A bound on $|f(z)+f(-z)-2f(0)|$ when $f:D(0,1)\to D(0,1)$ is holomorphic

Question

Let $f:D(0,1)\to D(0,1)$ be holomorphic. Is it true that $|f(z)+f(-z)-2f(0)|\le 2|z|^2$ ?

I can prove it if $f(0)=0$, because in that case, by Schwraz Lemma on $f$, we have $|f(z)|\le |z|$, so $\Big|\dfrac {f(z)+f(-z)}{2z}\Big|\le 1$ for $z\ne 0$ , so defining $g(z):=\dfrac {f(z)+f(-z)}{2z}$ for $z\ne 0$ and $g(0)=0$, we have $g$ is holomorphic in $D(0,1)$ and $|g(z)|\le 1$, so by Schwarz lemma, $|g(z)|\le z$, giving me $|f(z)+f(-z)|\le 2|z|^2$.

Now when $f(0)\ne 0$, I would want to work with $g(z):=\dfrac {f(z)-f(0)+f(-z)-f(0)}{2z}$ for $z\ne 0$ and $g(0)=0$. This $g$ is holomorphic. However, We no longer have $|f(z)-f(0)|\le |z|$ because no longer $|f(z)-f(0)|\le 1$, so we can't apply Schwarz to $f-f(0)$ first.

Rather we do have $\bigg|\dfrac {f(z)-f(0)}2 \bigg|\le 1$. So $|f(z)-f(0)|\le 2|z|$. This gives me $|g(z)|\le 2$, so

applying Schwarz on $g(z)/2 $ (rather than on $g(z)$ ) we get $|g(z)|\le 2|z|$, so that $|f(z)+f(-z)-2f(0)|\le 4|z|^2$, which is way off than what I want.

Please help.

Answer 1

Counterxample: $f(z)=\frac{2z^2-1}{2-z^2}, f(0)=-\frac{1}{2}, f(z)=f(-z), |f(1)-f(0)|=\frac{3}{2}>1$, so there is a $z$ close to $1$ for which the requested inequality fails, while $f(z)=B(z^2)$ for $B$ the disc automorphism $\frac{z-\frac{1}{2}}{1-\frac{1}{2}z}$

Answer 2

Evidently the answer is no. A simpler argument for $4|z|^2$: Let $$g(z)=f(z)+f(-z)-2f(0).$$Then $g(0)=g'(0)=0$, so $h(z)=g(z)/z^2$ is holomorphic, and maximum moduls shows $|h(z)|\le 4$.