In classical normed spaces the function \(t\mapsto( \Vert x_{0}+ty \Vert - \Vert x_{0} \Vert )/t\) from \(R\backslash\{0\}\) to R plays an important role in Gâteaux differentiability of the classical norm [23]. Motivated by this, we consider in an RN module \((E, \Vert \cdot \Vert )\) the mapping f defined by
$$f(t,y)=\frac{ \Vert x_{0}+ty \Vert - \Vert x_{0} \Vert }{t},\quad \forall t\in{L^{0}(\mathcal {F},R)}\backslash \{0\},\forall y\in E, $$
where \(x_{0}\) is a fixed element in \(S(E)\). It is feasible to define the Gâteaux differentiability of random norm via \(f(t,y)\) and establish its relation to random smoothness. It should be pointed out that some terminologies and properties in this section are closely related to [19], Section 5.
For any \(D\in\widetilde{\mathcal {F}}\) with \(P(D)>0\) and \(t_{1},t_{2}\) in \({L^{0}(\mathcal {F},R)}\) such that \(0< t_{1}< t_{2}\) on D, we can verify that
$$\begin{aligned} &I_{D}\frac{ \Vert x_{0}+t_{1}y \Vert - \Vert x_{0} \Vert }{t_{1}}\leq I_{D}\frac{ \Vert x_{0}+t_{2}y \Vert - \Vert x_{0} \Vert }{t_{2}}, \\ &I_{D}\frac{ \Vert x_{0}-t_{2}y \Vert - \Vert x_{0} \Vert }{-t_{2}}\leq I_{D}\frac{ \Vert x_{0}-t_{1}y \Vert - \Vert x_{0} \Vert }{-t_{1}} \end{aligned}$$
and
$$I_{D}\frac{ \Vert x_{0}-t_{1}y \Vert - \Vert x_{0} \Vert }{-t_{1}}\leq I_{D}\frac{ \Vert x_{0}+t_{1}y \Vert - \Vert x_{0} \Vert }{t_{1}} $$
as in the classical case, see [23] for details. Consequently, \(f(t,y)\) is nondecreasing in t in the sense that \(I_{D}f(t_{1},y)\leq I_{D}f(t_{2},y)\) for any \(t_{1},t_{2}\in{L^{0}(\mathcal {F},R)}\) and \(D\in\widetilde{\mathcal {F}}\) with \(P(D)>0\) such that \(t_{1},t_{2}\neq0\) on D and \(t_{1}< t_{2}\) on D. Besides, \(f(t,y)\) possesses the following properties as described in Lemma 4.1.
Lemmas 4.1(1) and 4.2 below are implied by [19], Lemma 5.1, and [19], Theorem 5.5(1), respectively. Here we present slightly different proofs to illustrate the typical stratification analysis on random normed modules and to keep self-contained.
Lemma 4.1
(1) \(\wedge\{I_{D}f(t,y):t\in L^{0}({\mathcal {F}},R),t>0\textit{ on }D\}=I_{D}\cdot\wedge\{f(s,y):s\in R,s>0\}\);
(2) \(\wedge\{I_{D}f(\xi s,y):s\in R,s>0\}=I_{D}\cdot\wedge\{ f(s,y):s\in R,s>0\},\forall\xi\in L^{0}({\mathcal {F}},R),\xi>0\)
on
D;
(3) \(\vee\{I_{D}f(t,y):t\in L^{0}({\mathcal {F}},R),t<0\textit{ on }D\}=I_{D}\cdot\vee\{f(s,y):s\in R,s<0\}\).
