# Extended Proof of the Theorem that a bounded analytic function is constant.

I am having trouble feeling convinced by my proof and more importantly - feeling confident in my working out. The question reads

(a) Let $f$ be an entire function such that there exist real constants $M$ and $N$ such that $|f(z)|<M|z|+N$ for all $z$. Prove that for any three pairwise different complex numbers $a,b,c$, $\frac{f(a)}{(a-b)(a-c)}+\frac{f(b)}{(b-a)(b-c)}+\frac{f(c)}{(c-a)(c-b)}=0$.

(b) Deduce that there are constants $A,B\in\mathbb{C}$ such that $f(z)=Az+B$ for all $z$.

So to begin, I realise that the LHS of the equality that I am required to prove is simply the sum of residues (without $2\pi i$) of the integral $\oint_{\Gamma}\frac{f(z)}{(z-a)(z-b)(z-c)}dz$. That is,

$$\oint_{\Gamma}\frac{f(z)}{(z-a)(z-b)(z-c)}dz=2\pi i\left(\frac{f(a)}{(a-b)(a-c)}+\frac{f(b)}{(b-a)(b-c)}+\frac{f(c)}{(c-a)(c-b)}\right),$$ for $\Gamma$ being the circle contour of radius $R$. Furthermore, for the equality to hold, singularities at $z=a,b,c$ must be inside $\Gamma$.

So the next chain of thought would be show that $$\oint_{\Gamma}\frac{f(z)}{(z-a)(z-b)(z-c)}dz=0.$$This will obviously yield the result I want. I relate to limits and the M-L Lemma to do this. So,

$$\left|\frac{f(z)}{(z-a)(z-b)(z-c)}dz\right|\leq\frac{MR+N}{(R-|a|)(R-|b|)(R-|c|)},$$ by the equality given in the question, reserve triangle inequality and since complex numbers $a,b,c$ lie inside the contour thus $|a|,|b|,|c|<R$. Thus by the M-L lemma we have that $$lim_{R\to\infty}\left|\oint_{\Gamma}\frac{f(z)}{(z-a)(z-b)(z-c)}dz\right|\leq \lim_{R\to\infty}\frac{(MR+N)2\pi R}{(R-|a|)(R-|b|)(R- |c|)}$$ Clearly the RHS of the inequality converges to $0$. Thus $$lim_{R\to\infty}\oint_{\Gamma}\frac{f(z)}{(z-a)(z-b)(z-c)}dz=0.$$But we know that $$lim_{R\to\infty}\oint_{\Gamma}\frac{f(z)}{(z-a)(z-b)(z-c)}dz=lim_{R\to\infty}2\pi i\left(\frac{f(a)}{(a-b)(a-c)}+\frac{f(b)}{(b-a)(b-c)}+\frac{f(c)}{(c-a)(c-b)}\right).$$ So $$2\pi i\left(\frac{f(a)}{(a-b)(a-c)}+\frac{f(b)}{(b-a)(b-c)}+\frac{f(c)}{(c-a)(c-b)}\right)=0$$$$\implies\left(\frac{f(a)}{(a-b)(a-c)}+\frac{f(b)}{(b-a)(b-c)}+\frac{f(c)}{(c-a)(c-b)}\right)=0.$$

With part (b), I begin by using the result I proved. Simplifying the expression I have that $$f(a)(b-c)+f(b)(c-a)+f(c)(a-b)=0.$$ It's clear that if I let $f(z)=Az+B$ and substitute it into the equation above then I am done. Is this valid though?

It's clear that if I let $f(z)=Az+B$ and substitute it into the equation above then I am done. Is this valid though?

No, that would be a petitio principii. That's the form you want to deduce.

But you have, for any three pairwise different $a,b,c \in \mathbb{C}$, that

$$\frac{f(a)}{(a-b)(a-c)} + \frac{f(b)}{(b-a)(b-c)} + \frac{f(c)}{(c-a)(c-b)} = 0.$$

Now, what if you rename the three, say you use $z = a$, $1 = b$, $0 = c$?

• Yes, I thought that would be wrong should be going forwards. Ok, I've renamed the 3 as you have stated - but I can't see where this leading. $f(z)=(f(1)-f(0))z+f(0)$. – Gustavo Louis G. Montańo Nov 4 '13 at 2:49
• Where does the $2$ come from? You should get $f(z) = f(1)z - f(0)(z-1) = (f(1)-f(0))z + f(0)$. – Daniel Fischer Nov 4 '13 at 2:53
• Sorry, typo, let me fix it. I see by setting those conditions the function is in the form that we want. Ok ! So I see why you made these substitutions. But I still can't concrete the conclusion. – Gustavo Louis G. Montańo Nov 4 '13 at 2:54
• What about, $f(z)=\frac{f(b)-f(c)}{b-c}z+\frac{f(c)b-f(b)c}{b-c}$? I think this is it. It also ties in with what you were trying to say :D ! – Gustavo Louis G. Montańo Nov 4 '13 at 3:03
• That's also correct. I find it easier to use a few standard points, less opportunity for errors. – Daniel Fischer Nov 4 '13 at 3:09