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?

Thank you to all for your time in advanced.


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 $z = a$, $1 = b$, $0 = c$?

  • $\begingroup$ 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. $f(z)=(f(1)-f(0))z+f(0)$. $\endgroup$ – Gustavo Louis G. Montańo Nov 4 '13 at 2:49
  • $\begingroup$ Where does the $2$ come from? You should get $f(z) = f(1)z - f(0)(z-1) = (f(1)-f(0))z + f(0)$. $\endgroup$ – Daniel Fischer Nov 4 '13 at 2:53
  • $\begingroup$ 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. $\endgroup$ – Gustavo Louis G. Montańo Nov 4 '13 at 2:54
  • $\begingroup$ 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 ! $\endgroup$ – Gustavo Louis G. Montańo Nov 4 '13 at 3:03
  • $\begingroup$ That's also correct. I find it easier to use a few standard points, less opportunity for errors. $\endgroup$ – Daniel Fischer Nov 4 '13 at 3:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.