Universal convexity and range problems of shifted hypergeometric functions

Theorem 1.2. Let a, b, c be positive numbers with $a \le b$ and $\delta = a + b - c \in (0,1)$ and $$\varepsilon = \left| \frac{\Gamma(c)\Gamma(-\delta)}{\Gamma(c-a)\Gamma(c-b)} \right| + 2^{1-\delta} \cdot \frac{\Gamma(c)\Gamma(\delta)}{\Gamma(a)\Gamma(b)}.$$ Then the following hold. - (I) The image of the unit disk $\mathbb{D}$ under the mapping $z_2F_1(a,b;c;z)$ is contained in $S_{\varepsilon}(\delta)$ . - (II) The image of the unit disk $\mathbb{D}$ under the mapping $z_2F_1(a,b;c;z^2)$ is contained in $S_{\varepsilon}^*(\delta)$ . When either c=a or c=b, we understand $\varepsilon=2^{1-\delta}$ in the above. Note that the above theorem does not tell univalence of the mappings. Indeed, Ponnusamy and Vuorinen [10, Thm 6.1] observed that $f(z)=z_2F_1(a,b;c;z)$ is not univalent for $a,b\in(0,2)$ and 0< c< ab/2. However, if f(z) is univalent and has a nice geometric property, we may make a stronger assertion. For a function f analytic on $\mathbb{D}$ and normalized by f(0) = f'(0) - 1 = 0, the order of convexity of f is defined by $$\kappa = \kappa(f) := \inf_{z \in \mathbb{D}} \operatorname{Re} \left( 1 + \frac{zf''(z)}{f'(z)} \right) \in [-\infty, 1].$$ It is known that f is convex, i.e. f is univalent on $\mathbb{D}$ and $f(\mathbb{D})$ is a convex domain if and only if $\kappa(f) \geq 0$ . It is also known that if $\kappa(f) \geq -1/2$ then f is univalent on $\mathbb{D}$ and $f(\mathbb{D})$ is convex in (at least) one direction, see [14] and [11, p.73]. Starlikeness of shifted hypergeometric functions was thoroughly studied by Küstner [3, 4]. However, convexity of shifted hypergeometric functions is seemingly difficult to establish except for the case when b=1, see [4, pp. 1368–1369]. Recently, one of the authors obtained the following result concerning the order of convexity. Theorem A. ([16, Thm.1.1]) For real parameters a, b and c satisfying $0 < a < 1 < b \le c < \min\{a+b, 1+a+b-ab\}$ , the order of convexity of the function $z_2F_1(a, b; c; z)$ is $$\kappa = \frac{c^2 - a^2 - b^2 + 3(a+b-c) - 2}{2(a+b-c)}.$$ In particular, $z_2F_1(a,b;c;z)$ is convex when $c^2-a^2-b^2+3(a+b-c)\geq 2$ . For convex functions, we can refine Theorem 1.2 in the following form.

Theorem 1.3. For positive numbers a, b and c with $\delta = a + b - c \in (0,1)$ , suppose that the shifted hypergeometric function $f(z) = z_2 F_1(a,b;c;z)$ is convex on $\mathbb{D}$ . Then the image $f(\mathbb{D})$ is contained in the sector $S = \{z : |\arg(z - B)| < \pi\delta/2\}$ and the boundary of S is contained in the union of asymptotic lines of the boundary of S, where S is optimal. Since the origin is contained in $f(\mathbb{D})$ , the inequality B < 0 must hold under the assumption of Theorem 1.3. Note also that $S \subset S_{-B}(\delta)$ . Ruscheweyh, Salinas and Sugawa in [12] extended the study of geometric properties of analytic functions on the unit disk to other disks and half planes containing the origin. They introduced the class of universally prestarlike functions of order $\alpha \leq 1$ in the domain $\Lambda$ (containing universally convex and universally starlike functions as the special cases when $\alpha = 0$ and $\alpha = 1/2$ , respectively). A general condition of a shifted hypergeometric function to be universally starlike is given in [12]. However, a similar condition for universal convexity is not known so far except for the case b = 1 (see Theorem C below) to the knowledge of the authors. (In [12], universal convexity is shown for functions of the form $(c/ab)[{}_2F_1(a,b;c;z)-1]$ .) In the present paper, we investigate the universal convexity of the shifted hypergeometric functions.