Proof
(1). It is clear that
$$\wedge\bigl\{ I_{D}f(t,y):t\in L^{0}({\mathcal {F}},R),t>0 \text{ on }D\bigr\} \leq I_{D}\cdot\wedge\bigl\{ f(s,y):s\in R,s>0\bigr\} . $$
For any fixed \(t\in L^{0}({\mathcal {F}},R)\) with \(t>0\text{ on }D\), denote \(D_{0}(t)=D\cap[t\geq1]\) and \(D_{n}(t)=D\cap[\frac{1}{n+1}\leq t<\frac{1}{n}],\forall n\geq1\), then \(D=\sum_{n=0}^{\infty}D_{n}(t)\). Since
$$I_{D_{n}(t)}f(t,y)\geq I_{D_{n}(t)}f\biggl(\frac{1}{n+1},y\biggr) \geq I_{D_{n}(t)}\cdot \wedge\bigl\{ f(s,y):s\in R,s>0\bigr\} ,\quad \forall n, $$
we obtain that
$$\begin{aligned} I_{D}f(t,y)&=\sum_{n=0}^{\infty}I_{D_{n}(t)}f(t,y) \geq\sum_{n=0}^{\infty}I_{D_{n}(t)}\cdot \wedge\bigl\{ f(s,y):s\in R,s>0\bigr\} \\ &= I_{D}\cdot\wedge\bigl\{ f(s,y):s\in R,s>0\bigr\} . \end{aligned}$$
The proofs of (2) and (3) are similar. □
Lemma 4.2
Denote
\(G_{+}(x_{0},y)=\wedge\{f(t,y):t\in R,t>0\},G_{-}(x_{0},y)=\vee\{ f(t,y):t\in R,t<0\}\). Then
\(G_{+}(x_{0},y)\)
and
\(G_{-}(x_{0},y)\)
satisfy the following:
-
(1)
\(G_{+}(x_{0},\xi y)=\xi G_{+}(x_{0},y),G_{-}(x_{0},\xi y)=\xi G_{-}(x_{0},y)\);
-
(2)
\(G_{+}(x_{0},y_{1}+y_{2})\leq G_{+}(x_{0},y_{1})+G_{+}(x_{0},y_{2})\),
where
\(x_{0}\in S(E),y,y_{1},y_{2}\in E,\xi\in L^{0}_{+}\).
Proof
(1). For any \(\xi\in L^{0}_{+}\), let \(A=[\xi>0]\), then by Lemma 4.1
$$\begin{aligned} G_{+}(x_{0},\xi y)&= \wedge \bigl\{ f(t,\xi y):t\in R,t>0 \bigr\} \\ &=\wedge \biggl\{ \frac{ \Vert x_{0}+t\xi y \Vert - \Vert x_{0} \Vert }{t}:t\in R,t>0 \biggr\} \\ &=\wedge \biggl\{ \frac{I_{A} \Vert x_{0}+t\xi y \Vert -I_{A} \Vert x_{0} \Vert }{t}:t\in R,t>0 \biggr\} \\ &= I_{A}\xi\cdot\wedge \biggl\{ I_{A}\cdot \frac{ \Vert x_{0}+t\xi y \Vert - \Vert x_{0} \Vert }{t\xi}:t\in R,t>0 \biggr\} \\ &= I_{A}\xi\cdot\wedge \biggl\{ I_{A}\cdot \frac{ \Vert x_{0}+t y \Vert - \Vert x_{0} \Vert }{t}:t\in R,t>0 \biggr\} \\ &= I_{A}\xi G_{+}(x_{0},y) \\ &= \xi G_{+}(x_{0},y). \end{aligned}$$
The proof of the other equality is similar.
