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12 votes
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An $L^1$ function but (really) no better?

There is a much more general result of Vallée-Poussin from which a negative answer to your question follows. Let $(X,\mu)$ be a measure space. We say that a family of function $\mathcal{F}\subset L^1(...
Piotr Hajlasz's user avatar
7 votes
Accepted

Weak concentration bounds for averages of independent random variables in Orlicz spaces

In general, the answer is no. Moreover, the answer is no even if \begin{equation} \phi(t)=t\ln(1+t). \tag{1} \end{equation} Indeed, suppose that $P(Z_i=0)=1-2p$ and $P(Z_i=b)=p=P(Z_i=-b)$ for all $...
Iosif Pinelis's user avatar
6 votes
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Looking for a reference for a version of the "The reverse Lebesgue dominated convergence theorem" for the Orlicz spaces

This is Theorem 1.4 in Bennet and Sharpley's "Interpolation of Operators" (page 3). It actually holds for Banach spaces of functions equipped with what they call "function norms" and not only Orlicz ...
Dirk's user avatar
  • 12.7k
5 votes
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Luxemburg norm as argument of Young's function: $\Phi\left(\lVert f \rVert_{L^{\Phi}}\right)$

The conjectured inequality does not hold. For a counterexample, consider $\Phi(t)=\max(t^2,t^3)$ and $\Omega=(0,1)$. Let $f=a\chi_{(0,b)}$ for $a,b\in (0,1)$. It can be calculated that $\|f\|_{L^\Phi}...
harfe's user avatar
  • 215
4 votes
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When do Orlicz norms tend to the uniform norm?

$\newcommand{\ep}{\varepsilon}\newcommand{\Si}{\Sigma}\newcommand{\R}{\mathbb R}$A natural generalization of the fact that the $p$-norm converges to the $\infty$-norm as $p\to\infty$ is as follows. ...
Iosif Pinelis's user avatar
3 votes

On the intersection of two Orlicz spaces

It's many years ago that I read it, but I think that some of the most general interpolation type results for Orlicz spaces were contained in O’Neil, Richard, Integral transforms and tensor products on ...
Martin Väth's user avatar
  • 1,869
3 votes
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Hoeffding to bound Orlicz norm

As I understand these notations, $\Psi_2(x)=\exp(x^2)-1$ and $\|Z\|_{\Psi}=\inf \{c>0 : \mathbb{E} (\Psi(|Z| / c)) \leqslant 1\}$. So the inequality $\|Z\|_{\Psi}\leqslant c$ means that $\mathbb{...
Fedor Petrov's user avatar
3 votes
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Reference Request: $L^p(x)$/(Musielak–Orlicz space) analogue of classical $L^p$ result

Yes this is true (in the variable exponent's case which was the objective of the question). See Theorem 3.4.12 of this book: "Lebesgue and Sobolev spaces with variable exponents" by "Lars Diening, ...
ABIM's user avatar
  • 5,405
2 votes
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Which Orlicz functions $f$ make the function $f^{-1}\left(\frac{\sum_{j=1}^s f(x_j)}{s}\right)$ convex?

It follows from Hardy--Littlewood--Polya (see section 3.16, page 86), that if for $x>0$ we have $f, f', f''>0$, and $f'/f''$ is concave, and $d\mu$ is a probability measure on the probability ...
Paata Ivanishvili's user avatar
2 votes

Young’s complement of $ x \mapsto x \, {\log^{+}}(x) $, $ N $-functions and Orlicz spaces

That space should be $L_{\exp}$. Check Bennett and Sharpley's "Interpolation of Operators" Chapter 4.6 or look in Rao and Ren's "Theory of Orlicz Spaces".
Dirk's user avatar
  • 12.7k
2 votes
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Hölder inequality between different Orlicz spaces

Yes, we can say so. Indeed, let us show that the conditions $f\in L^r$ and $g\in L^s\ln L$ imply $fg\in L\ln^t L$ for $t:=1/s$. Moreover, we shall show that the value $t=1/s$ here is optimal, as it ...
Iosif Pinelis's user avatar
2 votes
Accepted

Elementary convexity example

Here is how to remove the assumption that $p-2+\delta\ge0$. Let \begin{equation*} s:=p-1+\delta. \end{equation*} The conditions $p>1$ and $\delta>0$ imply $s\ge0$. No other conditions on $...
Iosif Pinelis's user avatar
2 votes
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Is the space of bounded $\psi_\infty$ Orlicz norm random variables equal to $L^\infty$?

For each real $k>0$, \begin{equation}E\psi_\infty(|X|/k)=\infty\,P(|X|>k)+P(|X|=k) \\ =\left\{\begin{aligned}\infty\text{ if } P(|X|>k)>0,\\ P(|X|=k)\le1\text{ if } P(|X|>k)=0. \end{...
Iosif Pinelis's user avatar
1 vote
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Improved bounds on $\lVert XY\rVert_{\psi_2}$ via concentration data of the (bounded) random variable $Y$?

This is not possible. Take $X \sim \mathcal{N}(0, 1)$, and $Z \sim \mathcal{N}(0,1)$, with $Y = \sqrt{B} Z$ when $|Z| < \sqrt{B}$ and $Y = B \mathop{sgn}(Z)$ otherwise. Clearly $\|X\|_{\psi_2} \leq ...
Jarosław Błasiok's user avatar
1 vote

Is the product of sub-Gaussian polynomials in $\mathbb{R}[x]/(x^n-1)$ sub-Gaussian?

I had known that Or (in general) via some stronger Orlicz-norm bound on $f(x)g(x)$ than the above $\psi_1$ bound? was plausibly related to the Bennett-Orlicz norm (or perhaps Bernstein-Orlicz norm), ...
Mark Schultz-Wu's user avatar
1 vote

Elementary convexity example

Here is a much simpler proof, actually of the more general fact that $$f(x):=x^p(1+\ln^+ x)^s$$ is convex in $x\ge0$ for any real $p\ge1$ and $s\ge0$, where $\ln^+ x:=\ln\max(1,x)$. For the left and ...
Iosif Pinelis's user avatar
1 vote
Accepted

Independent Sums and Orlicz Norms

For the equality $\|S_n\|_{L^2}=\|X_i\|_{L^2}$ you need the zero-mean condition -- that $EX_i=0$. Let $X,X_1,\dots,X_n$ be any random variables with the same norm: $\|X\|=\|X_1\|=\cdots=\|X_n\|$. ...
Iosif Pinelis's user avatar

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