If $ f(\mathbb{D})=D$ where $D$ is star-shaped domain, show that $f(\mathbb{D}_r)$ is a star-shaped domain. Let $f$ be holomorphic function in $\mathbb{D}$ which is injective with $f(\mathbb{D})=D$ and $f(0)=0$, where $D$ is a star-shaped domain. Show that $f(\mathbb{D}_r)$ is star-shaped domain where $\mathbb{D}_r=\{z:|z|<r\}, 0<r<1$.
I find it really difficult and I have no idea even on how to begin with this problem. I know that i want to show that for every point in $f(\mathbb{D}_r)$ the line segment with initial point at 0 and ending the chosen point is contained in $f(\mathbb{D}_r)$ but cannot use it properly.
Any hint or analytic solution would be very helpful as I want to understand the idea on how to solve this problem.

Let $g=f^{-1}$ and fix $w \in f(\mathbb{D}_r), 0<t<1$ and let $g(w)=z$ so $|z|<r, f(z)=w$. Pick $0<t<1$ and we need to show that there is $|z_t|<r, f(z_t)=tw$
Let $|z|<|z_1|<r$ for some fixed $z_1$ and consider the function $h(u)=g(tf(uz/z_1))$ for $|u|<1$ (which is well defined since $D$, the domain of $g$, is starlike wr to $0$ so since $f(uz/z_1)$ is in it by definition, then $tf(uz/z_1)$ is in it too!)
Note that $h(0)=0, |h(u)|<1$ (since the image of $g$ is the unit disc) so by Schwarz Lemma we have $|h(u)| \le |u|$ and if we let $u=z_1$ we claim that $h(z_1)=z_t$ the required value in $\mathbb{D}_r$ since $|z_t|=|h(z_1)| \le |z_1|<r$, while $z_t=g(tf(z))=g(tw)$ hence $f(z_t)=tw$ so we are done!
The same kind of proof shows that if $f$ convex (univalent, $f(0)=0, f(\mathbb D)$ convex domain), then $f(\mathbb{D}_r)$ is also convex for all $0<r<1$