(2). Notice that
$$\begin{aligned} & \wedge\bigl\{ f(2t,y_{1}):t\in R,t>0\bigr\} +\wedge\bigl\{ f(2t,y_{2}):t\in R,t>0\bigr\} \\ &\quad =\wedge\bigl\{ f(2t,y_{1})+f(2t,y_{2}):t\in R,t>0\bigr\} . \end{aligned}$$
(4)
In fact, for any positive numbers \(t_{1}\) and \(t_{2}\),
$$\begin{aligned} f(2t_{1},y_{1})+f(2t_{2},y_{2})& \geq \bigl[f(2t_{1},y_{1})+f(2t_{1},y_{2}) \bigr]\wedge \bigl[f(2t_{2},y_{1})+f(2t_{2},y_{2}) \bigr] \\ &\geq \wedge\bigl\{ f(2t,y_{1})+f(2t,y_{2}):t\in R,t>0\bigr\} . \end{aligned}$$
Consequently, the left-hand side of (4) is not less than the right-hand side, so that (4) can be easily verified. Then, by the following
$$\begin{aligned} f(t,y_{1}+y_{2})&= \frac{ \Vert x_{0}+t(y_{1}+y_{2}) \Vert - \Vert x_{0} \Vert }{t} \\ &\leq \frac{ \Vert x_{0}+2ty_{1} \Vert - \Vert x_{0} \Vert }{2t}+\frac{ \Vert x_{0}+2ty_{2} \Vert - \Vert x_{0} \Vert }{2t} \\ &= f(2t,y_{1})+f(2t,y_{2})\quad (\forall t\in R,t>0) \end{aligned}$$
it is easy to see that
$$\begin{aligned} &\wedge\bigl\{ f(t,y_{1}+y_{2}):t\in R,t>0\bigr\} \\ &\quad \leq\wedge \bigl\{ f(2t,y_{1})+f(2t,y_{2}):t\in R,t>0\bigr\} \\ &\quad =\wedge\bigl\{ f(2t,y_{1}):t\in R,t>0\bigr\} +\wedge\bigl\{ f(2t,y_{2}):t\in R,t>0\bigr\} , \end{aligned}$$
which implies that
$$G_{+}(x_{0},y_{1}+y_{2})\leq G_{+}(x_{0},y_{1})+G_{+}(x_{0},y_{2}). $$
□
The Gâteaux differentiability was defined for proper local functions from a random locally convex module to \(\bar{L}^{0}(\mathcal {F})\) in [19], Definition 5.2. Since a random locally convex module is a generalization of a random normed module and a random norm is \({L}^{0}\)-convex, it is easy to see that a random norm is also a proper local function. For random norms we present the following definition of Gâteaux differentiability, which is equivalent to that in [19], Definition 5.2. Under Definition 4.3 we can establish the relations among supporting functionals, points of random smoothness and Gâteaux differentiability of random norms.
Definition 4.3
Let \((E, \Vert \cdot \Vert )\) be an RN module, \(x_{0}\in S(E)\) and \(y\in E\). Then \(G_{-}(x_{0},y)\) and \(G_{+}(x_{0},y)\) are respectively called the left-hand and right-hand Gâteaux derivative of the random norm \(\Vert \cdot \Vert \) at \(x_{0}\) in the direction y. \(\Vert \cdot \Vert \) is called Gâteaux differentiable at \(x_{0}\) in the direction y if \(G_{-}(x_{0},y)=G_{+}(x_{0},y)\), in which case the common value of \(G_{-}(x_{0},y)\) and \(G_{+}(x_{0},y)\) is denoted by \(G(x_{0},y)\) and is called the Gâteaux derivative of \(\Vert \cdot \Vert \) at \(x_{0}\) in the direction y. If \(\Vert \cdot \Vert \) is Gâteaux differentiable at \(x_{0}\) in every direction \(y\in E\), then it is called Gâteaux differentiable at \(x_{0}\). Finally, if \(\Vert \cdot \Vert \) is Gâteaux differentiable at every point of the random unit sphere \(S(E)\), then \(\Vert \cdot \Vert \) is said to be Gâteaux differentiable.
Remark 4.4
It is obvious that \(G_{-}(x,y)\leq G_{+}(x,y), G_{-}(x,-y)=-G_{+}(x,y)\) and \(G_{-}(x,x)=G_{+}(x,x)=G(x,x)= \Vert x \Vert =I_{A_{x}}\) for any x in \(S(E)\) and y in E.
