of the solutions with $$n,n-1,n-2,…$$

anyway, can I get a hint?

I believe I have to start doing some multiplications and arrive at some fractions looking like $$\frac{1}{1-x}$$ but I don’t know how to get there

## If $\|f_n\|_{p}\leq n^{-2}, f_n\in L^{p},\forall n\in\mathbb{N}$, does pointwise convergence of $f_n$ follow for a.e. $x\in\mathbb{R}$?

Let $$p\in[1,\infty)$$ and $$(f_n)_{n\in\mathbb{N}}\subset L^{p}(\mathbb{R})$$ such that $$\|f_n\|_{p}\leq n^{-2}$$ for all $$n\in\mathbb{N}.$$ Does $$(f_n)_{n\in\mathbb{N}}$$ necessarily converge pointwise a.e.? (Proof or counterexample.)

$$\textbf{Attempt:}$$ I think one can construct a counterexample to this. I was considering the following construction done by Saz in this post: https://math.stackexchange.com/a/3132844/595519.

I provide the details below:

Consider the following sequence of intervals $$[-1,0], [0,1],\left[-2,-\tfrac{3}{2}\right], \left[-\tfrac{3}{2},-1\right], \left[-1,-\tfrac{1}{2}\right],\left[-\tfrac{1}{2},0\right],\left[0,\tfrac{1}{2}\right],\left[\tfrac{1}{2},1\right],\left[1,\tfrac{3}{2}\right],\left[\tfrac{3}{2},2\right],\left[-3,-\tfrac{8}{3}\right],\left[-\tfrac{8}{3},-\tfrac{7}{3}\right],\dots,\left[\tfrac{7}{3},\tfrac{8}{3}\right],\left[\tfrac{8}{3},3\right],\dots$$ Denote the $$n$$th interval in the sequence by $$I_n.$$ By construction, it follows that $$m(I_n)\rightarrow 0$$ as $$n\rightarrow\infty.$$ Now denote the characteristic function of the $$n$$th interval above by $$1_{I_n},$$ and put $$f_n(x)=n^{-3}\cdot m(I_n)^{-1/p}\cdot1_{I_n}(x)$$. It follows that $$\left(\int_{\mathbb{R}}|f_n(x)|^p\,dx\right)^{1/p}=n^{-3},$$ and it follows that $$\|f_n-0\|_{p}\rightarrow 0$$ as $$n\rightarrow\infty,$$ so $$f_n\rightarrow 0$$ in the $$L^p$$-norm.

However, I note that the sequence $$(f_n)_{n\in\mathbb{N}}$$ I have constructed above fails, since it actually does converge pointwise to $$f\equiv 0,$$ and even worse, it does so everywhere.

Is there any way to save this example, or should I scratch it and try something else? Finally, is there a way to use decaying exponentials in this problem to make the sequence decrease fast enough?

Thank you for time and appreciate any feedback.

## Show that $f_n \rightarrow 0$ in $C([0, 1], \mathbb{R})$

I was given the following problem and was wondering if I was on the right track.

Let $$f_n(x) = \frac{1}{n} \frac{nx}{1 + nx}, \: 0 \le x \le 1$$

Show that $$f_n \rightarrow 0$$ in $$C([0, 1], \mathbb{R})$$.

I have this theorem that I figured I could use:

$$f_k \rightarrow f$$ uniformly on A $$\iff$$ $$f_k \rightarrow f$$ in $$C_b$$.

In this case, $$C_b$$ is the collection of all continuous functions on $$[0,1]$$. So if I can prove the function is uniformly continuous, this would prove that $$f_n \rightarrow 0$$. Can I apply this theorem like this to prove what I want? Also, if I can, would using the Weierstrass M test be best to prove uniform convergence here?

Thanks

## How do i compute $f_n = 3f_{n-1} + 2\sqrt{2f_{n-1}^2 – 2}$ for n around 10^18?

So I have the reccurence $$f_n = \begin{cases} 3f_{n-1} + 2\sqrt{2f_{n-1}^2 – 2}, &n > 0\ 3, &n > 1\ \end{cases}$$ and I need to compute it in $$\lg(n)$$, for n as big as $$10^{18}$$. I tried to reduce it to a closed form equation but I don’t see how that could be achieved.