Posts by Alexandru Nemeth

SEQUENTIAL REGULARITY AND THE DIRECTIONAL DIFFERANTIABILITY OF CONVEX OPERATORS ARE EQUIVALENT
A.B. Németh

where $⩽$$⩽$<=\leqslant$⩽$ stands for the order relation induced by the cone $K$$K$KK$K$ in $E$$E$EE$E$.
Let $G$$G$GG$G$ be an operator from $H$$H$HH$H$ to $E$$E$EE$E$. The epigraph of $G$$G$GG$G$ is by definition the subset of the Chartesian product $H\times E$$H\times E$H xx EH \times E$H\times E$ given by

It is easy to check that the operator $F:H\to E$$F:H\to E$F:H rarr EF: H \rightarrow E$F:H\to E$ is convex $1f$$1f$1f1 f$1f$ and only if epi $F$$F$FF$F$ is a convex set in the vector space $H\times E$$H\times E$H xx EH \times E$H\times E$.

Suppose now that E is a locally convex space. Then E and equivalently its positive cone $K$$K$KK$K$ is called normal if it has a neighbourhood basis consisting of sets $U$$U$UU$U$ with the property $U=(U+K)\cap (U-K)$$U=(U+K)\cap (U-K)$U=(U+K)nn(U-K)U=(U+K) \cap(U-K)$U=(U+K)\cap (U-K)$.

The ordered locally convex space. E and equivalently its positive cone K is called (sequentially) regular (see e.g. (M1)) if every decreasing net (sequence) in $K$$K$KK$K$ is convergent.

Te observe that the term locally convex ordered regular space is used by various authors and papers in slightly different senses. Ve adopt here the above definition for the sake of simplicity. It coincides with that used by MCARTHUR (MI) when the cone K is closed. BORWEIN (B) uses for (sequentially) regular closed cones the term (countably) Daniell.

Consider the operator $G$$G$GG$G$ from the vector space $H$$H$HH$H$ to the oxdered locally convex space E. If for some $u$$u$uu$u$ and $h$$h$hh$h$ in $H$$H$HH$H$ there exists the limit

then it is called the directional derivative of $G$$G$GG$G$ at $u$$u$uu$u$ in the direction h .

If $F$$F$FF$F$ is a convex operator from the vector space $H$$H$HH$H$ to the locally convex ordered regular space $E$$E$EE$E$, then it has directional derivative at every point $u$$u$uu$u$ and in any direction h (see e.g. (B) Proposition 3.7 (c)). This follows from the observation that for fixed $u$$u$uu$u$ and $h$$h$hh$h$ the function $\phi$$\phi$varphi\varphi$\phi$ defined by
(1) $\phi (t)=t^{-1}(F(u+th)-F(u))$$\phi (t)=t^{-1}(F(u+th)-F(u))$quad varphi(t)=t^(-1)(F(u+th)-F(u))\quad \varphi(t)=t^{-1}(F(u+t h)-F(u))$\phi (t)=t^{-1}(F(u+th)-F(u))$
is increasing with $t$$t$tt$t$ in $R\mathrm{\setminus }\{0\}$$R\mathrm{\setminus }\{0\}$R\\{0}R \backslash\{0\}$R\mathrm{\setminus }\{0\}$ and bounded from below on $(0,\mathrm{\infty })$$(0,\mathrm{\infty })$(0,oo)(0, \infty)$(0,\mathrm{\infty })$ according the converdty of $F$$F$FF$F$. If only sequential regularity of $E$$E$EE$E$ is postulated, then the nozmality of $E$$E$EE$E$ is also used in proving the existence of the directional derivatives of a convex operator (see e.g. (B)). The normality can be assumed to be a mild condition on a cons since the most of important oxdered topological vector spaces have normal positive cones. In some cases regularity and sequential regularity are equivalent (N2). On the other hand if is known that in a Fréchet space every closed regular cone is normal (see (M2), Theorem 3). However, we can avoid the normality assumption in the proof of the result in the title. To show that this circuastance is consistent we show that there are ordered regular normed spaces which are not normal. We shall adapt for this an example due to BRECKNER and ORB $\mathrm{\$}(\genfrac{}{}{0ex}{}{10}{10})$$\mathrm{\$}(\genfrac{}{}{0ex}{}{10}{10})$$((10)/(10))\$\binom{10}{10}$\mathrm{\$}(\genfrac{}{}{0ex}{}{10}{10})$ used for other purposes in (BOI), Remark 2.4.3.

