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