Questions tagged [functional-inequalities]
For questions about proving and manipulating functional inequalities.
629 questions
0 votes
1 answer
65 views
Yet another nice functional inequality [duplicate]
Need to find out all the functions $f \colon \mathbb{R} \to \mathbb{R}$ such that $$f(x+y) \leq f(xy).$$ Since $f(x) \leq f(0)$ for every $x \in \mathbb{R}$ and $f(0) \leq f(-x^2)$, it follows that $f(...
- 629
-1 votes
1 answer
62 views
Hardy weighted inequality
In book A. Kufner, L. Maligranda and L-E. Persson in part of sufficiency of proof Hardy weighted inequality as $$ ||Hf||_{L_{q,u(x)}} \lesssim ||f||_{L_{p,v(x)}} $$ for $1 \leq p \leq q < \infty $ ...
1 vote
0 answers
50 views
A trick of symmetrization in proving concentration inequalities
Let $\mathscr{F}$ be a class of functions (each of the form $f: \mathcal{X} \rightarrow \mathbb{R}$ ), and let $\left(X_1, \ldots, X_n\right)$ be drawn from a product distribution $\mathbb{P}=\...
- 447
1 vote
1 answer
86 views
If $N=2$ then $\|u\|_{L^4}^2\leq C \|u\|_{L^2}\|Du\|_{L^2}$
I am studying self-studying PDEs and trying to understand a set of related papers where the inequality the following inequality appears: $N=2$ then $\|u\|_{L^4}^2\leq C \|u\|_{L^2}\|Du\|_{L^2}$. Here $...
- 87
3 votes
0 answers
132 views
Functional inequality involving function that is its own inverse
Let $f:[0,1]\to[0,1]$ be differentiable and equal to its own inverse, i.e. $f(f(x))=x$. Assume also that $f(1) = 0$. Does the following inequality hold: $$\left|2\int_0^1\frac{x}{f'(x)}\,dx\right|^3\...
- 2,654
2 votes
0 answers
153 views
Does the fractional Leibniz rule hold in Lorentz spaces?
Setting Consider the following setting: Let $d \in \mathbb{N}$. Lorentz space: $L^{p,q} = L^{p,q}(\mathbb{R}^d; \mathbb{R})$, $1<p<\infty$, $1\leq q \leq \infty$ (I assume familiarity with ...
- 954
8 votes
2 answers
430 views
Find all $f$ such that $f(x)f(y) \le f(xy)+x + y$
Find all: $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that: $$f(x)f(y)\le f(xy)+x+y, \forall x,y\in \mathbb{R}^+$$ Here is what I have done: Let $x,y \to 1,1$ then: $$f(1)^2\le f(1)+2 \implies f(1)\le 2$$ ...
- 757
0 votes
0 answers
57 views
Yet another functional inequality with asymptotics
Say that I have a continuous increasing function $f : [1, +\infty) \to \mathbb R^+$ which is asymptotically $f(x) = o(x^{1+\varepsilon})$ for any $\varepsilon > 0$ (confer the little-$o$ notation ...
- 457