Theorem 1.4. Let a, b and c be positive numbers with $a \leq b \leq c$ . If one of the conditions (I) and (II) below is satisfied, then the shifted hypergeometric function $f(z) = z_2F_1(a,b;c;z)$ is universally convex on the slit plane $\Lambda = \mathbb{C} \setminus [1,+\infty)$ : - (I) a < 1, $a + b \ge 1$ and $1 + a + b ab \le c \le 2$ ; - (II) $b \le 1$ and $2 \le c$ .

Lemma 2.1. Let $F_n$ (n = 1, 2, ...) be a sequence of functions in $\mathcal{T}$ . If $F_n$ converges to a function F on $\Lambda$ pointwise. Then $F \in \mathcal{T}$ .

Lemma 2.3. Let $F \in \mathcal{T}$ . Then $\text{Im} [zF(z)] \geq 0$ if Im z > 0.

Lemma 2.4. Let F(z) be analytic on $\Lambda$ . Then $F \in \mathcal{T}$ if and only if the following four conditions are fulfilled: - (i) F(0) = 1; - (ii) $F(x) \in \mathbb{R}$ for $x \in (-\infty, 1)$ ; - (iii) $\operatorname{Im} F(z) \geq 0$ for $\operatorname{Im} z > 0$ ; - (iv) $\limsup_{x \to +\infty} F(-x) \ge 0$ . <span id="page-4-0"></span>Remark 2.5. By the integral form of functions in $\mathcal{T}$ we have indeed F(x) is positive and non-decreasing in $-\infty < x < 1$ for every $F \in \mathcal{T}$ . In particular, the limit $\alpha = \lim_{x \to +\infty} F(-x)$ exists and satisfies $0 \le \alpha \le F(0) = 1$ . Note also that $\alpha = 1$ occurs only when $F(z) \equiv 1$ . We remark further that $G = (F - \alpha)/(1 - \alpha) \in \mathcal{T}$ if $\alpha < 1$ . Hence, $F = \alpha + (1 - \alpha)G$ . Note that the class $\mathcal{T}$ is convex; namely, $aF + (1 - a)G \in \mathcal{T}$ whenever $F, G \in \mathcal{T}$ and $0 \le a \le 1$ . In particular, $1 - a + aF \in \mathcal{T}$ for $F \in \mathcal{T}$ and $0 \le a \le 1$ because the constant function 1 belongs to $\mathcal{T}$ . It is a good position to give an account on the proof of Theorem C. Proof of Theorem C. Put $g(z) = z_2 F_1(a, 2; c; z)$ and recall $f(z) = z_2 F_1(a, 1; c; z)$ . Then z f'(z) = g(z) (see [4, p. 1368]) and thus $$\Psi(z) = 1 + \frac{zf''(z)}{f'(z)} = \frac{zg'(z)}{g(z)}.$$ Under the assumption, Theorem 1.8 in [12] implies that g is universally starlike; in other words, the function $\Psi(z) = zg'(z)/g(z)$ belongs to the class $\mathcal{T}$ . Hence, we conclude that $1 + zf''(z)/[2f'(z)] = (1 + \Psi(z))/2$ belongs to $\mathcal{T}$ , too. The next lemma asserts that a nontrivial transformation preserves the class $\mathcal{T}$ .

