I have the following problem:
Let $f,g$ be analytic functions when $|z|<2$. Let us define $D=max\{|f(z)|+|g(z)|: |z|\leq 1\}$. Let us suppose that $M=|f(w)|+|g(w)|$ for some $w$ with $|w|\leq 1$. Then I need to show that $|w|=1$.
So my Idea is the following:
Let us remark that $\Omega=\{|z|<2|\}$ is open and path connected, thus also connected. Furthermore we know that $f,g:\Omega \rightarrow \Bbb{C}$ are analytic. In addition we remark that $D=max\{|f(z)|:|z|\leq 1\}+max\{|g(z)|:|z|\leq 1\}$. Thus $$|f(w)|=sup_{|z|\leq 1}|f(z)|,~~~|g(w)|=sup_{|z|\leq 1}|g(z)|$$Now using the maximal priciple gives us that $f,g$ are constant, so the maximum is attaint everywhere in $\Omega$.
Furthermore we have the equalities that $\alpha f=|f|$ and $\beta g=|g|$ for some $\alpha, \beta$ modulus one numbers. This gives us $$|\alpha f+\beta g|\leq |f|+|g|$$ Since $f,g$ attains the maximum inside $\Omega$ also $|f|+|g|$ does, therefore we have that $$\alpha f(w)+\beta g(w)=|f(w)|+|g(w)|$$ which is a maximum for |f|+|g|.
Is this correct so far? But now I don't see how to finish so that I can conclude that $|w|=1$.