Let I denote the vector space of all polynomials with real caefficients defined on the interval $[-1,0]$$[-1,0]$[-1,0][-1,0]$[-1,0]$, equiped with the uniform norm. Let us consider the cone K defined by

The cone $K$$K$KK$K$ is regular since if $\left(K_h\right)$$\left(K_h\right)$(K_(h))\left(K_{h}\right)$\left(K_h\right)$ is a decreasing sequence in $K$$K$KK$K$, then it is necessarily a sequence of polynomials of a given finite oxder, sey $m$$m$mm$m$, and according the ordering introduced by $\mathbb{K}$$\mathbb{K}$K\mathbb{K}$\mathbb{K}$, their coefficients form decreasing non-negative sequences of real numbers. Since all the coefficients of the terms of order $>$$>$>>$>$ m are 0 , the sequence ( $x_a$$x_a$x_(a)x_{a}$x_a$ ) tends uniformly to a polynomial with non-negative coefficients and the regularity follows.

To prove that $K$$K$KK$K$ is not a nomal cone, consider the sequences $\left(x_n\right)$$\left(x_n\right)$(x_(n))\left(x_{n}\right)$\left(x_n\right)$ and $\left(y_n\right)$$\left(y_n\right)$(y_(n))\left(y_{n}\right)$\left(y_n\right)$ defined by

One has $0⩽x_n⩽y_n$$0⩽x_n⩽y_n$0 <= x_(n) <= y_(n)0 \leqslant x_{n} \leqslant y_{n}$0⩽x_n⩽y_n$ for each $n\in \mathbb{N}$$n\in \mathbb{N}$n inNn \in \mathbb{N}$n\in \mathbb{N}$. The sequence $\left(y_n\right)$$\left(y_n\right)$(y_(n))\left(y_{n}\right)$\left(y_n\right)$ tends uniformly to 0 , but $\left(x_n\right)$$\left(x_n\right)$(x_(n))\left(x_{n}\right)$\left(x_n\right)$ does not. This contradicts a well known ariterion exists. We shall show that $\underset{t⩽0}{lim}\zeta (t)$$\underset{t⩽0}{lim} \zeta (t)$lim_(t <= 0)zeta(t)\lim _{t \leqslant 0} \zeta(t)$\underset{t⩽0}{lim}\zeta (t)$ also exists and equals $v$$v$vv$v$. of normality (see (P), Proposition II. 1.3 or (EO1), Theoren 2.4.2). If we assume by contradiction that the filter of lower sections $M_{t_0}=\{\phi (t),t⩽t_0\},t_0>0$$M_{t_0}=\left\{\phi (t),t⩽t_0\right\},t_0>0$M_(t_(0))={varphi(t),t <= t_(0)},t_(0) > 0M_{t_{0}}=\left\{\varphi(t), t \leqslant t_{0}\right\}, t_{0}>0$M_{t_0}=\{\phi (t),t⩽t_0\},t_0>0$ does not converge to $v$$v$vv$v$, we can get The ain of the present note is to show that the directional dexivability of the convex operators with values in an ordered a neighbourhood $U$$U$UU$U$ of 0 in $E$$E$EE$E$ such that for every $n$$n$nn$n$ there exists a $t_n,0<t_n<1/n$$t_n,0<t_n<1/n$t_(n),0 < t_(n) < 1//nt_{n}, 0<t_{n}<1 / n$t_n,0<t_n<1/n$ such that locally convex space characterizes the sequentially regular spaces. Nore precisely we shall prove the following

THDORET. Let E be an ordered locally convex space. Then E is sequentially regular if and only if every convex operatos defined on a vector space end taking values in I has directional derivative at every point and in every direction.

The necessity of the theorem is in fact known and follows from the monotonity of the operator $\phi$$\phi$varphi\varphi$\phi$ dafined in (1). The only problem in this direction is that we must use only sequential regularity to conclude the existence of directional derivatives.

