For questions about proving and manipulating functional inequalities.

# Questions tagged [functional-inequalities]

519 questions

**22**

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 $\Omega\subset \subset \mathbb{R^{n}}$ a bounded convex set, and for any $1 \leq p \leq q$ such…

user100463

**20**

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

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**16**

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

- 4,794
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**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

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**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

- 2,346
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**13**

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

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**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

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**11**

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

- 489
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- 12

**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

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**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

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**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

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**8**

votes

**2**answers

### Prove that $0\le\int_0^1\log(u){\rm d}x+\frac1{2\pi^2}\int_0^1\frac1{u^2}\left(\frac{{\rm d}u}{{\rm d}x}\right)^2{\rm d}x$

I want to prove a special case of a functional inequality stated in a book. To be specific, let $u(x)$ being a positive, differentiable function on $[0,1]$ with unit mass (i.e., $\int_0^1 u(x){\rm d}x = 1$), then we want to show that
$$0 \leq…

Fei Cao

- 2,346
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**8**

votes

**1**answer

### $P(x)+P'''(x)\geq P'(x) + P''(x)$ then $P(x)>0 \, \forall x\in \mathbb R $

Let $P(x)$ be a polynomial function of real coefficients with the following property:
$$P(x)+P'''(x)\geq P'(x) + P''(x) $$
then, $$P(x)\geq0 \quad \forall x\in \mathbb R $$
I've tried writing the polynomial in its expanded form and cancelling terms,…

J. Doe

- 103
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**8**

votes

**1**answer

### The Functional Inequality $f(x) \ge x+1$, $f(x)f(y)\le f(x+y)$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function that satisfies the following conditons.
$$f(x)f(y)\le f(x+y)$$
$$f(x)\ge x+1$$
What is $f(x)$?
It is not to difficult to find that $f(0)=1$.
If $f(x)$ is differentiable, we can…

Chad Shin

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