Lemma 4.5
Let
\((E, \Vert \cdot \Vert )\)
be an RN module, \(x_{0}\in S(E)\)
and
\(f\in S(E^{\ast })\)
such that
\(P(D)>0\), where
\(D=A_{x_{0}}\cap A_{f}\). Then
\(I_{D}f\)
supports
\(U(E)\)
at
\(x_{0}\)
if and only if
$$ I_{D}G_{-}(x_{0},y)\leq I_{D}\operatorname{Re}f(y) \leq I_{D}G_{+}(x_{0},y),\quad\forall y\in E. $$
(5)
Proof
(⇒). By the following facts
$$\begin{aligned} &\operatorname{Re}I_{D}f(x_{0})=\vee\bigl\{ \operatorname{Re}I_{D}f(x):x\in U(E) \bigr\} =I_{D} \Vert f \Vert ^{\ast}=I_{D}, \\ &I_{D}f(x_{0})=\operatorname{Re}I_{D}f(x_{0})-i\operatorname{Re}I_{D}f(ix_{0}) \quad\text{and}\quad I_{D}f(x_{0})\leq I_{D} \Vert f \Vert ^{\ast} \Vert x_{0} \Vert =I_{D}, \end{aligned}$$
one can obtain that \(I_{D}f(x_{0})=I_{D}\). Since
$$I_{D}\operatorname{Re}f(ty)=I_{D}\operatorname{Re}f(x_{0}+ty)-I_{D}\operatorname{Re}f(x_{0}) \leq I_{D}\bigl( \Vert x_{0}+ty \Vert - \Vert x_{0} \Vert \bigr) $$
for any positive numbers t and any \(y\in E\), the following inequalities hold:
$$\begin{aligned} I_{D}\frac{ \Vert x_{0}-ty \Vert - \Vert x_{0} \Vert }{-t}&=-I_{D}\frac{ \Vert x_{0}+t(-y) \Vert - \Vert x_{0} \Vert }{t} \\ &\leq-I_{D}\frac{\operatorname{Re}f(t(-y))}{t} =I_{D} \operatorname{Re}f(y)\leq I_{D}\frac{ \Vert x_{0}+ty \Vert - \Vert x_{0} \Vert }{t}, \end{aligned}$$
which implies (5).
(⇐). Since \(G_{-}(x_{0},x_{0})= \Vert x_{0} \Vert =G_{+}(x_{0},x_{0})\), it follows that \(I_{D}\operatorname{Re}f(x_{0})=I_{D}\) by (5). Noticing that
$$\operatorname{Re}I_{D}f(x_{0})=I_{D}=\vee\bigl\{ \operatorname{Re}I_{D}f(x):x\in U(E)\bigr\} , $$
we complete the proof. □
Theorem 4.6
Let
\((E, \Vert \cdot \Vert )\)
be an RN module and
\(x_{0}\)
be a point in
\(S(E)\). Then the following statements hold:
-
(1)
\(x_{0}\)
is a point of random smoothness of
\(U(E)\)
if and only if
\(\Vert \cdot \Vert \)
is Gâteaux differentiable at
\(x_{0}\);
-
(2)
If
\(x_{0}\)
is a point of random smoothness of
\(U(E)\)
and
f
in
\(S(E^{\ast})\)
supports
\(U(E)\)
at
\(x_{0}\), then
\(I_{A_{f}}G(x_{0},y)=\operatorname{Re}I_{A_{f}}f(y)\).
