To show UC to zero, given an $\varepsilon$, you want to show that there exists an $N$ such that $|f_k(x)| < \varepsilon$ for all $k > N$ and all $x$ in the given domain
The key thing to note is that $\ln$ is monotone increasing, meaning each $f_i(x)$ is monotone decreasing. Notice that this means $\text{Sup}(f_k(x)) = f_k(1)$
Thus, it suffices to show that $f_k(1) \rightarrow 0$ as $k \rightarrow \infty$.
I want to prove that if the set $A$ is bounded then $f(A)$ is also bounded, where $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous.
I have done the following: We suppose that $f(A)$ is not bounded. So, there is a sequence $\{x_n\}_{n=1}^{\infty}$ in $f(A)$ with $\lim x_n=\infty$, or not?
Since $\{x_n\}_{n=1}^{\infty} \in f(A)$ we have that $\{f^{-1}(x_n)\}_{n=1}^{\infty} \in A$. Therefore, $\{f^{-1}(x_n)\}_{n=1}^{\infty}$ is bounded. From Bolzano-Weierstrass there is a convergent subsequence, let $\{f^{-1}(x_{n_k})\}_{k=1}^{\infty}$ with $f^{-1}(x_{n_k})\rightarrow y\in \mathbb{R}$.
That's an interesting idea you're trying to do @MaryStar. One problem I see is that $f$ might not be injective? That preimage set might be uncountable
Potentially salvageable though; let me think
@KasmirKhaan, that'll be where the supremum of all the function evals for all $x$ in the domain is at. If we want to show U.C. to zero, it suffices to show this maximum goes to zero as $k \rightarrow \infty$
By the same reasoning as in part A of the previous question
@SimpleArt have shown it also by an other way... Since $A$ is bounded there are the $a=\inf A$ and $b=\sup A$ and so $A\subseteq [a,b]$. Then $f(A)\subseteq f([a,b])$. Since the continuous image of a closed interval is a closed and bounded interval, we get the desired result...
Yeah, you'd need to find some other route, or maybe it's not UC at all. A good giveaway is when every f_k has an asymptote at some value, e.g. @KasmirKhaan
Is this the correct way to make this visualization of the derivative of sine more... rigorous? At least, for $u\in(0,\pi/2)$.
Borrowed from Proofs without words. To try to make this rigorous, I argued that when $u\pm\Delta u$ is in the first quadrant, that we have the following geometric...
If your set is bounded you can find a compact superset of it, since f is continuous it will send this into another compact set, which needs to be bounded
@AlessandroCodenotti This is closed and bounded, and so compact, right? And how do we know that f send a compact set into another compact set? How could we prove this?
How to prove: Continuous function maps compact set to compact set using real analysis?
i.e. if $f: [a,b] \rightarrow \mathbb{R}$ is continuous, then $f([a,b])$ is closed and bounded.
I have proved the bounded part. So now I need some insight on how to prove $f([a,b])$ is closed, i.e. $f([a,b])...
Random curiosity to anyone who might see this. If one had a function that gave the number of prime numbers between 0 and x how useful would that be for solving things like legendre's conjecture. Note. I'm only curious if such a function is useful not how it would be useful.
Update:
If one may recall the following series:
$$\sum_{n=0}^\infty\frac1{n^2+x}=\frac1{2x}+\frac\pi{2\sqrt x\tanh(\pi\sqrt x)}$$
Then differentiating both sides $k$ times yields
$$\sum_{n=0}^\infty\frac1{(n^2+x)^{k+1}}=\frac1{2x^{k+1}}+\frac{(-1)^k\pi}{2\times k!}\frac{d^k}{dx^k}\frac1{\s...
This forum brings together a broad enough base of mathematicians to collect a "big list" of equivalent forms of the Riemann Hypothesis...just for fun. Also, perhaps, this collection could include statements that imply RH or its negation.
Here is what I am suggesting we do:
Construct a more o...
Feel free to skip to the highlighted parts and the ending to see the formula in action.
Suppose we had a continuous and differentiable function that satisfied the following equation:
$$f(x,p)=f(x-1,p)+x^p,\quad f(0,p)=0$$
Differentiating with respect to $x$, we get
$$f'(x,p)=f'(x-1,p)+px^{p...
This post is one of my favorites
And I even found it being used to close more basic questions
Is this the correct way to make this visualization of the derivative of sine more... rigorous? At least, for $u\in(0,\pi/2)$.
Borrowed from Proofs without words. To try to make this rigorous, I argued that when $u\pm\Delta u$ is in the first quadrant, that we have the following geometric...
@robjohn Since you have answers on two of those posts, perhaps you will be willing to have a look and say whether it is better to merge or to close as duplicates in some direction: chat.stackexchange.com/transcript/2165/2017/1/5 Thanks!
Since I am trying to get the bounty room going, perhaps a reasonable thing could be to promote it here. So here is link to the relevant meta post and here is the room.
And I will also repost this here - in case some differential geometer is around (I guess Ted Shifrin is not yet back):
Suppose $$f(x)= \int_{0}^{x}tf(t)dt$$
I apply integration by parts on the Right hand side.
$$f(x)= \left[t\int_{lower?}^{upper?} f(t)dt\right]_0^x - \int_0^x\left(\int_{lower?}^{upper?}f(t)dt\right)dt$$
I cannot understand the lower bound and upper bound of the integrals (which I showed with qu...
Is this the correct way to make this visualization of the derivative of sine more... rigorous? At least, for $u\in(0,\pi/2)$.
Borrowed from Proofs without words. To try to make this rigorous, I argued that when $u\pm\Delta u$ is in the first quadrant, that we have the following geometric...
How to prove: Continuous function maps compact set to compact set using real analysis?
i.e. if $f: [a,b] \rightarrow \mathbb{R}$ is continuous, then $f([a,b])$ is closed and bounded.
I have proved the bounded part. So now I need some insight on how to prove $f([a,b])$ is closed, i.e. $f([a,b])...
This forum brings together a broad enough base of mathematicians to collect a "big list" of equivalent forms of the Riemann Hypothesis...just for fun. Also, perhaps, this collection could include statements that imply RH or its negation.
Here is what I am suggesting we do:
Construct a more o...
Suppose $$f(x)= \int_{0}^{x}tf(t)dt$$
I apply integration by parts on the Right hand side.
$$f(x)= \left[t\int_{lower?}^{upper?} f(t)dt\right]_0^x - \int_0^x\left(\int_{lower?}^{upper?}f(t)dt\right)dt$$
I cannot understand the lower bound and upper bound of the integrals (which I showed with qu...
