[proofplan]
The scalar $H^\infty$ transfer functions form a Banach algebra under multiplication, with submultiplicative norm. The small-gain hypothesis makes the loop product $G\Delta$ have norm less than one, so the inverse of $1+G\Delta$ is given by a stable Neumann series. We then derive the closed-loop maps from the feedback equations and check that each is a product of stable factors. The final sufficient condition follows from submultiplicativity.
[/proofplan]
[step:Invert the feedback denominator by a Neumann series]
Let
\begin{align*}
L=G\Delta.
\end{align*}
The scalar $H^\infty$ transfer functions form a Banach algebra for the $H^\infty$ norm, and multiplication is submultiplicative. Since $\|L\|_\infty<1$, the Neumann series
\begin{align*}
M=\sum_{k=0}^{\infty}(-L)^k
\end{align*}
converges in the stable transfer-function algebra. Therefore $M$ is stable and satisfies
\begin{align*}
(1+L)M=1
\end{align*}
and
\begin{align*}
M(1+L)=1.
\end{align*}
Thus $(1+G\Delta)^{-1}$ exists as a stable transfer function.
[/step]
[step:Derive and check the closed-loop maps]
Let $a$ and $b$ denote scalar stable external test input signals, with $a$ injected at the plant input and $b$ injected at the controller output. Let $y$ denote the scalar plant-output signal and let $u$ denote the scalar controller-output signal produced by these test inputs. The negative-feedback loop equations are
\begin{align*}
y=G(u+a)
\end{align*}
and
\begin{align*}
u=b-\Delta y.
\end{align*}
Substituting the second equation into the first gives
\begin{align*}
y=G(b-\Delta y+a).
\end{align*}
Thus
\begin{align*}
(1+G\Delta)y=G(a+b).
\end{align*}
Multiplying by
\begin{align*}
S=(1+G\Delta)^{-1}
\end{align*}
gives
\begin{align*}
y=SGa+SGb.
\end{align*}
The controller-side internal signal is
\begin{align*}
u=b-\Delta y=b-\Delta SGa-\Delta SGb.
\end{align*}
Since the interconnection is scalar, the factors commute. Also $1-G\Delta S=S$, because $S(1+G\Delta)=1$. Hence the transfer maps from $(a,b)$ to $(y,u)$ are combinations of $GS$, $\Delta GS$, and $S$.
The stable $H^\infty$ algebra is closed under multiplication. We have already proved that $S$ is stable, and $G$ and $\Delta$ are stable by hypothesis. Therefore $GS$, $\Delta S$, and $G\Delta S$ are stable. Hence all internal closed-loop maps in the displayed two-input test are stable, so the well-posed feedback interconnection is internally stable.
[/step]
[step:Derive the product-norm sufficient condition]
If
\begin{align*}
\|G\|_\infty\|\Delta\|_\infty<1,
\end{align*}
then submultiplicativity gives
\begin{align*}
\|G\Delta\|_\infty\le \|G\|_\infty\|\Delta\|_\infty<1.
\end{align*}
The first part of the theorem applies, so the feedback interconnection is internally stable.
[guided]
The loop product is $L=G\Delta$. Since $\|L\|_\infty<1$, the Neumann series $\sum_{k\ge 0}(-L)^k$ converges in the scalar $H^\infty$ Banach algebra. Its sum is a stable inverse for $1+L$, because multiplying the partial sums by $1+L$ telescopes to $1-(-L)^{N+1}$, and the last term tends to zero in norm. Thus the sensitivity $S=(1+G\Delta)^{-1}$ is stable.
To see the internal maps, let $a$ and $b$ be scalar stable test input signals, let $y$ be the plant-output response, and let $u$ be the controller-output response. The equations are $y=G(u+a)$ and $u=b-\Delta y$. Solving gives $(1+G\Delta)y=G(a+b)$, so $y=SGa+SGb$. Then $u=b-\Delta y=b-\Delta SGa-\Delta SGb$. Thus every map from the test inputs to the internal signals is a product or sum of $S$, $G$, and $\Delta$. These are stable, so the interconnection is internally stable. Finally, $\|G\Delta\|_\infty\le \|G\|_\infty\|\Delta\|_\infty$, so the product-gain hypothesis implies the small-gain hypothesis.
[/guided]
[/step]