Lemma 2.6. ([16, Lem.2.5]). Let $$F \in \mathcal{T}$$ . Then the function $\frac{1}{(1-z)F(z)}$ belongs to $\mathcal{T}$ . The behavior of the hypergeometric function ${}_{2}F_{1}(a,b;c;z)$ at z=1 is completely different according to the sign of a+b-c as follows (see also [1], [7], [8]):

Lemma 2.7. Let $a, b, c \in \mathbb{R}$ with $c \neq 0, -1, -2, ...$ (1) If c-a-b>0, the limit of ${}_2F_1(a,b;c;z)$ exists as $z\to 1$ in the domain $\Lambda=\mathbb{C}\setminus [1,+\infty)$ and it is given by $$_{2}F_{1}(a,b;c;1) = \frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}.$$ (2) If c - a - b = 0, as $z \to 1$ in $\Lambda$ $$_{2}F_{1}(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\log\frac{1}{1-z} + O(1).$$ (3) If c - a - b < 0, as $z \to 1$ in $\Lambda$ $$_{2}F_{1}(a,b;c;z) = \frac{\Gamma(c)\Gamma(a+b-c)}{\Gamma(a)\Gamma(b)}(1-z)^{c-a-b} + O(|1-z|^{c-a-b+\varepsilon})$$ <span id="page-5-3"></span>for some number $\varepsilon > 0$ . <span id="page-5-4"></span>By using this, we can show the following fact.

Lemma 2.8. Let $0 < a \le b \le c$ . Then (2.2) $$\lim_{x \to +\infty} \frac{{}_{2}F_{1}(a+1,b;c;-x)}{{}_{2}F_{1}(a,b;c;-x)} = 0.$$ If b < c. (2.3) $$\lim_{x \to +\infty} \frac{x_2 F_1(a+1,b+1;c+1;-x)}{{}_2F_1(a+1,b;c;-x)} = +\infty.$$ If furthermore $b \le a + 1$ in addition to b < c, (2.4) $$\lim_{x \to +\infty} \frac{{}_{2}F_{1}(a+1,b+1;c+1;-x)}{{}_{2}F_{1}(a+1,b;c;-x)} = 0.$$

Lemma 2.9. Let $0 < a \le b \le c$ and f denote the shifted hypergeometric function $z_2F_1(a,b;c;z)$ . Then $$\lim_{x \to -\infty} \frac{xf''(x)}{f'(x)} = -a.$$

Lemma 2.12. ([13, Lem.3.2]) Let $\Omega$ be an unbounded convex domain in $\mathbb{C}$ whose boundary is parametrized positively by a Jordan curve w(t) = u(t) + iv(t), 0 < t < 1, with $w(0^+) = w(1^-) = \infty$ . Suppose that $u(0^+) = +\infty$ and that v(t) has a finite limit as $t \to 0^+$ . Then $v(t) \leq v(0^+)$ for 0 < t < 1. The next two lemmas describe the properties of the ratio of two hypergeometric functions in different aspects.

Lemma 2.13. ([3, Thm.1.5], [15, p.337-339 and Thm.69.2]). If $-1 \le a \le c$ and $0 \le b \le c \ne 0$ , then the three functions $$\frac{{}_{2}F_{1}(a+1,b;c;z)}{{}_{2}F_{1}(a,b;c;z)}, \quad \frac{{}_{2}F_{1}(a+1,b+1;c+1;z)}{{}_{2}F_{1}(a,b;c;z)} \quad and \quad \frac{{}_{2}F_{1}(a+1,b+1;c+1;z)}{{}_{2}F_{1}(a+1,b;c;z)}$$ belong to the class $\mathcal{T}$ .

