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\noindent
In several papers a result called Segal's lemma appears. This
result is Lemma~1.4 in I.~E.~Segal's paper Construction of nonlinear
local quantum processes~I. in Annals of Math.~92 (1970) 462--481.
Here I describe a slightly modified and more detailed formulation
and proof of this result. I present a slighly different version of
Segal's proof. I finish this note with the original
formulation and proof of Segal's result.

\medskip\noindent
{\bf Segal's lemma.} Let $(M_i,\mu_i)$ $(i=1,\dots,n)$ be separable
measure spaces, and let $\bold T_i$ be an integral operator on
$L_p(M_i,\mu_i)$ with positive kernel $K_i$, i.e. let
$$
(\bold T_if)(x_i)=\int_{M_i} K_i(x_i,y_i)f(y_i)\mu_i(\,dy_i),
\qquad x_i\in M_i, \; y_i\in M_i, \quad i=1,\dots,n,
$$
where $f(\cdot)$ is a function in $L_p(M_i,\mu_i)$. Let $\bold T_i$ be
a contraction from $L_p(M_i,\mu_i)$ to $L_q(M_i,\mu_i)$, for certain given
$1\le p<\infty$ and $1\le q<\infty$ for each $i$. Then the
algebraic tensor product
$\bold T_1\times\dots\times\bold T_n$ is a contraction from
$L_p(M_1\times\cdots\times M_n,\mu_1\times\cdots\times\mu_n)$
to
$L_q(M_1\times\cdots\times M_n,\mu_1\times\cdots\times\mu_n)$
i.e.
$$
\align
&\left(\int_{M_1\times\cdots\times M_n}
\left|\left(\bold T_1\times\dots\times\bold T_n f)
(x_1,\dots,x_n)\right)\right|^q\mu_1(\,dx_1)\dots\mu_n(\,dx_n)\right)^{1/q} \\
&\qquad \le \left(\int_{M_1\times\cdots\times M_n}
\left|f(y_1,\dots,y_n)\right|^p\mu_1(\,dy_1)\dots\mu_n(\,dy_n)\right)^{1/p}
\endalign
$$
for all
$f(y_1,\dots,y_n)
\in L_p(M_1\times\cdots\times M_n,\mu_1\times\cdots\times\mu_n)$,
where
$$
\align
&(\bold T_1\times\cdots\times \bold T_n f)(x_1,\dots,x_n) \\
&\qquad =\int_{M_1\times\cdots\times M_n}K_1(x_1,y_1)\cdots K_n(x_n,y_n)
\,f(y_1,\dots,y_n) \mu_1(\,dy_1)\dots \mu_n(\,dy_n).
\endalign
$$

\medskip\noindent
{\it Proof.}\/ It is enough to prove the estimate of the Lemma for $n=2$,
because  then the lemma follows for general $n$ by simple induction.
It suffices to prove this inequality only for functions $f$ with the
additional property $f(y_1,y_2)\ge 0$ for all $(y_1,y_2)\in M_1\times M_2$.
Because of this additional condition we can omit the absolute value
in the subsequent calculations, since we are working with non-negative
functions.

Let us fix a function $f(y_1,y_2)\in L_p(M_1\times M_2,\mu_1\times\mu_2)$
with $y_1\in M_1$ and $y_2\in M_2$ such that $f(y_1,y_2)\ge0$ for all
$(y_1,y_2)\in M_1\times M_2$. Then the function $f(y_1,y_2)$ with a
fixed point $y_1\in M_1$ is a function in $L_p(M_2,\mu_2)$
for almost all $y_1\in M_1$. Hence we can define for all $y_1\in M_1$
the $L_p(M_2,\mu_2)$ valued measurable function  $F(y_1)$ on $M_1$
by the formula $(F(y_1))(y_2)=f(y_1,y_2)$ for all $y_1\in M_1$ and
$y_2\in M_2$, and $F(y_1)$ has the norm
$\|F(y_1)\|_p=\left(\int_{M_2}f(y_1,y_2)^p\mu_2(\,dy_2)\right)^{1/p}$. The
function $\|F(y_1)\|_p$, $y_1\in M_1$, is in the Banach space
$L_p(M_1,\mu_1)$, since
$\int\|F(y_1)\|_p^p\mu_1(\,dy_1)
=\int_{M_1\times M_2}f(y_1,y_2)^p \mu_1(\,dy_1)\mu_2(\,dy_2)<\infty$.
We also define the $L_q(M_2,\mu_2)$ valued function $G(x_1)$ for
all $x_1\in M_1$ by the formula
$(G(x_1))(y_2)=\int_{M_1} K_1(x_1,y_1)f(y_1,y_2)\mu_1(\,dy_1)$. Then
$$
(G(x_1))(y_2)=\int_{M_1}\!\! K(x_1,y_1)(F(y_1))(y_2)\mu_1(\,dy_1)
=\left(\int_{M_1}\!\! K_1(x_1,y_1)F(y_1)\mu_1(\,dy_1)\right)(y_2),
$$
i.e. $G(x_1)=\int_{M_1} K_1(x_1,y_1)F(y_1)\mu_1(\,dy_1)$
for all $x_1\in M_1$.

