An extension of Young's Inequality

Yi Fan · science forum beginner Joined: 17 May 2005 Posts: 1

I'm a graduate student in Polytechnic University. We found the following lemma holds (We proved). It can be regarded as an extension of the popular Young's Inequality under saturation. We would like your comments. Any suggestions are welcome. It is best to edit the following in a latex editor to have a readable format. Thank you.

\begin{Lemma}

For any integer $n\ge 2$ and any $a_{ij}\ge0, i,j=1,...,n$,
the following inequality holds:

\begin{equation}\sum_{i=1}^n\sum_{j=1,j\neq i}^n a_{ij}\left|z_i
sat(z_j)\right| \le\sum_{i=1}^n b_{n,i} z_i sat(z_i) \end{equation}

where $b_{n,i}$ is defined as
\begin{equation}
b_{n,i}=\sum_{j=1,j\neq i}^n\max\Big\{a_{ij},a_{ji}\Big\},i=1,...,n.
\end{equation}

\end{Lemma}

%Please see the proof as follows:

\begin{proof}

First notice that, given a pair of odd and non-decreasing functions
$(f,g)$, for any real $z_1$ and $z_2$, it holds:

\begin{eqnarray*}
|f(z_1)g(z_2)|+|f(z_2)g(z_1)|\le f(z_1)g(z_1)+f(z_2)g(z_2).
\end{eqnarray*}

Indeed, it suffices to use the fact that
$\left(f(z_1)-f(z_2)\right)\left(g(z_1)-g(z_2)\right)\ge0.$
Then, observing the identity
\begin{eqnarray*}
|f(z_1)g(z_2)+f(z_2)g(z_1)|=|f(z_1)g(z_2)|+|f(z_2)g(z_1)|,
\end{eqnarray*}

it follows that
\begin{eqnarray*}
|f(z_1)g(z_2)|+|f(z_2)g(z_1)|\le f(z_1)g(z_1)+f(z_2)g(z_2).
\end{eqnarray*}

The above facts can be applied to show that

\begin{eqnarray*}
\sum_{i=1}^n\sum_{j\neq i}^n
a_{ij}z_isat(z_j)&=&\sum_{j=2}^n\left[a_{1j}z_1sat(z_j)+a_{j1}z_jsat(z_1
)\right]\\&&+\sum_{j= 3}^n
\left[a_{2j}z_2sat(z_j)+a_{j2}z_jsat(z_2)\right]+\cdots
\\
&&+a_{n-1,n}z_{n-1}sat(z_n)+a_{n,n-1}z_{n}sat(z_{n-1})
\\
&\le&\sum_{j=2}^n \max\{a_{1j},a_{j1}\}
\left[z_1sat(z_1)+z_jsat(z_j)\right]
\\&&+\sum_{j=
3}^n\max\{a_{2j},a_{j2}\} \left[z_2sat(z_2)+z_jsat(z_j)\right]+\cdots
\\ &&+\max\{a_{n-1,n},a_{n,n-1}\}[z_{n-1}sat(z_{n-1})+z_{n}sat(z_{n})]
\\&=&\sum_{j\neq 1}^n\max\{a_{1j},a_{j1}\}z_1sat(z_1)
+\sum_{j\neq2}^n\max\{a_{2j},a_{j2}\}z_2sat(z_2)
\\&&
+\cdots+\sum_{j\neq n}^n\max\{a_{nj},a_{jn}\}z_nsat(z_n)
\\&=&
\sum_{i=1}^nb_{n,i}z_isat(z_i).
\end{eqnarray*}

\end{proof}

Dave Elliott · science forum beginner Joined: 30 May 2005 Posts: 7

The inequality for monotone functions used by Fan (his paper with
Jiang will appear in ACC 2005)
may not be well-known. Has anyone seen it in the literature?
-- Dave
%%%Quoting in LaTeX from the posted article:%%%%
First notice that, given a pair of odd and non-decreasing functions
$(f,g)$, for any real $z_1$ and $z_2$, it holds:

\begin{eqnarray*}
|f(z_1)g(z_2)|+|f(z_2)g(z_1)|\ le f(z_1)g(z_1)+f(z_2)g(z_2).
\end{eqnarray*}

Indeed, it suffices to use the fact that
$\left(f(z_1)-f(z_2)\right)\le ft(g(z_1)-g(z_2)\right)\ge0.$
Then, observing the identity
\begin{eqnarray*}
|f(z_1)g(z_2)+f(z_2)g(z_1)|=|f (z_1)g(z_2)|+|f(z_2)g(z_1)|,
\end{eqnarray*}

it follows that
\begin{eqnarray*}
|f(z_1)g(z_2)|+|f(z_2)g(z_1)|\ le f(z_1)g(z_1)+f(z_2)g(z_2).
\end{eqnarray*}

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