Proposition 3.1. Let a, b and c be positive numbers with $a \le b \le c$ and $a \le 1$ and put $f(z) = z_2 F_1(a, b; c; z)$ . If the inequality $$(3.1) (2-c)(c+ab-a-b-1) = (2-c)(1-a)(1-b) - (2-c)^2 \ge 0$$ holds, then <span id="page-10-1"></span> $$\frac{f''(z)}{f'(z)} = \frac{a+b-1}{1-z} + \left(1 - a - b + \frac{2ab}{c}\right) \Phi_1(z),$$ for a function $\Phi_1 \in \mathcal{T}$ . Proof. We first consider the special case when (1-a)(1-b)=0. Note that (3.1) then implies c=2. For instance, if a=1, then we can easily see that $f'(z)=(1-z)^{-b}$ and f''(z)/f'(z)=b/(1-z). Thus the conclusion is confirmed in this case. Therefore, in the following, we will assume that either $(1-a)(1-b)\neq 0$ . Since $a\leq 1$ , we have thus a<1. We set $\tau = ab/c$ (> 0) for the sake of brevity and make preliminary observations. When $c + ab - a - b - 1 \ge 0$ and $2 - c \ge 0$ , we have $$1 - a - b + 2\tau \ge 1 - a - b + 2(1 + a + b - c)/c = \frac{(1 + a + b)(2 - c)}{c} \ge 0.$$ When $c+ab-a-b-1 \le 0$ and $2-c \le 0$ , we have similarly $1-a-b+2\tau \le 0$ . Therefore, the assumption (3.1) implies that $c+ab-a-b-1, 1-a-b+2\tau, 1-b, (1-a)(1-b)-(2-c)$ and 2-c have the same sign (admitting to be 0). We next write $F(z) = {}_2F_1(a,b;c;z)$ , $G(z) = {}_2F_1(a+1,b;c;z)$ and $H(z) = {}_2F_1(a+1,b+1;c+1;z)$ for convenience. We infer from the contiguous relation (3.2) $$G(z) - F(z) = -\frac{b}{c} zH(z)$$ that (2.5) can now be rearranged as <span id="page-10-2"></span> $$\frac{zf''(z)}{f'(z)} = \frac{2 - c + (a + b - 1)z}{1 - z} + \frac{c - 2 + (1 - a)(1 - b)z}{(1 - z)(1 + \tau zH(z)/F(z))}$$ $$= \frac{(a + b - 1)z}{1 - z} + \frac{z}{1 - z} \frac{(1 - a)(1 - b) + (2 - c)\tau H(z)/F(z)}{1 + \tau zH(z)/F(z)}.$$ Hence, $$\frac{f''(z)}{f'(z)} = \frac{a+b-1}{1-z} + \frac{1-a-b+2\tau}{(1-z)M(z)},$$ where (3.3) $$M(z) = \frac{(1-a-b+2\tau)[1+\tau zw(z)]}{(1-a)(1-b)+(2-c)\tau w(z)} \quad \text{and} \quad w(z) = \frac{H(z)}{F(z)}.$$ We show now that $M \in \mathcal{T}$ , which yields that $\Phi_1(z) = 1/[(1-z)M(z)]$ belongs to $\mathcal{T}$ by Lemma 2.6. For this, it is sufficient to verify conditions (i)-(iv) in Lemma 2.4. Condition (i): M(0) = 1 is easy to check. Since $0 < a \le b \le c$ , we can apply Lemma 2.13 to obtain $w \in \mathcal{T}$ ; that is, w is represented by $$w(z) = \int_0^1 \frac{d\mu(t)}{1 - tz}, \quad z \in \Lambda,$$ for a probability measure µ on [0, 1]. We now check condition (ii). Since (1 − a)(1 − b) and 2 − c have the same sign and w(x) > 0 for x ∈ (−∞, 1), the denominator (1 − a)(1 − b) + (2 − c)τw(x) of M(x) is either positive for all x < 1 or negative for all x < 1. Hence, the denominator of M(x) does not vanish and thus M(x) ∈ R for x < 1, which confirms condition (ii). In order to verify condition (iii), we compute the imaginary part of M(z) as follows: $$\operatorname{Im} M(z) \left| (1-a)(1-b) + (2-c)\tau