Starlike and convex maps in Banach spaces

Lemma 1. Let $v(x, t) \colon B \times I \to B$ be holomorphic for each $t \in I = [0, 1]$ , v(0, t) = 0 and v(x, 0) = x. If $\lim_{t\to 0^+} [(x - v(x, t))/t] = w(x)$ exists and is holomorphic in B then $w \in \mathscr{S}_0(B)$ .

Lemma 2. Let $f: B \to Y$ be a biholomorphic map (holomorphic with a holomorphic inverse) of B onto an open set $f(B) \subset Y$ and let f(0) = 0. Let $F(x, t): B \times I \to Y$ be a holomorphic function of x for each $t \in I$ , F(x, 0) = f(x), F(0, t) = 0 and suppose $F(B, t) \subset f(B)$ for each $t \in I$ . Further, suppose $\lim_{t\to 0^+} [(F(x, 0) - F(x, t))/t] = F(x)$ exists and is holomorphic. Then F(x) = Df(x)(w(x)) where $w \in \mathscr{S}_0(B)$ (Df(x) is the Fréchet derivative of f at x).

Lemma 3. If $w \in \mathcal{S}_0(B)$ and $|\alpha| < 1$ then $\frac{1}{\alpha}w(\alpha x) \in \mathscr{S}_0(B)$ $\left(\frac{1}{\alpha}w(\alpha x)\right)$ is understood to be the limit value $$Dw(0)(x)$$ when $\alpha = 0$ . Further, if $x' \in T(x)$ , 0 < ||x|| < 1 then x'(w(x)) = 0 if and only if x'(Dw(0)(x)) = 0 and in this case $x'(1/\alpha \cdot w(\alpha x)) \equiv 0$ when $|\alpha| < 1/||x||$ .

Theorem 1. Suppose $f: B \rightarrow Y$ is starlike and that $f^{-1}$ is holomorphic on f(B) open $\subset Y$ . There exists $w \in \mathscr{S}(B)$ such that $$f(x) = Df(x)(w(x)).$$

Theorem 2. Let $f: B \to Y$ be holomorphic and f(0) = 0. Assume Df(x) has a bounded inverse for each $x \in B$ (hence $f^{-1}$ exists and is holomorphic in a neighborhood of each point of f(B)) and that for some $w \in \mathcal{S}(B)$ , f(x) = Df(x)(w(x)). Suppose further that for each r, 0 < r < 1, there exists M(r) such that $||[Df(x)]^{-1}|| \leq M(r)$ when $||x|| \leq r$ . Then f is starlike.

Lemma 4. If $f: B \rightarrow Y$ is a biholomorphic map of B onto a convex domain then $f(B_r)$ is convex for each r, $0 \le r \le 1$ .

Theorem 3. If $f: B \rightarrow Y$ is convex then (3) $$D^{2}f(x)(x, x) + Df(x)(x) = Df(x)(w(x))$$ where $w \in \mathcal{P}(B)$ .

Lemma 5. If $w \in Q_0(B)$ and $|\alpha| < 1$ then $1/\alpha \cdot w(\alpha x, \alpha y) \in Q_0(B)$ (the limit value at $\alpha = 0$ is Dw(0, 0)(x, y). Further, if $x' \in T(x)$ , $0 \neq x \in B$ , and $||y|| \leq ||x||$ then x'(w(x, y)) = 0 if and only if x'(Dw(0, 0)(x, y)) = 0. Proof. See the proof of Lemma 3.

Theorem 4. Suppose $f: B \to Y$ is convex. Then f(x) - f(y) = Df(x)(w(x, y)) where $w \in Q(B)$ .

Theorem 5. Suppose $f: B \to Y$ is holomorphic, Df(x) has bounded inverse for each $x \in B$ and that for some $w \in Q(B)$ , f(x) - f(y) = Df(x)(w(x, y)). Suppose further that for each r, 0 < r < 1, there exists M(r) > 0 such that $||[Df(x)]^{-1}|| \leq M(r)$ when $||x|| \leq r$ . Then f is convex.

Theorem 6. Let $f: B \to Y$ be convex, $0 \neq x \in B$ and let $x' \in T(x)$ . Then the hyperplane $\{y \in Y: x'[(Df(x))^{-1}(y)] = x'[(Df(x))^{-1}(f(x))]\}$ is a supporting hyperplane for the convex set $f(B_{||x||})$ . If $y \neq 0$ , x'(y) = 0 and ||x + ty|| = ||x|| for $0 < t < t_0$ , then $x'[(Df(x))^{-1}(f(x + ty))] = x'[(Df(x))^{-1}(f(x))]$ for $0 < t < t_0$ (i.e., f(x + ty) lies in the supporting hyperplane described above).