It is immediate that we can state the theorem considering directional derivability of the convex operators defined on the vector space $R$$R$RR$R$ only.
2. The proof of the theorem. We prove first the necessity, that is that if E is sequentially regular then every convex opcator $F$$F$FF$F$ from the vector space $H$$H$HH$H$ to $\mathbb{E}$$\mathbb{E}$E\mathbb{E}$\mathbb{E}$ has directional derivative at every point $u$$u$uu$u$ and in every direction $h$$h$hh$h$. Consider the function $\phi$$\phi$varphi\varphi$\phi$ derined on $R\mathrm{\setminus }\{0\}$$R\mathrm{\setminus }\{0\}$R\\{0}R \backslash\{0\}$R\mathrm{\setminus }\{0\}$ by (1). Since $\phi$$\phi$varphi\varphi$\phi$ is increasing with $t$$t$tt$t$ and lower bounded on $(0,\mathrm{\infty })$$(0,\mathrm{\infty })$(0,oo)(0, \infty)$(0,\mathrm{\infty })$, the sequence $(\rho (1/n))$$(\rho (1/n))$(rho(1//n))(\rho(1 / n))$(\rho (1/n))$ decreases $f\circ rn\to \mathrm{\infty }$$f\circ rn\to \mathrm{\infty }$f@rn rarr oof \circ r n \rightarrow \infty$f\circ rn\to \mathrm{\infty }$ and is lower bounded. Since I is sequentially regular,

Now, we can put together the sequences $(1/n)$$(1/n)$(1//n)(1 / n)$(1/n)$ and $\left(t_n\right)$$\left(t_n\right)$(t_(n))\left(t_{n}\right)$\left(t_n\right)$ in order to get a decreasing sequence $\left(s_m\right)$$\left(s_m\right)$(s_(m))\left(s_{m}\right)$\left(s_m\right)$ in $(0,\mathrm{\infty })$$(0,\mathrm{\infty })$(0,oo)(0, \infty)$(0,\mathrm{\infty })$ which converges to 0 . Then $\left(\phi \left(s_m\right)\right)$$\left(\phi \left(s_m\right)\right)$(varphi(s_(m)))\left(\varphi\left(s_{m}\right)\right)$\left(\phi \left(s_m\right)\right)$ is decreasing for $m\to \mathrm{\infty }$$m\to \mathrm{\infty }$m rarr oom \rightarrow \infty$m\to \mathrm{\infty }$ and lower bounded. It must be convergent since E is sequentially regular. But this contradicts (2) and (3). Hence

exists.
To prove the sufficence, i.e., that directional derivability. of the convex operators implies sequential regularity, we proceed by contradiction. We shall give indeed a constructive proof of the following proposition :

If the positive cone $K$$K$KK$K$ in $E$$E$EE$E$ contains a decreasing sequence of elements which is not convergent, then there exists a convex operator from $R$$R$RR$R$ to $E$$E$EE$E$ without directional derivative at $0\in R$$0\in R$0in R0 \in R$0\in R$ in the direction $l\in \mathrm{R}$$l\in \mathrm{R}$l inRl \in \mathrm{R}$l\in \mathrm{R}$.

An essential role in our construction has the assertion :
(a) If $K$$K$KK$K$ contains a decreasing sequence which is not convergent, then it contains a decreasing sequence $\left(x_n\right)$$\left(x_n\right)$(x_(n))\left(x_{n}\right)$\left(x_n\right)$ for which it exists a
sequence $\left(t_n\right)$$\left(t_n\right)$(t_(n))\left(t_{n}\right)$\left(t_n\right)$ of numbers with the property $0<t_n<1$$0<t_n<1$0 < t_(n) < 10<t_{n}<1$0<t_n<1$ for each $n$$n$nn$n$, such that

and $\left(x_n\right)$$\left(x_n\right)$(x_(n))\left(x_{n}\right)$\left(x_n\right)$ is not conversent.
Indeed, if $\left(y_n\right)$$\left(y_n\right)$(y_(n))\left(y_{n}\right)$\left(y_n\right)$ is a decreasing sequence in $K$$K$KK$K$, which is not convergent, then if we put

then $\left(x_n\right)$$\left(x_n\right)$(x_(n))\left(x_{n}\right)$\left(x_n\right)$ mill satisfy the condition of the assertion (a).
(b) Let $\left(x_a\right)$$\left(x_a\right)$(x_(a))\left(x_{a}\right)$\left(x_a\right)$ be a sequence with the property asserted in (a).