w(z) \right|^{2}$$ $$= (1-a-b+2\tau) \operatorname{Im} \left[ [1+\tau z w(z)] \left( (1-a)(1-b) + (2-c)\tau \overline{w(z)} \right) \right]$$ $$= (1-a-b+2\tau) \operatorname{Im} \left( (1-a)(1-b)\tau z w(z) - (2-c)\tau w(z) + (2-c)\tau^{2} z \left| w(z) \right|^{2} \right)$$ $$= (1-a-b+2\tau)\tau \operatorname{Im} \left( \int_{0}^{1} \frac{(1-a)(1-b)z - (2-c)}{1-tz} d\mu(t) + (2-c)\tau z \left| w(z) \right|^{2} \right)$$ $$= (1-a-b+2\tau)\tau y \left( \int_{0}^{1} \frac{(1-a)(1-b) - (2-c)t}{|1-tz|^{2}} d\mu(t) + (2-c)\tau \left| w(z) \right|^{2} \right),$$ for z = x + iy. Making use of the preliminary observations above, we see that (1 − a − b + 2τ )[(1 − a)(1 − b) − (2 − c)t] ≥ 0 for 0 ≤ t ≤ 1 and (1 − a − b + 2τ )(2 − c) ≥ 0, by which we verify condition (iii). By [\(2.3\)](#page-5-1) and the contiguous relation [\(3.2\)](#page-10-2), we have $$\lim_{x \to +\infty} xw(-x) = \lim_{x \to +\infty} \frac{xH(-x)}{G(-x) + bxH(-x)/c} = \lim_{x \to +\infty} \frac{cxH(-x)/G(-x)}{c + bxH(-x)/G(-x)} = \frac{c}{b},$$ and, in particular, w(−x) → 0 as x → +∞. Thus, when a, b 6= 1, condition (iv) is checked by computing $$\lim_{x \to +\infty} M(-x) = (1 - a - b + 2\tau) \lim_{x \to +\infty} \frac{1 - \tau x w(-x)}{(1 - a)(1 - b) + (2 - c)\tau w(-x)}$$ $$= \frac{(1 - a - b + 2\tau)(1 - a)}{(1 - a)(1 - b)} = \frac{1 - a - b + 2\tau}{1 - b} \ge 0.$$ Hence, we have now concluded that M(z) belongs to the class T , as required. <span id="page-11-0"></span>Proposition 3.2. First let a, b and c be positive numbers with a ≤ b ≤ 1 and c ≥ 2. Then, the pre-Schwarzian derivative of the shifted hypergeomegtric function f(z) = z2F1(a, b; c; z) is expressed by $$\frac{zf''(z)}{f'(z)} = a\Phi_2(z) - a$$ for a function Φ² ∈ T . Proof. We use the same notation as in the previous proof. In addition, set σ = (a − 1)b/c and σ ′ = (c − a − 1)b/c. For a while, we assume that 0 < a ≤ b < 1 < 2 < c. By the proof of Theorem 1.2 of [17] we have $$\frac{zf''(z)}{f'(z)} = -a + \frac{a(c-a-1)}{(1-a)(1-z)} + \frac{a}{1-a} \cdot \frac{2-c+(1-a)(b-1)z}{(1-z)(1+\sigma zH(z)/G(z))}$$ $$= -a + \frac{a}{1-z} \frac{1+(b-1)z-\sigma'zH(z)/G(z)}{1+\sigma zH(z)/G(z)}$$ $$= -a + \frac{a}{(1-z)M(z)},$$ where $M = M_1/M_2$ and $$M_1(z) = 1 + \sigma z w_1(z), \quad M_2(z) = 1 + (b-1)z - \sigma' z w_1(z), \quad w_1(z) = \frac{H(z)}{G(z)}.$$ We show now that $M \in \mathcal{T}$ , which implies that $\Phi_2(z) = 1/[(1-z)M(z)]$ belongs to $\mathcal{T}$ by virtue of Lemma 2.6. (Note that the proof of Theorem 1.2 of [17] was flawed [18] but the above formula is valid.) Lemma 2.13 implies that $$w_1(z) = \int_0^1 \frac{d\mu(t)}{1 - tz}, \quad z \in \Lambda,$$ for a probability measure $\mu$ on [0,1]. In order to show that $M \in \mathcal{T}$ , we only need to check conditions (i)-(iv) in Lemma 2.4. Condition (i) follows by