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.

When does $|R_n|/|P_n| \rightarrow 0$ imply $|R_n/{\sim}| / |P_n/{\sim}| \rightarrow 0$ for an equivalence relation $\sim$?

Here is, maybe, a not very well-posed question:

Let $ P_n$ be a sequence of sets such that their sizes are non-decreasing.

Let $ R_n \subset P_n$ be a sequence of sets such their sizes are also non-decreasing.

Let $ \sim$ be an equivalence relation (the same relation) on each set $ P_n$ .

Assume that $ \frac{|R_n|}{|P_n|} \rightarrow 0$ as $ n \rightarrow \infty$ .

Question: Do we know under what hypotheses (on the $ P_n$ s, the equivalence relation $ \sim$ , $ R_n$ s, etc) does $ \frac{|R_n/\sim|}{|P_n/\sim|} \rightarrow 0$ ? If it is known, then I would love a resource.

Here is an example where the proportion of equivalence classes doesn’t go to zero even when the actual proportion of the subsets goes to zero:

Take $ R_n = \{2^i| 1 \leq i \leq n\}$ and $ P_n = \{\text{non-zero even numbers} \leq 2^n\}$ .

Define $ x \sim y$ if and only if $ x \equiv y \mod 4$ .

Then we know that $ \frac{|R_n|}{|P_n|} \rightarrow 0$ . But, $ |P_n/{\sim}| =|R_n/{\sim}| = 2$ for $ n > 1$ .

Hence, $ \frac{|R_n/\sim|}{|P_n/\sim|} \rightarrow 1$ .

Proof expression $ ((\neg C \vee B) \wedge ( A \rightarrow C) \wedge (B \rightarrow D) )\rightarrow \neg A \vee D $ by rules of interference


Task

Proof expression $ $ ((\neg C \vee B) \wedge ( A \rightarrow C) \wedge (B \rightarrow D) )\rightarrow \neg A \vee D $ $

Attempt to proof

$ $ 1. B \text{ (premise) } $ $ $ $ 2. B \rightarrow D (\text{premise}) $ $ $ $ 3. A \rightarrow D (\text{premise}) $ $ $ $ 4. D (\text{modus ponens}) $ $ $ $ 5.(\neg A \vee D) Add $ $

Is this correct?

Minimizing sequence $\Rightarrow$ Palais-smale sequence

Set $ F:\mathbb{R}^n\rightarrow \mathbb{R}$ a $ C^2$ -function that is bounded from below. Set $ x_n$ a minimizing sequence, i.e. $ F(x_n)\rightarrow \alpha = \inf F$ . I want to prove that under the assumption of $ \Vert D^2 F\Vert$ is bounded, $ x_n$ is a palais-smale sequence, i.e. $ F(x_n)\rightarrow \alpha = \inf F$ and $ \Vert DF(x_n)\Vert\rightarrow 0$ . I have come across the variational Ekeland principle, that maybe is useful here… however I am not sure and stucked. Any help is welcome! Thanks in advance

Prove that: $\mathbb{K}[x]$ has a zero divisor $\Rightarrow$ $\mathbb{K}$ has a zero divisor


The problem

I’ve been trying to solve the above problem. There seems to already be some work regarding zero divisors in polynomial rings over here, but I’m not sure it is applicable to me, because it looks specifically for some $ r \in \mathbb{K}$ such that for a zero divisor $ F \in \mathbb{K}[x]$ holds $ F \cdot r = 0$ .

I, on the other hand, am looking to prove that if $ \mathbb{K}[x]$ has zero divisors, it implies $ \mathbb{K}$ has zero divisors.

My ideas so far

I’ve thought of the following: Let $ p(x), q(x) \in \mathbb{K}[x], \quad p(x) \neq 0 \neq q(x), \quad p(x) \cdot q(x) = 0$ .

Since the product of $ p(x)$ and $ q(x)$ is zero, it shouldn’t matter which $ x \in \mathbb{K}$ you choose. The product should still be zero regardless. So, freely choosing a single $ x \in \mathbb{K}$ , we should be able to reduce both $ p(x)$ and $ q(x)$ to a value in $ \mathbb{K}$ , with the equation still holding. As such, we’d have found our zero divisors in $ \mathbb{K}$ .

Unfortunately, we have to watch out so that $ x$ isn’t a zero of either $ p(x)$ or $ q(x)$ , or else we’d reduce one of them to $ 0$ and as such we wouldn’t be finding a zero divisor. This leaves me stuck.

Obviously, for my approach I’d have to show that there is an $ x \in \mathbb{K}$ , so that neither $ p(x)$ nor $ q(x)$ reduce to $ 0$ .

Does $g_n \rightarrow 0$ converge weakly?

This is where I am stuck while solving another problem.

Let $ T:L^1 \rightarrow X$ be an operator such that $ T|_{L^2(\mu)}$ is compact.

Suppose $ f_n$ be a sequence in $ L^1$ such that $ f_n \rightarrow 0$ weakly. Then set $ g_n=f_n\mathbb{1}_{A_n}$ where $ A_n=\{x : |f_n(x)|<M\}$ for some fixed $ M$ .

Then how can we say that $ \|Tg_n\| \rightarrow 0$ in norm?

It would be very nice if we could say $ g_n \rightarrow 0$ weakly.

Thanks for the help!

What will be $\lim\limits_{x \rightarrow 0^+} \left ( 2 \sin {\frac {1} {x}} + \sqrt x \sin {\frac {1} {x}} \right )^x.$

Evaluate

$ $ \lim\limits_{x \rightarrow 0^+} \left ( 2 \sin {\frac {1} {x}} + \sqrt x \sin {\frac {1} {x}} \right )^x.$ $

I tried by taking log but it wouldn’t work because there are infinitely many points in any neighbourhood of $ 0$ where $ \ln \left ( \sin {\frac {1} {x}} \right )$ doesn’t exist. How to overcome this situation?

Any help will be highly appreciated.Thank you very much for your valuable time.

Proof of $ f:I_n \rightarrow I_m \Rightarrow n = m$

I am trying to improve my proof-writing skills.

Would the following proof be correct for the $ f:I_n \rightarrow I_m \Rightarrow n = m$ bit?

Problem:

Prove that the notion of number of elements of a nonempty finite set is a well defined concept. More precisely, prove that there exists a bijection $ f:I_n \rightarrow I_m $ if and only if $ n = m$ .

Attempt:

First prove $ f:I_n \rightarrow I_m \Rightarrow n = m$ .

Assume that $ f $ is a bijective function such that $ f:I_n \rightarrow I_m $

By definition, since $ f$ is injective, $ \forall a, b \in I_n, f(a) = f(b) \Rightarrow a = b $ , where $ f(a), f(b) \in I_m $ . Therefore every element in $ I_m$ corresponds to at most one element in $ I_n$ . $ \quad (1)$

Also by definition, since $ f$ is surjective, $ \forall b \in I_m, \exists a \in I_n $ . That is every element in $ I_m$ corresponds to at least one element in $ I_n$ . $ \quad (2)$

Now if $ n > m$ , then by $ (2)$ some element in $ I_n$ corresponds to an element in $ I_m$ which is already mapped to. This cannot be true.

Likewise, if $ n < m$ then by $ (1)$ some element in $ I_n$ corresponds to more than one element in $ I_m$ . Again, this cannot be true.

Therefore $ n = m$ .