Then there exists a strictly decreasing sequence $\left(a_n\right)$$\left(a_n\right)$(a_(n))\left(a_{n}\right)$\left(a_n\right)$ of positive real numbers $a_n$$a_n$a_(n)a_{n}$a_n$ such that $lim{a}_n=0$$lim{a}_n=0$lima_(n)=0\lim a_{n}=0$lim{a}_n=0$ and so as to bave
(4) $(a_{n-1}-a_n)(a_n{x}_n-a_{n+1}{x}_{n+1})⩽(a_n-a_{n+1})(a_{n-1}{x}_{n-1}-a_n{x}_n)$$\left(a_{n-1}-a_n\right)\left(a_n{x}_n-a_{n+1}{x}_{n+1}\right)⩽\left(a_n-a_{n+1}\right)\left(a_{n-1}{x}_{n-1}-a_n{x}_n\right)$(a_(n-1)-a_(n))(a_(n)x_(n)-a_(n+1)x_(n+1)) <= (a_(n)-a_(n+1))(a_(n-1)x_(n-1)-a_(n)x_(n))\left(a_{n-1}-a_{n}\right)\left(a_{n} x_{n}-a_{n+1} x_{n+1}\right) \leqslant\left(a_{n}-a_{n+1}\right)\left(a_{n-1} x_{n-1}-a_{n} x_{n}\right)$(a_{n-1}-a_n)(a_n{x}_n-a_{n+1}{x}_{n+1})⩽(a_n-a_{n+1})(a_{n-1}{x}_{n-1}-a_n{x}_n)$
for every $n⩾2$$n⩾2$n >= 2n \geqslant 2$n⩾2$.
We put $a_1=2,a_2=1$$a_1=2,a_2=1$a_(1)=2,a_(2)=1a_{1}=2, a_{2}=1$a_1=2,a_2=1$ and shall show that if $a_1,\dots ,a_n,n⩾2$$a_1,\dots ,a_n,n⩾2$a_(1),dots,a_(n),n >= 2a_{1}, \ldots, a_{n}, n \geqslant 2$a_1,\dots ,a_n,n⩾2$ were choosen, then $a_{n+1}$$a_{n+1}$a_(n+1)a_{n+1}$a_{n+1}$ can be determined in order to have $0<a_{n+1}<min\{2^{-n+3},a_n\}$$0<a_{n+1}<min\left\{2^{-n+3},a_n\right\}$0 < a_(n+1) < min{2^(-n+3),a_(n)}0<a_{n+1}<\min \left\{2^{-n+3}, a_{n}\right\}$0<a_{n+1}<min\{2^{-n+3},a_n\}$ and

Indeed, according the special feature of $\left(x_n\right)$$\left(x_n\right)$(x_(n))\left(x_{n}\right)$\left(x_n\right)$ there exists $0<t_{n-1}<<1$$0<t_{n-1}<<1$0 < t_(n-1)<<10<t_{n-1}< <1$0<t_{n-1}<<1$ such that $x_n⩽t_{n-1}{x}_{n-1}$$x_n⩽t_{n-1}{x}_{n-1}$x_(n) <= t_(n-1)x_(n-1)x_{n} \leqslant t_{n-1} x_{n-1}$x_n⩽t_{n-1}{x}_{n-1}$ and hence, in order to realise (5) it suffices to show that there exists arbitrarily small positive solution in $a_{n+1}$$a_{n+1}$a_(n+1)a_{n+1}$a_{n+1}$ of the inequality

But this inequality is equivalent with

where the right hand side term is a positive real number. Hence $a_{n+1}$$a_{n+1}$a_(n+1)a_{n+1}$a_{n+1}$ can be choosen to satisfy our requirements. Now (5) implies

wherefrom it follows the relation (4).
(c) Let us define the operator $F:R\to E$$F:R\to E$F:R rarr EF: R \rightarrow E$F:R\to E$ by

where $\left(x_n\right)$$\left(x_n\right)$(x_(n))\left(x_{n}\right)$\left(x_n\right)$ and $\left(a_n\right)$$\left(a_n\right)$(a_(n))\left(a_{n}\right)$\left(a_n\right)$ are the sequences satisfying the concitions in (b). Then the opexator $F$$F$FF$F$ is convex.

Let us define the affine operators $F_n:R\to E,n=0,1,2,\dots$$F_n:R\to E,n=0,1,2,\dots$F_(n):R rarr E,n=0,1,2,dotsF_{n}: R \rightarrow E, n=0,1,2, \ldots$F_n:R\to E,n=0,1,2,\dots$, with $E_0(t)=0,t\in R$$E_0(t)=0,t\in R$E_(0)(t)=0,t in RE_{0}(t)=0, t \in R$E_0(t)=0,t\in R$,

and

for $n⩾2$$n⩾2$n >= 2n \geqslant 2$n⩾2$. The operators $F_n$$F_n$F_(n)F_{n}$F_n$ are obviously convex, hence the epigraphs

are convex.
We shall show that
(7)