definition. We show now (ii). The point is to see $M_2(x) = 1 + (b-1)x - \sigma' x w_1(x) \neq 0$ for x < 1. Since $$xw_1(x) = \int_0^1 \frac{x}{1-tx} d\mu(t)$$ is non-decreasing in $-\infty < x < 1$ , b-1 < 0 and $\sigma' > 0$ , the function $M_2(x)$ is decreasing in x < 1. we need only to show that $M_2(1^-) \ge 0$ . By Lemma 2.10, $$\lim_{x \to 1^{-}} M_{2}(x) = b - \sigma' \lim_{x \to 1^{-}} \frac{H(x)}{G(x)}$$ $$= b - \frac{(c - a - 1)b}{c} \cdot \frac{c}{\max\{b, c - a - 1\}}$$ $$= \max\{1 + a + b - c, 0\} \ge 0$$ and condition (ii) has been verified. Next we show condition (iii). As before, by making use of Lemma 2.3, we have $$\operatorname{Im} M(z) |1 + (b - 1)z - \sigma' z w_1(z)|^2$$ $$= \operatorname{Im} \left[ [1 + \sigma z w_1(z)] \left( 1 + (b - 1)\overline{z} - \sigma' \overline{z w_1(z)} \right) \right]$$ $$= (1 - b)y + (\sigma + \sigma') \operatorname{Im} \left[ z w_1(z) \right] + (b - 1)\sigma |z|^2 \operatorname{Im} w_1(z)$$ > 0 for z = x + iy with y > 0, since all the coefficients 1 - b, $\sigma + \sigma' = (c - 2)b/c$ , $(b - 1)\sigma$ are positive. Finally, we check (iv) in Lemma 2.4. By (2.4), we have $$M(-x) = \frac{M_1(-x)}{M_2(-x)} = \frac{1/x - \sigma w_1(-x)}{1/x + (1-b) + \sigma' w_1(-x)} \to 0$$ as $x \to +\infty$ and thus (iv) is confirmed. Thus the assertion has been proved when $a \le b < 1$ and 2 < c. The general case is verified by using, for example, the approximation $a_n = a - 1/n$ , $b_n = b - 1/n$ , $c_n = c + 1/n$ (n: large enough) with the compactness property of the class $\mathcal{T}$ (see Lemma 2.1). Now the proof is complete. Proof of Theorem 1.4. We will employ the same symbols as in the proof of Proposition 3.1. Thus f(z) means the shifted hypergeometric function $z_2F_1(a,b;c;z)$ . By Theorem B, it is enough to prove that the function $1 + zf''(z)/[2f'(z)] = (1 + \Psi(z))/2$ belongs to the class $\mathcal{T}$ , where $\Psi(z) = 1 + zf''(z)/f'(z)$ . By convexity of $\mathcal{T}$ , it suffices to show that $\Psi$ belongs to $\mathcal{T}$ . Thus we only need to verify conditions (i)–(iv) in Lemma 2.4 for $\Psi$ . At this point, we note that (iv) is guaranteed by Lemma 2.9, which says that $\Psi(-x) \to 1 - a \ge 0$ as $x \to +\infty$ . Case (I): Note that the assumptions of Proposition 3.1 are satisfied in this case. Thus the function $\Psi$ is expressed as $$\Psi(z) = 1 + \frac{(a+b-1)z}{1-z} + (1-a-b+2\tau)z\Phi_1(z), \quad z \in \Lambda.$$ Conditions (i) and (ii) are clearly satisfied. Condition (iii) is easily verified by using Lemma 2.3. Case (II): We make use of Proposition 3.2. All the assumptions of the proposition are satisfied in this case. Recall the formula there: $$\Psi(z) = 1 + \frac{zf''(z)}{f'(z)} = 1 - a + a\Phi_2(z)$$ $\Box$ for some $\Phi_2 \in \mathcal{T}$ . Hence $\Psi \in \mathcal{T}$ by convexity of $\mathcal{T}$ .

