Question: Let $f$ be analytic in a neighborhood of $\overline{\mathbb{D}}$ (the closed unit disk) such that $\int_{|z|=1}|f(z)-z||dz|\leq\frac{1}{100}$. Prove that $f$ has at least one zero in $\mathbb{D}$.
Thoughts: I feel like this should use Maximum Modulus principle, but I can't seem to get that argument to work out. I've seen some similar questions, but nothing that gives us an assumption like the integral here. I've thought maybe something like Rouche or Schwarz may help, but I continue to get nowhere. Any help is greatly appreciated! Thank you.
Best Answer
Let $f(z)=\sum a_kz^k$.
Note that $\int_{|z|=1}|f(z)-z||dz|=\int_{|z|=1}|f(z)/z^k-1/z^{k-1}||dz| \ge$
$|\int_{|z|=1}(f(z)/z^k-1/z^{k-1})dz|$,
and this last integral is $2\pi |a_{k-1}|$ if $k \ge 1, k \ne 2$ and $2\pi |a_1-1|$ if $k=2$, so one gets $|a_k| \le 1/(200 \pi)$ if $k \ne 1$ and $|a_1| \ge 1-1/(200\pi)$.
But now one can apply Rouche on say $|z|=1/2$ to $f, a_1z$ as a simple majorization shows that $|f-a_1z| <|a_1|/2$ for $|z|=1/2$ and we are done; one can of course do much better than $1/2$ in finding a $c<1$ st $f$ has a zero with $|z|<c$ if needed