Introduction
If \(\displaystyle{\left(S_{i},\cdot_{i}\right)\,,1\leq i\leq 3}\) are normed \(\displaystyle{\mathbb{R}}\)  spaces and
\(\displaystyle{f:S_{1}\longrightarrow S_{2}\,,g:S_{2}\longrightarrow S_{3}}\) are bounded linear operators, then
the function \(\displaystyle{g\circ f: S_{1}\longrightarrow S_{3}}\) is linear (easy) and bounded cause:
\(\displaystyle{\forall\,s_1\in S_{1}: (g\circ f)(s_1)_{3}=g(f(s_1))\leq g\cdot f(s_1)_{2}\leq g\,f\cdot s_1_{1}}\) .
Since \(\displaystyle{g\circ f=\sup\,\left\{(g\circ f)(s_1)_{3}: s_{1}\in S_{1}\,,s_{1}_{1}\leq 1\right\}}\), we get :
\(\displaystyle{g\circ f\leq g\cdot f}\) .
Solution
Since the normed \(\displaystyle{\mathbb{R}}\) spaces \(\displaystyle{\left(X,\cdot\right)\,,\left(Y,\cdot\right)}\) are isomorphic,
there exists \(\displaystyle{f\in GL\,(X,Y)}\), which means that \(\displaystyle{f:X\longrightarrow Y}\) is a linear and bounded operator
and \(\displaystyle{11}\) and onto \(\displaystyle{Y}\) with \(\displaystyle{f^{1}\in \mathbb{B}\,(Y,X)}\) .
Therefore, the set \(\displaystyle{\left\{T\cdot T^{1}: T\in GL\,(X,Y)\right\}}\) is nonempty subset of \(\displaystyle{\mathbb{R}}\)
which is lower bounded by zero because \(\displaystyle{T\cdot T^{1}\geq 0\,,\forall\,T\in GL\,(X,Y)}\) and thus :
\(\displaystyle{d\,(X,Y)\in\mathbb{R}\cap\left[0,+\infty\right)}\) . Let now \(\displaystyle{T\in GL\,(X,Y)}\). Then,
\(\displaystyle{T\circ T^{1}=Id_{Y}: Y\longrightarrow Y}\) and according to the introduction :
\(\displaystyle{T\cdot T^{1}\geq T\circ T^{1}=Id_{Y}=\sup\,\left\{y_{Y}: y\in Y\,,y_{Y}\leq 1\right\}=1}\), which means
that the real number \(\displaystyle{1}\) is a lower bound of the set
\(\displaystyle{\left\{T\cdot T^{1}: T\in GL\,(X,Y)\right\}}\), so:
\(\displaystyle{d\,(X,Y)=\inf\,\left\{T\cdot T^{1}: T\in GL\,(X,Y)\right\}\geq 1}\) .
Note
If \(\displaystyle{\left(X,\cdot\right)=\left(Y,\cdot\right)}\), then \(\displaystyle{d\,(X,Y)=d\,(X,X)=1}\), because the set
\(\displaystyle{GL\,(X,X)}\) contains the identity function of \(\displaystyle{X\,,Id_{X}: X\longrightarrow X}\) and also :
\(\displaystyle{Id_{X}\cdot Id_{X}^{1}=Id_{X}\cdot Id_{X}=1\cdot 1=1}\), that is :
\(\displaystyle{d\,(X,X)=\min\,\left\{T\cdot T^{1} : T\in GL\,(X,X)\right\}=1}\) .