Definition 1. ([12, Def.1.4]) A function holomorphic in $\Lambda = \mathbb{C} \setminus [1, +\infty)$ is called universally convex if it maps every half-plane or disk in $\Lambda$ containing the origin onto a convex domain. We denote by $\mathcal{P}$ the class of analytic functions P with P(0) = 1 and $\operatorname{Re} P > 1/2$ on $\mathbb{D}$ . The following formula is known as the Herglotz representation: <span id="page-2-1"></span> $$P(z) = \int_{\partial \mathbb{D}} \frac{d\mu(\zeta)}{1 - \bar{\zeta}z}, \quad z \in \mathbb{D},$$ for a (Borel) probability measure $\mu$ on the unit circle $\partial \mathbb{D}$ . (In what follows, we will surpress "Borel" because we will consider only Borel measures here.) Similarly, we consider the class $\mathcal{T}$ of those analytic functions F with F(0) = 1 which admit the representation (1.1) $$F(z) = \int_0^1 \frac{d\mu(t)}{1 - tz}, \quad z \in \Lambda,$$ for a probability measure $\mu$ on the interval [0,1]. Such a function F is expanded in the form of power series $F(z) = c_0 + c_1 z + c_2 z^2 + \ldots$ on |z| < 1 for a totally monotone (or completely monotone) sequence $\{c_n\}$ with $c_0 = 1$ , that is, $\Delta^k c_n \geq 0$ for all $k, n \geq 0$ , where $\Delta^0 c_n = c_n$ and $\Delta^{k+1} c_n = \Delta^k c_n - \Delta^k c_{n+1}$ for $k, n \geq 0$ . Conversely, for a totally monotone sequence $\{c_n\}$ with $c_0 = 1$ , the function $F(z) = \sum c_n z^n$ is known to be in the class $\mathcal{T}$ (see [15] for details). An analytic characterization of universally convex functions is given in terms of the class $\mathcal{T}$ as follows. Theorem B. ([12, Thm.1.6]) An analytic function f on $\Lambda$ with f(0) = 0 and f'(0) = 1 is universally convex if and only if 1 + zf''(z)/(2f'(z)) belongs to the class $\mathcal{T}$ . We note that a normalized analytic function f on $\mathbb{D}$ is convex if and only if 1 + zf''(z)/(2f'(z)) belongs to the class $\mathcal{P}$ . Since $\mathcal{T} \subset \mathcal{P}$ , we see that a universally convex function is always convex. Let $f(z) = z + a_2 z^2 + a_3 z^3 + \dots$ be a universally convex function. Then, by a formal computation, we have for $1 + zf''(z)/(2f'(z)) = 1 + c_1 z + c_2 z^2 + \dots$ $$c_1 = a_2$$ , $c_2 = 3a_3 - 2a_2^2$ , and $c_3 = 6a_4 - 9a_2a_3 + 4a_2^3$ . Since $\{c_n\}$ with $c_0 = 1$ is totally monotone, we have countably many inequalities such as $$0 \le c_1 = a_2 \le 1$$ and $0 \le c_1 - c_2 = a_2 - 3a_3 + 2a_2^2 \le 1 - c_1 = 1 - a_2$ . For the shifted hypergeometric function $f(z) = z_2 F_1(a, b; c; z) = z + \alpha z^2 + \alpha \beta z^3 + \dots$ with $\alpha = ab/c$ and $\beta = (a+1)(b+1)/[2(c+1)]$ , we have the necessary conditions for universal convexity of f such as $$0 \le \alpha \le 1$$ , $3\beta \le 1 + 2\alpha$ and $\alpha(2 + 2\alpha - 3\beta) \le 1$ . It is, of course, not practical to check all the inequalities $\Delta^k c_n \geq 0$ to obtain a sufficient condition for universal convexity of f. Since $zf'(z) = z_2F_1(a, 2; c; z)$ for $f(z) = z_2F_1(a, 1; c; z)$ , we have the following result immediately from Ruscheweyh et al. [12, Thm.1.8]. Theorem C. Let a and c be positive numbers with $a \le 1$ and $2 \le c$ . Then the shifted hypergeometric function $f(z) = z_2 F_1(a, 1; c; z)$ is universally convex. Since the theorem does not appear in this form in the literature, we will include a brief account of the proof after Remark 2.5 below. In the present paper, the authors extend it in the following way.