For questions about proving and manipulating functional inequalities.
Questions tagged [functional-inequalities]
628 questions
26
votes
0 answers
Gradient Estimate - Question about Inequality vs. Equality sign in one part
$\DeclareMathOperator{\diam}{diam}\newcommand{\norm}[1]{\lVert#1\rVert}\newcommand{\abs}[1]{\lvert#1\rvert}$For $u \in C^{1}(\overline{\Omega})$, for $\Omega\Subset \mathbb{R^{n}}$ a bounded convex set, and for any $1 \leq p \leq q$ such that…
user100463
21
votes
5 answers
If $f$ is continuous and $\,f\big(\frac{1}2(x+y)\big) \le \frac{1}{2}\big(\,f(x)+f(y)\big)$, then $f$ is convex
Let $\,\,f :\mathbb R \to \mathbb R$ be a continuous function such that $$
f\Big(\dfrac{x+y}2\Big) \le \dfrac{1}{2}\big(\,f(x)+f(y)\big) ,\,\, \text{for all}\,\, x,y \in \mathbb R,
$$
then
how do we prove that $f$ is convex that…
Souvik Dey
- 8,535
20
votes
0 answers
Smallest $c$ such that $f'0$ and $f''' \le f.$
Let $f: \mathbb{R} \to \mathbb{R}$ be a $C^3$ function such that $f,f',f'',f'''>0$ and $f''' \le f.$ What is the smallest $c$ such that we can guarantee $f'1.$ On the other hand, I managed to show $c =…
Display name
- 5,310
17
votes
2 answers
Weighted Poincare Inequality
I'm trying to prove a result I found in a paper, and I think I'm being a bit silly.
The paper claims the following:
By the Poincare inequality on the unit square $\Omega \subset \mathbb{R}^2$ we have that
$$\int_{\Omega} f(x)^2 dx \leq C…
Il-Bhima
- 559
16
votes
0 answers
How to prove this polynomial inequality?
How can we prove the following?
If $\frac{dP_{n}}{dz}|_{z=z_{0}}=0$ then $|P_{n}(z_{0})|<2$ for all $n>1$, where $P_{n}(z)\equiv P_{n-1}^{2}+z$ and $P_{1}\equiv z$
$z$ is in the complex plane.
It appears that…
Jerry Guern
- 2,764
15
votes
11 answers
If $f(x)\leq f(f(x))$ for all $x$, is $x\leq f(x)$?
If I have $f(x)\leq f(f(x))$ for all real $x$, can I deduce $x\leq f(x)$?
Thank you.
JSCB
- 13,698
- 15
- 67
- 125
13
votes
1 answer
prove a challenging inequality or find a counterexample to it
Suppose $\mathcal{M}_1$ represents the space of smooth probability density functions with unit mean, whose support is contained in $[0,\infty)$ (or $\mathbb{R}_+$). Define the following functional $$\mathrm{J}(f):= \int_0^\infty x\frac{(f')^2}{f}…
Fei Cao
- 2,988
12
votes
1 answer
Prove that $\int_0^1|f''(x)|dx\ge4.$
Let $f$ be a $C^2$ function on $[0,1]$. $f(0)=f(1)=f'(0)=0,f'(1)=1.$ Prove that
$$\int_0^1|f''(x)| \, dx\ge4.$$
Also determine all possible $f$ when equality occurs.
Christmas Bunny
- 5,016
11
votes
2 answers
How prove this function inequality $xf(x)>\frac{1}{x}f\left(\frac{1}{x}\right)$
Let $f(x)$ be monotone decreasing on $(0,+\infty)$, such that
$$0\dfrac{1}{x}f\left(\dfrac{1}{x}\right),\qquad\forall x\in(0,1).$$
My ideas:
Since
$f(x)$ is monotone…
math110
- 94,932
- 17
- 148
- 519
10
votes
1 answer
A curious norm related to the L¹ norm
If $f \in C^0([0,1])$, one can define
$$\Vert f \Vert_? = \sup_{J \subset [0,1]} \left\lvert \int_J f \right\rvert,$$
where $J$ runs among all subintervals of $[0,1]$.
This is a norm on $C^0([0,1])$ (and Lebesgue's density theorem shows that this…
PseudoNeo
- 10,358
10
votes
3 answers
Continuous function satisfying $ f\left(\dfrac{x+t}{2}\right) \le f(x) + f(t)$ inequality must be $0$
Let $f$ a real function defined and continuous on $[0,1]$ such that
$$f(0)=f(1)=0$$
$$ f\left(\dfrac{x+t}{2}\right) \le f(x) + f(t)$$ for all $x,t$
prove that $f$ is zero.
My try was proving first that f is nonnegative (no problem) then using the…
daiski
- 169
10
votes
1 answer
Can we show that the determinant of this matrix is non-zero?
Consider the following symmetric matrix
$M=
\begin{bmatrix}
f(x) & f(2x) & \dots & f(nx)\\
f(2x) & f(4x) & \dots & f(2nx)\\
\vdots & \vdots & \dots & \vdots\\
f(nx) & f(2nx) & \dots & f(n^2x)
\end{bmatrix}$,
where $f(x):…
KRL
- 1,272
10
votes
1 answer
Solve $f (x + y) + f (y + z) + f (z + x) \ge 3f (x + 2y + 3z)$
Find all functions $f : \mathbb{R} \to \mathbb{R}$ which satisfy :
$f (x + y) + f (y + z) + f (z + x) ≥ 3f (x + 2y + 3z)$
for real $x,y,z$.
Attempt at solution:
I have tried plugging in $x = -y$ and $x = -z$. This does not seem to be getting me…
user424290
9
votes
1 answer
A unsolved puzzle from Number Theory/ Functional inequalities
The function $g:[0,1]\to[0,1]$ is continuously differentiable and
increasing. Also, $g(0)=0$ and $g(1)=1$. Continuity and
differentiability of higher orders can be assumed if necessary. The
proposition on hand is the following:
If for…
Juanito
- 2,482
- 12
- 25
9
votes
0 answers
Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that $\alpha\,f(yz)+\beta\,f(zx)+\gamma\,f(xy)\geq f(x+y+z)$ for all $x,y,z\in\mathbb{R}$.
Let $\alpha,\beta,\gamma$ be three real numbers. Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that
$$\alpha\,f(yz)+\beta\,f(zx)+\gamma\,f(xy)\geq f(x+y+z)$$ for all $x,y,z\in\mathbb{R}$.
Remarks.
If $\alpha=\beta=\gamma=0$, then any…
Batominovski
- 50,341