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Abstract

Let ${\mathcal S}$ denote the family of all univalent functions $f$ in the unit disk $\ID$ with the normalization $f(0)=0= f'(0)-1$. There is an intimate relationship between the operator $P_f(z)=f(z)/f'(z)$ and the Danikas-Ruscheweyh operator $T_f:=\int_{0}^{z}(tf'(t)/f(t))\,dt$. In this paper we mainly consider the univalence problem of $F=P_f$, where $f$ belongs to some subclasses of ${\mathcal S}$. Among several sharp results and non-sharp results, we also show that if $f\in {\mathcal S}$, t

Results & Lemmas (15)

Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.

Theorem 1. Theorem 1. If f ∈S⋆(β), then Pf ∈U in the disk |z| < 1/(1 + p 2(1 −β)). The result is sharp (as for univalence) as the function z/(1…
Theorem 1. If f ∈S⋆(β), then Pf ∈U in the disk |z| < 1/(1 + p 2(1 −β)). The result is sharp (as for univalence) as the function z/(1 −z)2(1−β) shows.
Corollary 1. Corollary 1. If f ∈S⋆, then Pf ∈U ∩S⋆in the disk |z| < √ 2 −1. The result is sharp (as for univalence) as the Koebe function z/(1 −z)2…
Corollary 1. If f ∈S⋆, then Pf ∈U ∩S⋆in the disk |z| < √ 2 −1. The result is sharp (as for univalence) as the Koebe function z/(1 −z)2 shows.
Corollary 2. Corollary 2. If f ∈S⋆(1/2), then Pf ∈U ∩S⋆in the disk |z| < 1/2. The result is sharp as the function z/(1 −z) shows.
Corollary 2. If f ∈S⋆(1/2), then Pf ∈U ∩S⋆in the disk |z| < 1/2. The result is sharp as the function z/(1 −z) shows.
Corollary 3. Corollary 3. If f ∈S⋆(1/2) such that f ′′(0) = 0, then Pf is starlike in the disk |z| < r2, where r2 ≈0.543689 is the root of the equation…
Corollary 3. If f ∈S⋆(1/2) such that f ′′(0) = 0, then Pf is starlike in the disk |z| < r2, where r2 ≈0.543689 is the root of the equation φ2(r) = 0, where φ2(r) = r3 + r2 + r −1.
Corollary 4. Corollary 4. Let f belong to either S⋆(1/2) or C(−1/2), such that f ′′(0) = 0. Then F ∈U in the disk |z| < 1/ √ 3.
Corollary 4. Let f belong to either S⋆(1/2) or C(−1/2), such that f ′′(0) = 0. Then F ∈U in the disk |z| < 1/ √ 3.
Theorem 2. Theorem 2. If f ∈G(α) for some α ∈(0, 1], then Pf is starlike in the disk |z| < 1 + α − p α(1 + α).
Theorem 2. If f ∈G(α) for some α ∈(0, 1], then Pf is starlike in the disk |z| < 1 + α − p α(1 + α) .
Corollary 5. Corollary 5. If f ∈G, then Pf is starlike in the disk |z| < 2 − √ 2 ≈0.585786. The same reasoning gives as in Corollary 3 the following.
Corollary 5. If f ∈G, then Pf is starlike in the disk |z| < 2 − √ 2 ≈0.585786 . The same reasoning gives as in Corollary 3 the following.
Corollary 6. Corollary 6. If f ∈G(α) such that f ′′(0) = 0 and for some α ∈(0, 1], then Pf is starlike in |z| < r4(α), where r4(α) is the root in the…
Corollary 6. If f ∈G(α) such that f ′′(0) = 0 and for some α ∈(0, 1], then Pf is starlike in |z| < r4(α), where r4(α) is the root in the interval (0, 1] of the equation φ4(r) = 0, φ4(r) = r4 −αr3 −(2 + α)r2 −αr + 1 + α.
Corollary 7. Corollary 7. If f ∈G such that f ′′(0) = 0, then Pf is starlike in |z| < r4, where r4 ≈0.64731 is the root in the interval (0, 1] of the…
Corollary 7. If f ∈G such that f ′′(0) = 0, then Pf is starlike in |z| < r4, where r4 ≈0.64731 is the root in the interval (0, 1] of the equation r4 −r3 −3r2−r+2 = 0.
Theorem 3. Theorem 3. If f ∈G(α) for some α ∈(0, 1], then F ∈U in the disk |z| < r5(α), where r5(α) = q −α+√ (1+α)2+1 2.
Theorem 3. If f ∈G(α) for some α ∈(0, 1], then F ∈U in the disk |z| < r5(α), where r5(α) = q −α+√ (1+α)2+1 2 .
Theorem 2 Theorem 2, one has z F(z) −1 = − αω(z) 1 + α −ω(z) and, using this relation, we find that UF(z) = − αω(z) 1 + α −ω(z) + α(1 + α)zω′(z) (1 +…
Theorem 2, one has z F(z) −1 = − αω(z) 1 + α −ω(z) and, using this relation, we find that UF(z) = − αω(z) 1 + α −ω(z) + α(1 + α)zω′(z) (1 + α −ω(z))2 = α[(1 + α)(zω′(z) −ω(z)) + ω2(z)] (1 + α −ω(z))2
Corollary 8. Corollary 8. If f ∈G, then F belongs to the class U in the disk |z| < q√ 5−1 2 ≈ 0.78615.
Corollary 8. If f ∈G, then F belongs to the class U in the disk |z| < q√ 5−1 2 ≈ 0.78615.
Theorem 4. Theorem 4. Let f ∈S with a2 = f ′′(0)/2!. Then F belongs to U in the disk |z| < r6(|a2|), where r6(|a2|) is the root of the equation φ5(r)…
Theorem 4. Let f ∈S with a2 = f ′′(0)/2!. Then F belongs to U in the disk |z| < r6(|a2|), where r6(|a2|) is the root of the equation φ5(r) = 0 that lies in the interval (0, 1), where φ5(r) = (a+1−1 4b2)r10−(5a+5−5 4b2)r8+(19a+10−19 4 b2)r6+(9a−10−9 4b2)r4+5r2−1 with b = |a2| and a = 2π2−12 3 ≈2.57974.
Theorem 1.1 Theorem 1.1] ∞ X n=1  n n + 1 2 |cn(f)|2 ≤2π2 −12 3 = a. By (10), we obtain zf ′(z) f(z) −1 = ∞
Theorem 1.1] ∞ X n=1  n n + 1 2 |cn(f)|2 ≤2π2 −12 3 = a. By (10), we obtain zf ′(z) f(z) −1 = ∞
Corollary 9. Corollary 9. Let f ∈S with f ′′(0) = 0, and a = 2π2−12 3. Then F belongs to U in the disk |z| < r6, where r6 ≈0.360794 is the root of the…
Corollary 9. Let f ∈S with f ′′(0) = 0, and a = 2π2−12 3 . Then F belongs to U in the disk |z| < r6, where r6 ≈0.360794 is the root of the equation (a + 1)r10 −5(a + 1)r8 + (19a + 10)r6 + (9a −10)r4 + 5r2 −1 = 0, that lies in the interval (0, 1).
Function classes studied:

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