An upper bound will be given on 
$\|G(x_1)\|_p=\left(\int_{M_2} [(G(x_1))(y_2)]^p\mu_2(\,dy_2)\right)^{1/p}$
for a fixed point $x_1\in M_1$. In this estimate the following result
will be applied. If ($M,\mu)$ is a separable measure space, $\bold X$ is
a Banach space, $H(y_1)$, $y_1\in M$, is a measurable non-negative
function on $M$, $U(y_1)$, $y_1\in M$, is an $\bold X$ valued
measurable function on $M$, then
$\left\|\int_{M} H(y_1) U(y_1)\mu(\,dy_1)\right\|\le 
\int_{M} H(y_1) \|(U(y_1))\|\mu(\,dy_1)$.

With the choice $(M,\mu)=(M_1,\mu_1)$, $\bold X=L_p(M_2,\mu_2)$,
$H(y_1)=K_1(x_1,y_1)$, $U(y_1)=F(y_1)$ for all $y_1\in M_1$ this
result yields that
$$
\align
\|G(x_1)\|_p&=\left\|\int_{M_1} K_1(x_1,y_1)F(y_1)\,\mu_1(dy_1)\right\|_p \\
&\le\int_{M_1} K_1(x_1,y_1)\|F(y_1)\|_p\mu_1(\,dy_1)=(\bold T_1(\|F\|_p)))(x_1).
\endalign
$$

This inequality together with the contraction property of
$\bold T_1$ yields the following estimate on the
$L_q(M_1,\mu_1)$-norm of the function $\|G(x_1)\|_p$ with arguments
$x_1\in M_1$ by the $L_p(M_1,\mu_1)$-norm of the function
$\|F(y_1)\|_p$ with arguments $y_1\in M_1$. 
$$
\left\| \|G(\cdot)\|_p\right\|_q
\le\left\|\bold T_1(\|F(\cdot)\|_p)\right\|_q
\le\left\| \|F(\cdot)\|_p\right\|_p.
$$

Let us also observe that since
$\|F(y_1)\|_p=\left(\int_{M_2}|f(y_1,y_2)|^p\mu_2(\,dy_2)\right)^{1/p}$
for  almost all $y_1\in M_1$
$$
\align
\left\| \|F(\cdot)\|_p\right\|_p&=
\left(\int_{M_1}\left(\int_{M_2}|f(y_1,y_2)|^p
\mu_2(\,dy_2)\right)\mu_1(\,dy_1)\right)^{1/p} \\
&=\left(\int_{M_1\times M_2} |f(y_1,y_2)|^p\mu_1(\,dy_1)\mu_2(\,dy_2)\right)^{1/p}
=\|f(y_1,y_2)\|_p.
\endalign
$$

Define the function
$u(x_1,y_2)=\int_{M_1} K_1(x_1,y_1)f(y_1,y_2)\mu_1(\,dy_1)$
which equals \hfill\break
$(G(x_1))(y_2)$ for all $x_1\in M_1$ and $y_2\in M_2$,
and let us estimate the number
$$
S(u(\cdot))=\left(\int _{M_1\times M_2}\left|\left(\int_{M_2}
K_2(x_2,y_2)u(x_1,y_2)\mu_2(\,dy_2)\right)\right|^q
\mu_1(\,dx_1)\mu_2(\,dx_2)\right)^{1/q}.
$$
The application of the contraction property of $\bold T_2$ for
$f(x_2)=u(x_1,x_2)$ with a fixed $x_1\in M_1$ yields the estimate
$$
\align
&\int_{ M_2}\left|\left(\int_{M_2}
K_2(x_2,y_2)u(x_1,y_2)\mu_2(\,dy_2)\right)\right|^q\mu_2(\,dx_2)
\le\left(\int_{M_2} |u(x_1,x_2)|^p\mu_2(\,dx_2)\right)^{q/p} \\
&\qquad=\left(\int_{M_2} |(G(x_1))(x_2)|^p\mu_2(\,dx_2)\right)^{q/p}
=\|G(x_1)\|_p^q.
\endalign
$$

