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Put $\alpha =-\log |c|>0$. Since $f$ has no zeros and the unit disc is simply connected, it has a holomorphic logarithm. Thus $F=-\log f$ is holomorphic with $\mathrm{Re}F>0$ and $\mathrm{Re}F(0)=\alpha$. The Cayley transform between the half-plane and disk givesThe Schwarz lemma yields $|{\varphi }^{\prime }(0)|\le 1$, so the Carathéodory derivative bound for the right half-plane gives $|F^{\prime }(0)|\le 2\alpha$. Since $f^{\prime }=-F^{\prime }f$, the sharp answer isEquality in the Schwarz lemma holds exactly when $\varphi (z)=\omega z$, $|\omega |=1$. Solving for $F$ and exponentiating gives exactly the extremizersTheir logarithmic real part is $\alpha \mathrm{Re}[(1+\omega z)/(1-\omega z)]>0$, so each is indeed a zero-free map into the unit disc with the prescribed value at zero.

For a fixed $w\in \mathbb{D}$, put $r=|w|$. Apply the Harnack inequality on the unit disk to the positive harmonic function $\mathrm{Re}F$ and then exponentiate its negative. The zero-free Schwarz lemma givesFor $w\ne 0$ both endpoints are attained: take $\omega w=r$ for the lower endpoint and $\omega w=-r$ for the upper endpoint in the displayed extremizers. At zero both are $|c|$. If the question is interpreted as allowing $w$ to vary as well, the sharp infimum and supremum over all admissible pairs $(f,w)$ are respectively $0$ and $1$, approached as $r\uparrow 1$; neither value is attained inside the unit disc. An individual function need not have these global extremal values: the constant function $f=c$ is a simple example.

The Harnack inequality on the unit disk states that a nonnegative harmonic function satisfiesOne proof applies the Poisson integral on the unit disk on a circle of radius $R>|z|$: the Poisson kernel on the circle lies between $(R-|z|)/(R+|z|)$ and its reciprocal. The mean value property for harmonic functions identifies the mean boundary value with $u(0)$; let $R\uparrow 1$.

A harmonic conjugate on the simply connected unit disc makes $H=u+iv$ holomorphic. The Carathéodory derivative bound for the right half-plane gives $|H^{\prime }(0)|\le 2u(0)$. The Cauchy-Riemann equations give $|H^{\prime }|=|\nabla u|$, so $|\nabla u(0)|\le 2u(0)$. For a fixed $z$, use the disk automorphismThe chain rule and conformality multiply the gradient length by $|{\psi }^{\prime }(0)|$. Applying the estimate at zero to $u\circ \psi$ proves the sharp gradient bound for positive harmonic functions:Sharpness follows from $u(\zeta )=\mathrm{Re}[(1+\zeta )/(1-\zeta )]$, which has equality at zero, and then composing with ${\psi }^{-1}$. These extremizers are positive multiples of the Poisson kernel on the circle with its boundary pole fixed at one point.

For a positive harmonic function on the unit disc, choose a global harmonic conjugate and apply the Carathéodory derivative bound for the right half-plane to $u+iv$. This gives $|\nabla u(0)|\le 2u(0)$ by the Cauchy-Riemann equations. A disk automorphism sending zero to $z$ has derivative modulus $1-|z{|}^2$, so the chain rule gives the displayed bound. Composing a boundary-pole Poisson kernel with a disk automorphism attains equality, proving the coefficient is optimal at every point.

For a zero-free holomorphic function $f:\mathbb{D}\to \mathbb{D}$, put $c=f(0)$ and $\alpha =-\log |c|$. A holomorphic logarithm makes $-\log f$ have positive real part. The Carathéodory derivative bound for the right half-plane yields the displayed sharp bound. Equality functions are $f(z)=c\exp [-2\alpha \omega z/(1-\omega z)]$, $|\omega |=1$. The Harnack inequality on the unit disk also yields the sharp pointwise bounds $|c{|}^{(1+r)/(1-r)}\le |f(w)|\le |c{|}^{(1-r)/(1+r)}$ for $r=|w|$. Aligning $\omega w=\pm r$ attains the two endpoints.