$\epsilon$-normals to convex sets I am reading the book Variational Analysis and
Generalized Differentiation I by B. Mordukhovich. On page 6 it is stated that the following inclusion:
$$
\hat{N}_{\varepsilon }\left( \bar{x};\Omega \right) \supset \hat{N}\left( 
\bar{x};\Omega \right) +\varepsilon \mathbb{B}^{\ast }.
$$
$\mathbb{B}^{\ast }$ denotes the closed unit ball in the dual space $X^{\ast
}$, and if $\Omega $ is convex, then for any $\varepsilon \geq 0$ we have: 
$$
\hat{N}_{\varepsilon }\left( \bar{x};\Omega \right) =\{x^{\ast }\in X^{\ast
}\mid \langle x^{\ast },x-\bar{x}\rangle \leq \varepsilon \Vert x-\bar{x}
\Vert \text{ whenever }x\in \Omega \}.
$$
Furthermore $\hat{N}\left( \bar{x};\Omega \right) :=\hat{N}_{0}\left( \bar{x}
;\Omega \right) $. Mordukhovich says that for convex set $\Omega $ the above
inclusion holds as equality. Unfortunately, I can't see why the reverse
inclusion holds.  I would be very grateful for the advice.
 A: Proposition 1. 
(Boris S. Mordukhovich, Variational Analysis and Generalized Differentiation I, page 5)
Let $\Omega$ be a nonempty convex set in a real Banach space $X$. Given $\bar{x}\in \Omega$ and $\varepsilon\geq 0$. Then
$$
\widehat{N}_\varepsilon(\bar{x}; \Omega):=
\{x^*\in X^*| \limsup_{x\overset{\Omega}{\rightarrow}\bar{x}}\frac{\langle x^*, x-\bar{x}\rangle}{\|x-\bar{x}\|}\leq \varepsilon\}
$$
is convex and closed in the norm topology of $X^*$. Moreover, if $X$ is reflexive then it is weak$^*$ closed in $X^*$.
Prososition 2. 
(Boris S. Mordukhovich, Variational Analysis and Generalized Differentiation I, Proposition 1.3)
Let $\Omega$ be a nonempty convex set in a real Banach space $X$. Then
$$
\widehat{N}_\varepsilon(\bar{x}; \Omega)=\{x^*\in X^*| \langle x^*, x-\bar{x}\rangle\leq \varepsilon\|x-\bar{x}\|
\; \text{whenever}\; x\in\Omega\}
$$
for any $\varepsilon\geq 0$ and $\bar{x}\in \Omega$. In particular, $\widehat{N}(\bar{x}; \Omega)$
agrees with the normal cone of convex analysis, i.e.
$$
\widehat{N}(\bar{x}; \Omega)=\{x^*\in X^*| \langle x^*, x-\bar{x}\rangle\leq 0
\; \text{whenever}\; x\in\Omega\}.
$$
By using Proposition 1. and Proposition 2. we obtain the following result.
Proposition 3. 
If $\Omega$ is a nonempty convex subset in a real Banach and reflexive space $X$ then
$$
\widehat{N}_\varepsilon(\bar{x}; \Omega)= \widehat{N}(\bar{x}; \Omega)+\varepsilon \mathbb{B}^*.
$$
Proof.
$(\supset)$ Suppose that $x_0^*\in \widehat{N}(\bar{x}; \Omega)$ and $u^*\in \varepsilon \mathbb{B}^*$. Then, it
follows from Proposition 2. that
$$
\langle x_0^*, x-\bar{x}\rangle\leq 0 \quad \forall x\in \Omega.
$$
Hence
\begin{equation*}
\begin{array}{lll}
\langle x_0^*+u^*, x-\bar{x}\rangle&=&\langle x_0^*, x-\bar{x}\rangle+\langle u^*, x-\bar{x}\rangle\\
&\leq&0+\|u^*\|\|x-\bar{x}\|\\
&\leq& \varepsilon\|x-\bar{x}\|
\end{array}
\end{equation*}
for all $x\in \Omega$. This implies that $x_0^*+u^*\in \widehat{N}_\varepsilon(\bar{x}; \Omega)$.
Therefore $\widehat{N}_\varepsilon(\bar{x}; \Omega)\supset\widehat{N}(\bar{x}; \Omega)+\varepsilon \mathbb{B}^*$.
$(\subset)$ Let $\widehat{N}^*:=\widehat{N}(\bar{x}; \Omega)+\varepsilon \mathbb{B}^*$.
Since $X$ is reflexive, it follows from Proposition 2. that $\widehat{N}(\bar{x}; \Omega)$ is convex and weak$^*$
closed in $X^*$.
Moreover $\varepsilon \mathbb{B}^*$ is convex and weak$^*$ compact in $X^*$.
Hence $\widehat{N}^*$ is nonempty ($0\in \widehat{N}^*$), weak$^*$ closed and convex in $X^*$.
Suppose that there exists $x^*\in X^*$ such that
$$
x^*\in \widehat{N}_\varepsilon(\bar{x}; \Omega) \; \text{and} \; x^*\notin \widehat{N}^*.
$$
By the separation theorem (see W. Rudin, Functional Analysis, Theorem 3.4(b))
there exists $x\in X$ such that
$$
\langle x^*, x\rangle>\sup_{f^*\in \widehat{N}^*}\langle f^*, x\rangle
$$
It follows from the above inequality that
$$
\begin{cases}
\langle x^*, x\rangle>\langle f_0^*, x\rangle \quad \forall f_0^*\in \widehat{N}(\bar{x}; \Omega),&\\
\langle x^*, x\rangle>\langle f_1^*, x\rangle \quad \forall f_1^*\in \varepsilon\mathbb{B}^*.&
\end{cases}
$$
Since $\widehat{N}(\bar{x}; \Omega)$ is cone, we have
$$
\begin{cases}
0\geq\langle f_0^*, x\rangle \quad \forall f_0^*\in \widehat{N}(\bar{x}; \Omega),&\\
\langle x^*, x\rangle>\varepsilon\|x\|.&
\end{cases}
$$
By Proposition 2. $\widehat{N}(\bar{x}; \Omega)$ agrees with normal cone in convex analysis and so
$$
\begin{cases}
x\in (\widehat{N}(\bar{x}; \Omega))^*=T(\bar{x}; \Omega)=\overline{\text{cone}(\Omega-\bar{x})},\\
\langle x^*, x\rangle>\varepsilon\|x\|,&
\end{cases}
$$
where $T(\bar{x};\Omega)$ is the tangent cone of $\Omega$ at $\bar{x}$.
Then, there exist $\{t_k\}\subset\mathbb{R}^+$ and $\{x_k\}\subset\Omega$ such that $t_k(x_k-\bar{x})\rightarrow x$.
Hence, for sufficiently large $k$ we have
$$
\langle x^*,t_k(x_k-\bar{x}) \rangle>\varepsilon\|t_k(x_k-\bar{x})\|
$$
or equivalently
$$
\langle x^*,x_k-\bar{x} \rangle>\varepsilon\|x_k-\bar{x}\|.
$$
This implies that $x^*\notin \widehat{N}_\varepsilon(\bar{x}; \Omega)$, which is an absurd.