Hence
$$
S(u(\cdot))\le \left(\int_{M_1}\|G(x_1)\|_p^q\mu_1(\,dx_1)\right)^{1/q}
=\|\,\|G(\cdot)\|_p\|_q\le\|\,\|F(\cdot)\|_p\|_p
=\|f(y_1,y_2\|_p.
$$
On the other hand,
$$
\align
S(u(\cdot))&=\biggl(\int _{M_1\times M_2}\left|\left(\int_{M_1\times M_2}
K_2(x_2,y_2)K_1(x_1,y_1)f(y_1,y_2)
\mu_1(\,dy_1)\mu_2(\,dy_2)\right)\right|^q  \\
&\qquad\qquad \mu_1(\,dx_1)\mu_2(\,dx_2)\biggr)^{1/q}
=\|(\bold T_1\times \bold T_2)f(y_1,y_2)\|_q,
\endalign
$$
and the statement of the lemma holds.

\bigskip\noindent
Here I write down Segal's result in its original form. I present
both its formulation and its proof.

\medskip\noindent
{\bf LEMMA 1.4.} Let $M_i$ $(i=1,\dots,n)$ be separable
measure spaces, and let $\bold T_i$ be an integral operator on
$L_p(M_i))$ with positive kernel $K_i$, which is a contraction
from $L_p(M_i)$ to $L_q(M_i)$, for certain given $p$ and $q$
for each $i$. Then the algebraic tensor product
$\bold T_1\times\dots\times\bold T_n$ is a contraction from
$L_p(M_1\times\dots\times M_n)$ to $L_q(M_1\times\dots M_n)$.

\medskip\noindent
{\it PROOF}.\/ It suffices by associativity to treat the case $n=2$.
Now if $\bold B$ is any separable Banach space, and if $L_p(M_1,\bold B)$
denotes the space of all strongly measurable $\bold B$-valued
functions $F$ on $M_1$ which are $p$-th power integrable, with
the norm $\|F\|=\left(\int\|F(x)\|^p \,dx\right)^{1/p}$, then
the operator $\bold T'_1\colon F\to G$, where
$G(x)=\int K_1(x, y)F(y)\,dy$, exists and is a contraction from
$L_p(M_1,\bold B)$ to $L_q(M_1,\bold B)$. For the mapping
$y\to K_1(x,y)F(y)$ is easily seen to be strongly measurable
from $M_1$ to $\bold B$, for each $x$; and
$$
\|G(x)\|\le \int K_1(x,y) \|F(y)\|\,dy,
$$
i.e., $\|G(\cdot)\|\le\bold T_1(\|F(\cdot)\|)$, so that
$\left\| \|G(\cdot)\|_{\bold B}\right\|_q
\le\left\| \bold T_1(\|F(\cdot)\|)_{\bold B}\right\|_q
\le\left\| \|F(\cdot)\|_{\bold B}\right\|_p$. This shows
the absolute integrability of the integral defining $G(x)$ almost
everywhere, and gives the estimate $\|\bold T'_1\|\le1$.

In addition, the operation $\bold T_2''$ from $L_q(M_1,\bold B)$ to
$L_q(M_2,\bold B')$, where $\bold B'$ is another separable Banach
space and $\bold T_2$ is a contraction from $\bold B$ to $\bold B'$,
defined by the equation $(\bold T_2''F)(x)=\bold T_2F(x)$,
$F\in L_q(M_1,\bold B)$, is easily seen to be a contraction. Now
taking $\bold B$ as $L_p(M_2)$ and $\bold B'$ as $L_q(M_2)$,
and making the natural identifications of $L_p(M_1,\bold B)$
with $L_p(M_1\times M_2)$ and of $L_q(M_1,\bold B')$ with
$L_q(M_1\times M_2)$ which are justified by the Fubini theorem, it
follows that the contraction $\bold T_2''\bold T_1'$ extends the
algebraic tensor product $\bold T_1\times\bold T_2$; the latter is
therefore a contraction, as stated.




\bye

