Real Analysis – Is Multiplication on ? a Lipschitz Function?

lipschitz-functionsreal-analysis

This question is related to this unanswered question on the site. My question is specifically about the special case when $n=1$:

Let $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ be the function defined as $f(x,y)=xy$. Is $f$ Lipschitz?

An unsuccessful attempt and some thoughts:

By definition, IF it is Lipschitz, one has a uniform estimate like
$$
|xy-uv|\le K|(x,y)-(u,v)|=K\sqrt{(x-u)^2+(y-v)^2}\tag{1}
$$
It suffices to prove a squared version of (1):
$$
(xy)^2+(uv)^2-2(xyuv)\le K \big(x^2+y^2+u^2+v^2-2(xu+yv)\big)\tag{2}
$$
I see no hope to get an $K$ for (2).

On the other hand, if we fix a value of $y$ and consider the map $x\mapsto xy$, then this map is Lipschitz with a Lipschitz constant depending on the magnitude of $y$. This seems to suggest that $f$ cannot be Lipschitz.

Best Answer

Not globally, no. You can compare $|x y_1-xy_2|=|x||y_1-y_2|$ while $\sqrt{(x-x)^2+(y_2-y_1)^2}=|y_2-y_1|$ so any Lipschitz constant would need to be at least as large as the largest possible value of $x$. Same for $y$.

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Are max entropy distributions under a norm constraint Lipschitz with respect to moving the constraint

The conjecture does not hold. Here is a counterexample (found numerically):

$p=(0.34545229, 0.14548291, 0.5090648),p'=(0.35411664, 0.02645345, 0.61942991),\epsilon=.05$.

Python code for the simulation is below:

import cvxpy as cp
import numpy as np
from scipy.spatial.distance import cdist

def F(p,eps=.05):
    n=len(p)
    x = cp.Variable(n)
    ent=cp.sum(cp.entr(x))
    constraints=[x>=0,cp.sum(x)==1,cp.sum(cp.abs(x-p)**2)<=eps**2]
    prob = cp.Problem(cp.Maximize(ent),constraints)
    prob.solve()
    return x.value
#randomly sampled distributions on simplex 
X=[]
#corresponding value of f(p) for each distribution p
Y=[]
for _ in range(1000):
    p=np.random.rand(3)
    p=p/np.sum(p)
    X.append(p)
    Y.append(F(p))
X=np.array(X)
Y=np.array(Y)
#the fraction of pairs for which the bound is violated by more than .001
np.mean(cdist(X,X)+10**-3<cdist(Y,Y))

In this simulation,about .3% of pairs violated the conjectured inequality by more than .001.

Is this operator between $\ell^{25}$ and $\ell^{12}$ continuous

Note that for all $n \in \mathbb{N}$ and for all $a,b \in \mathbb{R}$ we have $a^n-b^n = (a - b) \ ( \ a^{n-1} \ + \ a^{n-2}b \ + \ a^{n-3}b^2 \ + \ \ldots \ + \ b^{n-1} \ )$. Let $x \in \mathcal{l}^{25}$ be fixed and let $y \in \mathcal{l}^{25}$ be such that $\|y-x\|_{25} \leq 1$. Thus for all $n$ we have $|x_n| \leq \|x\|_{25}$ and $|y_n| \leq \|y-x\|_{25}+\|x\|_{25} \leq 1+\|x\|_{25}$.

Let us compute $$\begin{align*} \|Tx-Ty\|_{12}^{12}&= \sum\limits_{n \in \mathbb{N}}|x_n^{2018}-y_n^{2018}|^{12} \\ &= \sum\limits_{n \in \mathbb{N}}|(x_n - y_n) \ ( \ x_n^{2017 } \ + \ x_n^{2018-2}y_n \ + \ x_n^{2018-3}y_n^2 \ + \ \ldots \ + \ y_n^{2018-1} \ )|^{12} \\ &= \sum\limits_{n \in \mathbb{N}}|(x_n - y_n) |^{12} |( \ x_n^{2017 } \ + \ x_n^{2018-2}y_n \ + \ x_n^{2018-3}y_n^2 \ + \ \ldots \ + \ y_n^{2018-1} \ )|^{12} \\ &\leq \left( \sum\limits_{n \in \mathbb{N}}|(x_n - y_n)|^{12 \cdot 3} \right)^{\frac 1 3} \left( \sum\limits_{n \in \mathbb{N}} |( x_n^{2017 } + x_n^{2018-2}y_n + x_n^{2018-3}y_n^2 + \ldots + y_n^{2018-1} )|^{12 \cdot \frac 3 2} \right)^{\frac 2 3} \end{align*}$$

Now $\big( \sum\limits_{n \in \mathbb{N}}|(x_n - y_n)|^{12 \cdot 3} \big)^{\frac 1 3}=\|x-y \|_{36}^{12} \leq \|x-y\|_{25}^{12}$.

Finally: $$\big( \sum\limits_{n \in \mathbb{N}} |( \ x_n^{2017 } \ + \ x_n^{2018-2}y_n \ + \ x_n^{2018-3}y_n^2 \ + \ \ldots \ + \ y_n^{2018-1} \ )|^{12 \cdot \frac 3 2} \big)^{\frac 2 3} \leq \big( \sum\limits_{n \in \mathbb{N}} |x_n|^{2017 \cdot 12 \cdot \frac 3 2} \ + \ |x_n|^{2016 \cdot 12 \cdot \frac 3 2}|y_n|^{12 \cdot \frac 3 2} \ + \ |x_n|^{2015 \cdot 12 \cdot \frac 3 2}|y_n|^{2 \cdot 12 \cdot \frac 3 2} \ + \ \ldots \ + \ |y_n|^{2017 \cdot 12 \cdot \frac 3 2} \ )|^{12 \cdot \frac 3 2} \big)^{\frac 2 3} \leq C(\|x\|_{25})$$ since $y \in B_{\mathcal{l}_{25}}(x,1)$. Thus we found an estimate as:

$$\|Tx-Ty\|_{12} \leq \big( C(\|x\|_{25}) \big) ^{\frac{1}{12}} \|x-y\|_{25}$$

for all $y \in B_{\mathcal{l}_{25}}(x,1)$

Best Answer

Related Solutions

Are max entropy distributions under a norm constraint Lipschitz with respect to moving the constraint

Is this operator between $\ell^{25}$ and $\ell^{12}$ continuous

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