# Tagged Questions

For questions about proving and manipulating functional inequalities.

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$\DeclareMathOperator{\diam}{diam}\newcommand{\norm}[1]{\lVert#1\rVert}\newcommand{\abs}[1]{\lvert#1\rvert}$For $u \in C^{1}(\overline{\Omega})$, for $\Omega\subset \subset \mathbb{R^{n}}$ a bounded ...
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### If $f(x)\leq f(f(x))$ for all $x$, is $x\leq f(x)$?

If I have $f(x)\leq f(f(x))$ for all real $x$, can I deduce $x\leq f(x)$? Thank you.
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### submodular-like functions on $\mathbb{R}$

The definition of submodularity led me to define a type of function on numbers (I suppose ordered rings, but consider the reals for now) as $f(x) + f(y) \ge f(x+y) + f(xy)$. Such functions exist: for ...
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### If $f(0)=f(1)=1$ and $|f(a)-f(b)| < |a-b|$ then $|f(a)-f(b)| < \frac{1}{2}$

Problem: $f$ be a function on $[0,1]$ such that $f(0)=f(1)=1$ and $f(a)-f(b) < |a-b|$ for all $a$ not equal to $b$. Prove that $|f(a)-f(b)| < \frac{1}{2}$. My attempt: Things I observed are ...
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### When does $f\left [g\left (x \right ) \right ]=g\left [ f\left ( x \right ) \right ]$ have roots.

I met a interesting equation: $$\sin\left [\cos\left (x \right ) \right ]=\cos\left [\sin\left (x \right ) \right ]$$ (And of course, the equation has no roots). So, Let $f\left ( x \right )$ ...
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### Inequality $\Gamma(\alpha+x)\Gamma(\beta+y)\leq C(\alpha; \beta)\Gamma(x+y-1)$

I need to prove $$\Gamma(\alpha+x)\Gamma(\beta+y)\leq C(\alpha; \beta)\Gamma(x+y),$$ for $x>\alpha$ and $y>\beta$ with $0<\alpha,\beta \leq \frac{1}{2}$ constants, and $C(\alpha, \beta)$ is a ...
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### Proving an inequality involving discrete variables

I'm trying to show that the following inequality holds $$\frac{1-x^{n}}{1-x^{n+1}}\geq\frac{\sum_{i=0}^{n-2}x^{i}(1-x_{1}^{n-(i+1)})}{\sum_{i=0}^{n-1}x^{i}(1-x_{1}^{n-i})},$$ where $n$ is a ...
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### E(X)*E(1/X) <= $(a+b)^2$/4ab

I've worked on the following problem and have a solution (included below), but I would like to know if there are any other solutions to this problem, particularly more elegant solutions that apply ...
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Are there two constants $C_1$, $C_2>1$ such that for large enough $n$ $$\frac{p_{n+1}^2-p_{n+1}^{-C_1}}{p_{n+1}^2-1}>\frac{2 C_2 p_{n+1} \log p_{n+1}-1}{2 C_2 p_{n} \log p_{n}-1} \left(\frac{p_n ... 1answer 79 views ### Functional inequality and one identity I'm a high school student from Bonn, Germany and I have to solve the following problem: If g:R \rightarrow R  is a function with the property g(ab)-ag(b)\leq bg(a), for all real numbers a and b, ... 1answer 43 views ### Finding appropriate function In order to obtain an estimate I'm wondering if there is a positive function f:\mathbb{R}\to \mathbb{R}_+ such that f(x) < |x| for |x| large enough (say |x|\ge C_0>0) so that the ... 0answers 44 views ### Find how many such complex numbers exist Let f:\mathbb{C}\to\mathbb{C} be defined by f(z)=z^2+iz+1. How many complex numbers z are there such that \text{Im}(z)>0 and both the real and the imaginary parts of f(z) are integers ... 0answers 62 views ### Convex functions: bounding the difference Suppose you are given a convex function f: R^d \rightarrow R. Let us say you are given x,x' \in R^d and there are x_1, x_2, \ldots, x_n \in R^d such that$$\sum_{i=1}^n (x_i - x') = x - x'.$$... 2answers 2k views ### Inequality involving Square Root \sqrt{3x-x^2}<4-x I know I can't simply square both sides of an inequality. I have narrowed down the possible values of x => x belongs to [0,3] because the expression inside the square root ... 2answers 193 views ### Finding a function satisfying a certain inequality This is a continuation of this post where I tried to find a function f(n) that would satisfy the induction step of an inductive argument and it was shown that such function does not exist. Trying ... 1answer 234 views ### Why are so many inequalities and estimate in PDEs? I want to study PDE, but when I read some PDE books, I notice that there are so many inequalities there and always lots of estimates. I really want to know why before I continue my study. I mean, ... 1answer 112 views ### Does this inequality have any solutions in \mathbb{N}? Does this (number-theoretic) inequality have any solutions x \in \mathbb{N}?$$\frac{\sigma_1(x)}{x} < \frac{2x}{x + 1}$$Notice that we necessarily have x > 1. 1answer 84 views ### Functional inequality \sum_{1\le i<j\le n}f(x_i+x_j)\ge \frac{n(n-1)}{2}f(a_1x_1+a_2x_2+…+a_nx_n) Let n\in\mathbb N, n\ge 2. Does there exist a set of non-zero real numbers a_1, a_2,..,a_n with this condition: If function f: \mathbb R \rightarrow \mathbb R satisfies the inequality$$...
[Khinchine Inequality] Let $a_1,\ldots,a_n\in R$, $\varepsilon_1,\ldots,\varepsilon_n$ be i.i.d. Rademacher random variables: $P(\varepsilon_i=1)=P(\varepsilon_i=-1)=0.5$, and $0<p<\infty$. Then ...