Theorem 7. Let $X = L(\mu)$ (the space of complex valued integrable functions on S). If $f: B \to Y$ is convex then f(x) - f(0) is linear.

Corollary 1. If $X = \angle^1$ and $f: B \to Y$ is convex then f(x) - f(0) is linear.

Theorem 8. Let $X = L^{\infty}(\mu)$ (the space of essentially bounded, complex valued, measurable functions on S) and suppose $f: B \to Y$ is convex. If $x, y \in B$ where x(s) = y(s) for all $s \in E \subset S$ and $\mu(E) > 0$ then $(Df(0))^{-1}(f(x))(s) = (Df(0))^{-1}f(y)(s)$ for almost all $s \in E$ . Proof. We may assume Y=X, f(0)=0 and Df(0)=I (otherwise, replace f by $(Df(0))^{-1}(f(x)-f(0))$ ). We will show that if $\mu(E)>0$ , then for almost all $s\in E$ we have $f(x)(s)=f(xx_E)(s)$ (where $x_E$ is the characteristic function of E). We accomplish this by showing, $(Df(x))^{-1}D^nf(x)(y,\dots,y)(s)=0$ a.e. on E when y(s)=0 on E ( $n\geq 2$ ) and $D^2f(x)(x_E,y)=0$ when y=0 on E. An induction argument will then show $D^nf(x)(x_E,y,y,\dots,y)=\lim z\to 0$ $[D^{n-1}f(x+zy)(x_E,y,y,\dots,y)-D^{n-1}f(x)(x_E,y,y,\dots,y)]/z=0$ for $n\geq 3$ when y=0 on E and using the fact that $D^nf(x)$ is n-linear and symmetric (writing $y=yx_E+y(1-x_E)$ ) we have $$D^n f(x)(y, y, \dots, y) = D^n f(x)(yx_E, yx_E, \dots, yx_E) + D^n f(x)(y(1 - x_E), y(1 - x_E), \dots, y(1 - x_E))$$ for all $n \ge 1$ when E is measurable. Hence for $s \in E$ we have $$D^n f(0)(y, y, \dots, y)(s) = D^n f(0)(yx_E, yx_E, \dots, yx_E)(s)$$ a.e. on $E$ . Therefore, $$f(x)(s) = \left[\sum_{k=1}^{\infty} \frac{1}{k!} D^k f(0)(x, x, \dots, x)\right](s)$$ $$= \left[\sum_{k=1}^{\infty} \frac{1}{k!} D^k f(0)(x, x, \dots, x)(s)\right]$$ $$\sum_{k=1}^{\infty} \frac{1}{k!} D^k f(0)(xx_E, xx_E, \dots, xx_E)(s)$$ $$= f(xx_E)(s) \text{ for almost all } s \in E.$$ Let E be a set of positive measure, let $0 < \varepsilon < r < 1$ and $$(7) x = \alpha x_E + (1 - x_E)u, u \in X$$ where $|\alpha| = r$ and $||u|| \leq r - \varepsilon$ . Let $y \in B$ satisfy $$(8) y(s) = 0 if s \in E.$$ Then $||x + \beta y|| = ||x|| = r$ for $|\beta|$ sufficiently small. Let x' be the real continuous linear functional on X given by $$x'(v) = \operatorname{Re} \int_{E} \left( \frac{\overline{\alpha}}{r\mu(E)} \right) v d\mu$$ . Then $x' \in T(x)$ and $x'(\beta y) = 0$ so Theorem 6 applies. Since arg $\beta$ is arbitrary we may therefore conclude $$(9) (Df(x))^{-1}D^2f(x)(y, y)(s) = 0 a.e. on E.$$ But $(Df(x))^{-1}D^2f(x)(y, y)$ is an analytic function of $\alpha$ for $|\alpha| < 1$ (E, u, and y) remain fixed) which satisfies (9) when $|\alpha| > ||u||$ . Hence (9) holds when $|\alpha| < 1$ and ||u|| < 1. For arbitrary $x \in B$ , we may approximate $xx_E$ on E by simple functions to see that (9) holds for all $x \in B$ if y(s) = 0 on E. An induction argument shows that (9) holds when $D^2f(x)(y, y)$ is replaced by $D^nf(x)(y, y, \dots, y)$ where $n \ge 2$ and y = 0 on E. We now wish to show $D^2 f(x)(x_E, y) = 0$ for all $x \in B$ and $y \in X$ satisfying y(s) = 0 if $s \in