$\def\sp{\kern.4mm}\def\bbN{\mathbb N}\def\bbZ{\mathbb Z}\def\bbR{\mathbb R}$Here is the answer to David Roberts' residual query: is a convenient diffeomorphism necessarily MB-smooth? No. A counterexample follows:

Let $E$ be the vector space of all real (two-sided) sequences $x=\langle\sp x_i:i\in\bbZ\sp\rangle$ for which $x_i=0$ for sufficiently small $i$ , topologized so that we get a linear homeomorphism $E\to\bbR\,^{\bbN}\times\bbR\,^{(\bbN)}$ defined by $x\mapsto(u,v)$ where $u_i=x_{i-1}$ and $v_i=x_{-i}$ for $i\in\bbN$ . Then define $f:E\to E$ by $x\mapsto y$ where $y_i=x_i$ for $i\in\bbZ\setminus\{\sp 0\sp\}$ and $y_0=x_0+\sum_{i\in\bbN}(x_i\cdot x_{-i})$ . The inverse is given by $y\mapsto x$ where $y_i=x_i$ for $i\in\bbZ\setminus\{\sp 0\sp\}$ and $x_0=y_0-\sum_{i\in\bbN}(y_i\cdot y_{-i})$ . Since the duality map $\bbR\,^{\bbN}\times\bbR\,^{(\bbN)}\to\bbR$ is a discontinuous but bornological, and hence conveniently smooth bilinear map, the assertion follows.

$\def\sp{\kern.4mm}\def\bbN{\mathbb N}\def\bbZ{\mathbb Z}\def\bbR{\mathbb R}$Here is the answer to David Roberts' residual query: is a convenient diffeomorphism necessarily MB-smooth? No. A counterexample follows:

Let $E$ be the vector space of all real (two-sided) sequences $x=\langle\sp x_i:i\in\bbZ\sp\rangle$ for which $x_i=0$ for sufficiently small $i$ , topologized so that we get a linear homeomorphism $E\to\bbR\,^{\bbN}\times\bbR\,^{(\bbN)}$ defined by $x\mapsto(u,v)$ where $u_i=x_{i-1}$ and $v_i=x_{-i}$ for $i\in\bbN$ . Then define $f:E\to E$ by $x\mapsto y$ where $y_i=x_i$ for $i\in\bbZ\setminus\{\sp 0\sp\}$ and $y_0=x_0+\sum_{i\in\bbN}(x_i\cdot x_{-i})$ . The inverse is given by $y\mapsto x$ where $y_i=x_i$ for $i\in\bbZ\setminus\{\sp 0\sp\}$ and $x_0=y_0-\sum_{i\in\bbN}(y_i\cdot y_{-i})$ . Since the duality map $\bbR\,^{\bbN}\times\bbR\,^{(\bbN)}\to\bbR$ is a discontinuous but bornological, and hence conveniently smooth bilinear map, the assertion follows.