$2\left|f'(0)\right|=\sup_{z,\omega \in \mathbb{D}} \left|f(z)-f(\omega)\right|$ implies that $f$ is linear. 
Suppose that $f:\mathbb{D}\to \mathbb{C}$ is holomorphic with
   $$2\left|f'(0)\right|=\sup_{z,\omega \in \mathbb{D}} \left|f(z)-f(\omega)\right|$$
  Prove that $f$ is linear.

My attempt
Suppose that 
$$f(z)=\sum_{n=0}^\infty a_n z^n$$
Then $2\left|f'(0)\right|=2|a_1|$, and 
\begin{align*}
\sup_{z,\omega \in \mathbb{D}} \left|f(z)-f(\omega)\right|
&\ge \sup_{|z|=1} \left|f(z)-f(-z)\right| \\
&=2|a_1|\sup_{|z|=1} \left|1+a_3z^2+a_5z^4+\cdots\right| \\
&\ge 2|a_1|
\end{align*}
where we use the maximum modulus principle.
But I got stuck to show that the equality holds precisely when $a_2=a_3=\cdots=0$. Any hints would be highly appreciated.
 A: Assume wlog $f'(0)=a_1=1$ so $|f(z)-f(w)| \le 2$
Let $f(\mathbb D)=U$ open; consider the convex open hull $K$ of $U$. In other words, we define inductively $U_1=U, U_2$ the union of segments with endpoints in $U_1$, $U_3$ the union of segments with endpoints in $U_2$ etc and $K=\cup U_k$ convex since any two points in $K$ appear in some $U_m$ hence the segment joining them is in $U_{m+1}$ hence in $K$. 
It is easy to verify that indeed $U_k$ hence $K$ open (so being convex it is simply connected) and that the diameter $d(U_k) \le 2$ by induction using that in a convex quadrilateral, any interior segment is not larger than the maximum of the sides and diagonals, hence $d(K)\le 2$. 
But now if $g$ is the unique Riemann map $g:\mathbb D \to K, g(0)=0, g'(0)>0$, it immediately follows by Schwarz Lemma applied to $g^{-1}(f(z))$ that $g'(0) \ge 1$ and by the diameter property $g'(0) \le 1$, hence $g'(0)=1, f=g$ convex univalent and $U=K$
It is well known and not hard to prove by symmetrization that the area of $K$ is at most $\pi d(K)^2/4=\pi$
But by the usual integral formula since $f(z)=z+\sum_{k\ge 2}{a_kz^k}$ univalent, we get that the area of $K$ is $\int_{\mathbb D}|f'(z)|^2dxdy=\pi(1+\sum_{k\ge 2}{k|a_k|^2})$ hence $a_k=0, k \ge 2$ and we are done!
A: Let, $\displaystyle f(z) = \sum\limits_{k=0}^{\infty} a_kz^k$ for $z \in \mathbb{D}$. Then defining $d_r := \operatorname{diam}f(r\mathbb{D})$ we see that $d_r/r$ is a non decreasing function in $r$ (this follows by applying the max-modulus principle to $\displaystyle \frac{f(z) - f(wz)}{z}$ on the disk $z \in r\mathbb{D}$ where, $|w| = 1$). Therefore, letting $r \to 0^+$ we see that $$2|a_1| = 2|f'(0)| = \limsup\limits_{r \to 0^+} \frac{d_r}{r} \le d_1. \tag{1}$$ Then the equality $2|a_1| = d_1$ implies $d_r/r = 2|a_1|$ for all $r \in [0,1)$. 
Also, from Schwarz lemma we must have $$|f(z) - f(-z)| \le \frac{d_r}{r}|z|, \, \text{ for } z \in r\mathbb{D} \tag{2}$$ and in particular $2|f'(0)| = 2|a_1| \le d_r/r$ (letting, $|z| \to 0^+$). I.e., the equality $2|a_1| = d_r/r$ corresponds to equality in Schwarz lemma. Hence, we have $$f(z) - f(-z) = \frac{d_r}{r}z = 2a_1z \tag{3}$$ (wlog, we can assume that the unimodular constant is $1$ and in particular $a_1$ is a real number).
Now, consider the function $g(\theta) := |f(e^{i\theta}z) - f(-z)|^2$ where, we have fixed a $z \in \partial r\mathbb{D}$ (i.e., $|z| = r$). Then from $(2)$ we know that $g(\theta)$ is maximized when $\theta = 0$. In particular we must have $g'(0) = 0$.
Now, substituting from relation $(3)$ we note that \begin{align*}g'(\theta) &= \frac{d}{d\theta}\left|f(e^{i\theta}z) - f(z) + 2a_1z\right|^2 \\&= \frac{d}{d\theta} \left[\left(f(e^{i\theta}z) - f(z) + 2a_1z\right) \left(\overline{f(e^{i\theta}z) - f(z) + 2a_1z}\right) \right] \\&= 2 \Re \left[ ie^{i\theta}zf'(e^{i\theta}z)\left(\overline{f(e^{i\theta}z) - f(z) + 2a_1z}\right) \right]. \tag{4}\end{align*}
That is $\displaystyle g'(0) = -2|z|^2a_1\Im \left[f'(z)\right] = 0$ for all $|z| = r$, and hence $\Im \left[f'(z)\right] = 0$ for $|z| = r$ implies $f'(z) \equiv a_1$ in $\mathbb{D}$. That is $f(z) = a_0 + a_1z $ is a linear function. $\square$