E$ . Let x and y be given by (7) and (8) and set $h = \beta y + t\alpha x_E$ where $|\beta| = t^{1/2}$ . Then for t sufficiently small, $||x + h||\sqrt{1 + t^2} ||x||$ and $$w\left(x, \frac{1}{\sqrt{1+t^2}}(x+h)\right) = (Df(x))^{-1}(f(x) - f\left(\frac{1}{\sqrt{1+t^2}}\right)(x+h)\right)$$ $$= -v - (Df(x))^{-1}(D^2f(x)(v,v)) - \cdots$$ where $$v = \left(\frac{1}{\sqrt{1+t^2}} - 1\right)x + \frac{1}{\sqrt{1+t^2}} \cdot h$$ . With x' as above, applying Theorem 4 we obtain $$\begin{aligned} 0 & \leq x' \Big( w \Big( x, \frac{1}{\sqrt{1+t^2}} (x+h) \Big) \Big) \ & = & \operatorname{Re} \Big( \frac{-irt\beta}{\mu(E)} \! \int_{\scriptscriptstyle E} (Df(x))^{-1} D^2 f(x) (x_{\scriptscriptstyle E}, \, y) d\mu \Big) \ & + o(t^2). \end{aligned}$$ Since $|\beta| = t^{1/2}$ and arg $\beta$ is arbitrary, we conclude (10) $$\int_{E} (Df(x))^{-1} D^{2} f(x)(x_{E}, y) d\mu = 0 \text{ when } y = 0 \text{ on } E.$$ Let $F \subset S$ , F measurable, and $\mu(F) > 0$ . Equation (10) implies $$\int_{F \cap E} (Df(x))^{-1} D^2 f(x)(x_E, y) d\mu = 0.$$ Also, $$\begin{align} \int_{F-E}\!\!L_x(x_{\!\scriptscriptstyle E},y)d\mu &= \int_{F-E}\!\!\left[L_x(yx_{\!\scriptscriptstyle F},x_{\!\scriptscriptstyle E})\right. \ &+ \frac{1}{2}L_x(x_{\!\scriptscriptstyle E}+y(1-x_{\!\scriptscriptstyle F}),x_{\!\scriptscriptstyle E}+y(1-x_{\!\scriptscriptstyle F})) -\frac{1}{2}L_x(x_{\!\scriptscriptstyle E},x_{\!\scriptscriptstyle E}) \ &- \frac{1}{2}L_x(y(1-x_{\!\scriptscriptstyle F}),y(1-x_{\!\scriptscriptstyle F}))\right]\!d\mu = 0 \end{array}$$ for the first term is 0 by (10) applied to F-E and the other 3 terms are 0 by (9) applied to F-E. Hence we conclude $(Df(x))^{-1}D^2f(x)(x_E, y)(s) = 0$ a.e. on S so $D^2f(x)(x_E, y) = 0$ . Again taking S to be the positive integers with $\mu$ the counting measure we obtain the following corollary.

Corollary 2. If $X = \angle^{\infty}$ and $f: B \to Y$ is convex then f(x) - f(0) = Df(0)(g(x)) where $g(x) = (g_1(x_1), g_2(x_2), \cdots)$ and $g_k(x_k) = x_k + a_{2k}x_k^2 + \cdots$ maps $|x_k| < 1$ onto a convex domain. 4. Discussion. The techniques used in this paper may also be used to extend the concepts of close-to-convexity and spirallikeness to Banach spaces. There are at least two different ways of extending the analytic condition for close-to-convexity to Banach spaces and these apparently lead to different classes of functions. Results concerning these concepts are as yet incomplete and will be discussed in a later paper.

Definition 1. A holomorphic map $f: B \to Y$ is starlike if f is one-to-one, f(0) = 0 and $(1 - t)f(B) \subset f(B)$ for all $t \in I$ .

Definition 2. If $f: B \rightarrow Y$ is a biholomorphic map of B onto a convex domain, we say that f is convex. If X=C, it is well known than a function f holomorphic in |z|<1 satisfying $f'(0)\neq 0$ is convex if and only if $\text{Re}\left[zf''(z)/f'(z)+1\right]>0$ in |z|<1. The analogous condition for a general Banach space is necessary but not sufficient.