Proof
(1). (⇒). Assume by way of contradiction that \(\Vert \cdot \Vert \) is not Gâteaux differentiable at \(x_{0}\). Then there exists \(y_{0}\) in E such that \(G_{-}(x_{0},y_{0})< G_{+}(x_{0},y_{0})\), namely, \(P(M)>0\), where \(M=[G_{-}(x_{0},y_{0})< G_{+}(x_{0},y_{0})]\). Since \(I_{A_{x_{0}}^{c}}G_{-}(x_{0},y_{0})=I_{A_{x_{0}}^{c}}G_{+}(x_{0},y_{0})=I_{A_{x_{0}}^{c}} \Vert y_{0} \Vert \) and \(I_{A_{y_{0}}^{c}}G_{-}(x_{0},y_{0})=I_{A_{y_{0}}^{c}}G_{+}(x_{0},y_{0})=0\), one knows that \(M\subset A_{x_{0}}\) and that \(M\subset A_{y_{0}}\).
Take an arbitrary s in \(L^{0}({\mathcal {F}}, R)\) such that
$$ G_{-}(x_{0},y_{0})< s< G_{+}(x_{0},y_{0}) \quad\text{on }M. $$
(6)
Define \(W=\{ry_{0}:r\in L^{0}({\mathcal {F}}, R)\}\) and \(f_{s}(ry_{0})=I_{M}rs,\forall r\in L^{0}({\mathcal {F}}, R)\), then it is clear that W is an \(L^{0}({\mathcal {F}}, R)\)-submodule of E, that \(f_{s}\) is an \(L^{0}\)-linear function on W, and that
$$f_{s}(ry_{0})=I_{M}rs\leq I_{M}rG_{+}(x_{0},y_{0})=I_{M}G_{+}(x_{0},ry_{0}),\quad \forall r\in L^{0}_{+}. $$
For any \(r\in L^{0}({\mathcal {F}}, R)\backslash L^{0}_{+}\), denote \(D=M\cap[r<0]\), then \(f_{s}(-I_{D}ry_{0})=-I_{D}rs\geq -I_{D}rG_{-}(x_{0},y_{0})=G_{-}(x_{0},-I_{D}ry_{0})=-G_{+}(x_{0},I_{D}ry_{0})\), and hence \(f_{s}(I_{D}ry_{0})\leq G_{+}(x_{0},I_{D}ry_{0})=I_{D}G_{+}(x_{0},ry_{0})\). Combining the fact that \(f_{s}(I_{M\backslash D}ry_{0})=I_{M\backslash D}rs\leq I_{M\backslash D}G_{+}(x_{0},ry_{0})\), we can see that \(f_{s}(ry_{0})\leq I_{D}G_{+}(x_{0},ry_{0})+I_{M\backslash D}G_{+}(x_{0},ry_{0})=I_{M}G_{+}(x_{0},ry_{0})\). Thus
$$f_{s}(ry_{0})\leq I_{M}G_{+}(x_{0},ry_{0}),\quad \forall r\in L^{0}({\mathcal {F}}, R). $$
By Theorem 2.1 there exists an \(L^{0}\)-linear function \(x_{s}^{\ast }:E\rightarrow L^{0}({\mathcal {F}}, R)\) such that \(x_{s}^{\ast}| _{W}=f_{s}\) and \(x_{s}^{\ast}(y)\leq G_{+}(x_{0},y),\forall y\in E\).