and
(8) : $(t,F_n(t))\in$$\left(t,F_n(t)\right)\in$(t,F_(n)(t))in\left(t, F_{n}(t)\right) \in$(t,F_n(t))\in$ epi $F_{n+1}$$F_{n+1}$F_(n+1)F_{n+1}$F_{n+1}$ if $t\in [a_{n+1},\mathrm{\infty })$$t\in \left[a_{n+1},\mathrm{\infty }\right)$t in[a_(n+1),oo)t \in\left[a_{n+1}, \infty\right)$t\in [a_{n+1},\mathrm{\infty })$
for $n=1,2,\dots$$n=1,2,\dots$n=1,2,dotsn=1,2, \ldots$n=1,2,\dots$. Indeed, we have
$F_n(t)-F_{n+1}(t)=(\frac{a_n{x}_n-a_{n+1}{x}_{n+1}}{a_n-a_{n+1}}-\frac{a_{n+1}{x}_{n+1}-a_{n+2}{x}_{n+2}}{a_{n+1}-a_{n+2}})(t-a_{n+1})$$F_n(t)-F_{n+1}(t)=\left(\frac{a_n{x}_n-a_{n+1}{x}_{n+1}}{a_n-a_{n+1}}-\frac{a_{n+1}{x}_{n+1}-a_{n+2}{x}_{n+2}}{a_{n+1}-a_{n+2}}\right)\left(t-a_{n+1}\right)$F_(n)(t)-F_(n+1)(t)=((a_(n)x_(n)-a_(n+1)x_(n+1))/(a_(n)-a_(n+1))-(a_(n+1)x_(n+1)-a_(n+2)x_(n+2))/(a_(n+1)-a_(n+2)))(t-a_(n+1))F_{n}(t)-F_{n+1}(t)=\left(\frac{a_{n} x_{n}-a_{n+1} x_{n+1}}{a_{n}-a_{n+1}}-\frac{a_{n+1} x_{n+1}-a_{n+2} x_{n+2}}{a_{n+1}-a_{n+2}}\right)\left(t-a_{n+1}\right)$F_n(t)-F_{n+1}(t)=(\frac{a_n{x}_n-a_{n+1}{x}_{n+1}}{a_n-a_{n+1}}-\frac{a_{n+1}{x}_{n+1}-a_{n+2}{x}_{n+2}}{a_{n+1}-a_{n+2}})(t-a_{n+1})$.
According the relation (4),

and hence

and

From the second relation we have (7) while from the first it follows (8).

We shall show that
(9)

To this end, we verify first that

Consider the order relation $\prec$$\prec$-<\prec$\prec$ in the vector space $\mathrm{R}\times \mathrm{E}$$\mathrm{R}\times \mathrm{E}$RxxE\mathrm{R} \times \mathrm{E}$\mathrm{R}\times \mathrm{E}$ induced by the cone $\{0\}\times K$$\{0\}\times K${0}xx K\{0\} \times K$\{0\}\times K$, with $K$$K$KK$K$ the positive cone in $E$$E$EE$E$. Then the relation (7) can be written as

Suppose that $t\in [a_{k+1},a_k)$$t\in \left[a_{k+1},a_k\right)$t in[a_(k+1),a_(k))t \in\left[a_{k+1}, a_{k}\right)$t\in [a_{k+1},a_k)$ if $k>1$$k>1$k > 1k>1$k>1$ and that $t\in [1,\mathrm{\infty })$$t\in [1,\mathrm{\infty })$t in[1,oo)t \in[1, \infty)$t\in [1,\mathrm{\infty })$ if $k=1$$k=1$k=1k=1$k=1$. Let $k>m$$k>m$k > mk>m$k>m$. Then using successively the relation above for $n=m,m+1,\dots ,k-1$$n=m,m+1,\dots ,k-1$n=m,m+1,dots,k-1n=m, m+1, \ldots, k-1$n=m,m+1,\dots ,k-1$, we get, since $t<a_{n+1}$$t<a_{n+1}$t < a_(n+1)t<a_{n+1}$t<a_{n+1}$, the relations

Since $t\in [a_{k+1},a_k)$$t\in \left[a_{k+1},a_k\right)$t in[a_(k+1),a_(k))t \in\left[a_{k+1}, a_{k}\right)$t\in [a_{k+1},a_k)$ for $k>1$$k>1$k > 1k>1$k>1$ and $t\in [1,0)$$t\in [1,0)$t in[1,0)t \in[1,0)$t\in [1,0)$ for $k=1$$k=1$k=1k=1$k=1$, we have $F(t)=F_k(t)$$F(t)=F_k(t)$F(t)=F_(k)(t)F(t)=F_{k}(t)$F(t)=F_k(t)$ and the above relation yields