An alternative approach: (inspired by Conrad's solution)

Let us denote $\displaystyle N(r) := \frac{1}{\pi r^2} \int_{r \mathbb{D}} |f'(z)|^2\,dx\,dy$ for $r \in [0,1]$.
Now, note that $\lim\limits_{r \to 0^+} N(r) = |f'(0)|^2 > 0$ (since, $f'(0) \neq 0$ otherwise it is trivial), i.e., since $f$ is locally injective near origin by the Area formula we have $$\frac{\text{Area}(f(\mathbb{rD}))}{\pi r^2} = N(r) = \frac{1}{\pi r^2}\int_{\mathbb D} |f'(z)|^2 \,dx\,dy = \sum_{k=1}^{\infty} k|a_k|^2r^{2k-2}$$ for all $r$ small enough. Therefore, $N(r)$ is strictly increasing for small $r > 0$ unless $a_k = 0$ for all $k \ge 2$, i.e., $f$ is linear.
Coming back to the problem if we assume $f$ is not linear then for small $r > 0$, $$|f'(0)|^2 = N(0) < N(r) = \frac{\text{Area}(f(\mathbb{rD}))}{\pi r^2} \le \frac{\pi d_r^2}{4\pi r^2} = |f'(0)|^2 \tag{5}$$ where, the second inequality in $(5)$ is due to the isodiametric inequality followed by the equality established in eqn $(1)$. Contradiction! 
Hence, $f$ must be linear. $\square$
A: One can also prove the result withouth resorting to geometric methods and instead using $2$ complex variables:
Define $$F(z,w)=\begin{cases}\frac{f(z)-f(w)}{z-w} &z\neq w
\\ f'(0)&\text{otherwise}\end{cases}$$
It is easy to see that $F(z,w)\in A(\mathbb{D}^2)$ (the algebra of holomorphic functions on the polydisc continuous up to the boundary).
Now, for functions in $A(\mathbb{D}^2)$ a particularly strong version of the maximum principle holds: the maximum of $|F(z,w)|$ is obtained on $\mathbb T^2$ (i.e. $\{(z,w)\in\mathbb D^2: |z|=|w|=1\}$). This version of the maximum principle implies that $2|f'(0)|\le \max_{z,w\in \partial \mathbb{D}}|f(z)-f(w)|$. If equality holds the maximum principle implies that $F$ is constant proving the claim.
The claimed version of the maximum principle is proved in "function theory in the polydiscs" by W. Rudin. There is also an elementary proof of the result: by the Cauchy integral formula it follows that, given $g\in A(\mathbb{D}^2)$ and $(z_0,w_0):|z_0|,|w_0|<1$ we have
$$|g(z_0,w_0)|\le \frac{1}{d(z_0,\partial \mathbb{D})d(w_0,\partial \mathbb{D})}\max_{(z,w)\in \mathbb T^2} |g(z,w)|$$
Since the formula holds for any $g\in A$, it holds for $g^n$ too, so
$$|g(z_0,w_0)|\le K(z_0,w_0)^{\frac1n}\max_{(z,w)\in \mathbb T^2} |g(z,w)|$$
taking $n\to \infty$ we get the claim.