Without loss of generality, suppose that E is a complex RN module. Let
$$g_{s}(y)=x_{s}^{\ast}(y)-ix_{s}^{\ast}(iy). $$
It is easy to show that \(g_{s}:E\rightarrow L^{0}({\mathcal {F}}, C)\) is \(L^{0}\)-linear and \(\operatorname{Re}g_{s}=x_{s}^{\ast}\), from which it follows that \(\operatorname{Re}g_{s}(y)\leq G_{+}(x_{0},y),\forall y\in E\). Then we obtain
$$ \operatorname{Re}g_{s}(y)\leq G_{+}(x_{0},y)\leq\frac{ \Vert x_{0}+y \Vert - \Vert x_{0} \Vert }{1}\leq \Vert y \Vert ,\quad \forall y\in E $$
and
$$ \operatorname{Re}g_{s}(y)=-\operatorname{Re}g_{s}(-y)\geq-G_{+}(x_{0},-y)=G_{-}(x_{0},y), $$
which together with the fact that \(\Vert g_{s} \Vert ^{\ast}= \Vert \operatorname{Re}g_{s} \Vert ^{\ast}\) yields that \(g_{s}\in E^{\ast}\), \(\Vert g_{s} \Vert ^{\ast}\leq1\) and \(G_{-}(x_{0},x_{0})\leq \operatorname{Re}g_{s}(x_{0})\leq G_{+}(x_{0},x_{0})\). Thus \(\operatorname{Re}g_{s}(x_{0})=I_{A_{x_{0}}}\) and so \(\Vert I_{A_{x_{0}}}g_{s} \Vert ^{\ast }=I_{A_{x_{0}}}\). Namely, \(I_{A_{x_{0}}}g_{s}\) supports \(U(E)\) at \(x_{0}\).
Therefore, for any \(s_{1}\) and \(s_{2}\) satisfying (6) and \(s_{1}I_{M}\neq s_{2}I_{M}\), there exist two corresponding \(g_{s_{1}}\) and \(g_{s_{2}}\) in \(E^{\ast}\) such that both \(I_{A_{x_{0}}}g_{s_{1}}\) and \(I_{A_{x_{0}}}g_{s_{2}}\) support \(U(E)\) at \(x_{0}\) and \(\Vert I_{A_{x_{0}}}g_{s_{1}} \Vert ^{\ast}= \Vert I_{A_{x_{0}}}g_{s_{2}} \Vert ^{\ast}=I_{A_{x_{0}}}\). Noticing that
$$\begin{aligned} &I_{A_{x_{0}}}\operatorname{Re}g_{s_{1}}(y_{0})=I_{A_{x_{0}}}x_{s_{1}}^{\ast }(y_{0})=I_{A_{x_{0}}}f_{s_{1}}(y_{0})=I_{M}s_{1} \\ & I_{A_{x_{0}}}\operatorname{Re}g_{s_{2}}(y_{0})=I_{A_{x_{0}}}x_{s_{2}}^{\ast }(y_{0})=I_{A_{x_{0}}}f_{s_{2}}(y_{0})=I_{M}s_{2}, \end{aligned}$$
we derive that \(I_{A_{x_{0}}}\operatorname{Re}g_{s_{1}}\neq I_{A_{x_{0}}}\operatorname{Re}g_{s_{2}}\), which is a contradiction to Proposition 3.6.
(⇐). Suppose that \(\Vert \cdot \Vert \) is Gâteaux differentiable at \(x_{0}\), f in \(S(E^{\ast})\) supports \(U(E)\) at \(x_{0}\). Then it is obvious that \(A_{f}\subset A_{x_{0}}\). By Lemma 4.5 we have
$$ I_{A_{f}}\operatorname{Re}f(y)=I_{A_{f}}G(x_{0},y),\quad \forall y\in E. $$
(7)
If \(f_{1}\) and \(f_{2}\) in \(S(E^{\ast})\) both support \(U(E)\) at \(x_{0}\) and \(P(A_{f_{1}f_{2}})>0\), then \(I_{A_{f_{1}f_{2}}}\operatorname{Re}f_{1}=I_{A_{f_{1}f_{2}}}\operatorname{Re}f_{2}\), which implies that \(I_{A_{f_{1}f_{2}}}f_{1}=I_{A_{f_{1}f_{2}}}f_{2}\). By Proposition 3.6
\(x_{0}\) is a point of random smoothness of \(U(E)\).
(2). It is clear by (7). □
Remark 4.7
It should be pointed out that further development of random smoothness under Definition 3.5 confronted some obstacles, one of which is how to establish the connection between random smoothness of an RN module E and classical smoothness of the abstract \(L^{p}\) space derived from E. It is an interesting problem and attracts some attention in different literature [19, 24].