10:25 AM
There is a new tag called , even tag-excerpt and tag-wiki were created.
7

If we count the palindromes among first $n$ numbers in all natural bases $b>1$, but ignore the one-digit-palindromes (ones that fill up the black triangle on the picture), Which base $b$ will contain the most palindromes? The $x$-axis represents the numbers, and the $y$-axis repr...

11:01 AM
Another new tag is , probably created by Leilva.
4

The is the standard version of Young's inequality $$ab \leq \frac{a^p}{p}+\frac{b^q}{q}$$ for $a,b,p,q >0$ and $$\frac{1}{p}+\frac {1}{q}=1$$ But there is another formula called generalization of Young inequality. Why it's a generalization? let $f$ denote a real-valued, continuous and strictl...

2

According to wikipedia, "let $f$ denote a real-valued, continuous and strictly increasing function on [0, c] with c > 0 and f(0) = 0. Let $f^{−1}$ denote the inverse function of $f$. Then, for all a ∈ [0, c] and b ∈ [0, f(c)]," $$ab \le \int_0^a f(x)\,dx + \int_0^b f^{-1}(x)\,dx$$ I am wonderi...

5

Recall that the convolution of two functions is given by $$f*g(y)=\int f(x)g(y-x)dx.$$ The well known inequality known as Young's inequality, say that $$\|f*g\|_r\leq\|f\|_p\cdot\|g\|_q$$ provided $\frac 1p + \frac 1q = 1 + \frac 1r$ and $1\le p,q,r\le\infty$. Obvious implications is that $... 12 Is there a geometric interpretation of Young's inequality, $$ab \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}$$ with$\dfrac{1}{p}+\dfrac{1}{q} = 1$? My attempt is to say that$ab$could be the surface of a rectangle, and that we could also say that:$\dfrac{a^{p}}{p}=\displaystyle \int_{0}^{a}x^{p-1...

5

I was doing some problems from Rudin's Principles of Mathematical Analysis and came across a problem in which he asks you to prove Hölder's inequality via Young's inequality: If $u$ and $v$ are nonnegative real numbers, and $p$ and $q$ are positive real numbers such that $\displaystyle \frac{... 7 Let$x, y, z \geqslant 0$and let$p, q, r > 1$be such that $$\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1.$$ How can one show that under these hypotheses we have $$xyz \leqslant \frac{x^p}{p} + \frac{y^q}{q} + \frac{z^r}{r}$$ with equality if and only if$x^p = y^q = z^r$, using twice the s... 1 I need to prove that if$1 < p < \infty$and$a, b \geqslant 0$then $$ab \leqslant \frac{a^{p}}{p} + \frac{b^{q}}{q}$$ where$\frac 1p+\frac 1q=1$. I fix$b$and maximize the function$f(a) = ab - \frac{a^{p}}{p}$, but the maximum I find is$b^{q}$with$q = \frac{1}{p-1}$. I have no idea how... 7 I am trying to understand how sharp Young's inequality for convolution is. The inequality says$||f \ast g||_r \leq ||f||_p ||g||_q$where as$1/p+1/q = 1+1/r$. Actually, there are a couple of papers (for example: Sharpness in Young's inequality for convolution) talking about the case of$p, q>1...

2

I am trying to prove Young's Inequality by considering the function $$h(u) = \frac{u^p}{p} + \frac{C^q}{qu^q}$$ for $C,u>0$ and $p,q >1$. We also require $$\frac{1}{p}+\frac{1}{q}=1$$ so that $pq=p+q$. I starting by using straightforward calculus to show that $h$ has a global minimum at $u=C^{\fr... 12 Part of an exercise to prove Holder's inequality in Rudin involves proving Young's Inequality... That is, given$\frac{1}{p}+\frac{1}{q} = 1$, prove $$ab \leqslant \frac{a^p}{p} + \frac{b^q}{q}.$$ Here's my attempt at a proof: Let $$f(x) = \frac{x^p}{p} + \frac{b^q}{q} -bx$$ then,$\$f...

1

I am referring to the inequality: Young's inequality The standard version for increasing functions. I read the article of Young and also a generalization of this claim in Hardy, Littlewood and Polya's Inequalities. But I don't see that Young proves rigorously his claim, and in Zygmund's Trigono...

0

There are certainly many inequalities which are rather important and useful and which appear frequently in various areas in mathematics (AM-GM, Jensen, Cauchy-Schwarz, etc.) The question I want to ask is whether some of them would be also useful as tags on this site. And if yes, for which of them...

11:19 AM
I have posted the above question as a reaction to the creation of and . Mainly because I do not want that we get into similar situation as with , where after the discussion on meta lead to removal of the tag, we had to bump about 70 questions.

7 hours later…
5:52 PM
Previously I have collected some comment on renaming (removing) tags without bumping here:

### Removing tags without bumping (by merging)

Jan 7 at 13:50, 41 minutes total – 7 messages, 1 user, 0 stars

Bookmarked Jan 7 at 21:32 by Martin Sleziak

I should probably add also this exchange with quid:
"Mods can rename tags." They actually cannot literally rename a tag (or I never found out how to do this). But via judicious merging an equivalent effect can be achieved without much effort and bumping. — quid Jan 17 at 2:26
@quid The linked answer by Jeff Atwood explicitly says: "ask here on a meta for a moderator to do it as a rename or merge which does not bump questions". Maybe I misunderstood the answer. This wording can be probably understood in two ways: A) He uses both words rename and merge as two possible names for the same thing. B) He mentions two possible actions by moderators. (But does not explain the difference between them.) I interpreted the answer as B), but from your comment I see that he probably mean A). — Martin Sleziak Jan 17 at 2:34
@quid I found a meta.SE post which confirms what you said: How to rename a tag? I hope that after the edits, my answer is more-or-less correct. (As I never was a moderator on a SE site, the answer is based on what I read on various meta sites and on what I've heard from diamond users - not on the firsthand experience.) — Martin Sleziak Jan 17 at 10:57
Thank you, for the update, and, sorry for the nit-picking. But then I know you like to be precise, so I hope it was not out-of-place. — quid Jan 17 at 11:23
And perhaps also links to two relevant questions:
2

Can you change the name of a tag? I recently came across the tag sage, which I think should be renamed to sagemath to be more clear. Is it possible to change the name of a tag? If so, how?

36

What do I need to do to rename a tag? I see some people post a question here and ask for a tag to be renamed. Is there some other way?