3 fixed error pointed out by KConrad

Here's a proof via character theory; in some ways it's a recasting of the Cauchy proof, but it has the advantage of being purely algebraic. In particular, this proof will work over any algebraically closed complete valued field, for example .(some mild care must be taken over a field of positive characteristic when one takes the limit $k\to\infty$).

Let $$f(z)=\sum_{n=0}^\infty a_nz^n$$ be a bounded entire function; we wish to show that $a_n=0$ for $n>0$. We use the fact that $$\sum_{\zeta \text{ a k-th rooth of unity}} \zeta^n$$ equals $0$ k$if$k$divides$n$and equals$k$0$ otherwise. So in particular setting $g_n(z)=f(z)/z^n$ we have for $k>n$ that

$$\frac{1}{k}\sum_{\zeta \text{ a k-th rooth of unity}} g_n(r\zeta)=a_n+\sum_{m=1}^\infty a_{km+n}r^{km}.$$ The sum on the right tends to zero with $k$, as the $a_{km+n}$ decrease very rapidly by the entireness of $f$. So we have $$a_n=\lim_{k\to\infty}\frac{1}{k}\sum_{\zeta \text{ a k-th rooth of unity}} g_n(r\zeta).$$ But taking the limit as $|r|\to \infty$ and using the boundedness of $f$, this is zero for $n>0$.

I say that this is a recasting of the Cauchy integral proof because the character sum we use is essentially a Riemann sum for the integral of $g_n$ around a circular contour. Indeed, one may prove the Cauchy integral formula for circular contours this way.

2 grammar

Here's a proof via character theory; in some ways it's a recasting of the Cauchy proof, but it has the advantage of being purely algebraic. In particular, this proof will work over any algebraically closed complete valued field, for example.

Let $$f(z)=\sum_{n=0}^\infty a_nz^n$$ be a bounded entire function; we wish to show that $a_n=0$ for $n>0$. We use the fact that $$\sum_{\zeta \text{ a k-th rooth of unity}} \zeta^n$$ equals $0$ if $k$ divides $n$ and equals $k$ otherwise. So in particular setting $g_n(z)=f(z)/z^n$ we have

$$\frac{1}{k}\sum_{\zeta \text{ a k-th rooth of unity}} g_n(r\zeta)=a_n+\sum_{m=1}^\infty a_{km+n}r^{km}.$$ The sum on the right tends to zero with $k$, as the $a_{km+n}$ decrease very rapidly by the entireness of $f$. So we have $$a_n=\lim_{k\to\infty}\frac{1}{k}\sum_{\zeta \text{ a k-th rooth of unity}} g_n(r\zeta).$$ But taking the limit as $|r|\to \infty$ and using the boundedness of $f$, this is zero for $n>0$.

I say that this is a recasting of the Cauchy integral proof because the character sum we use is essentially a Riemann sum for the integral of $g_n$ around a circular contour. Indeed, one may prove the Cauchy integral formula for circular contours this way.

1

Here's a proof via character theory; in some ways it's a recasting of the Cauchy proof, but it has the advantage of being purely algebraic. In particular, this proof will work over any algebraically closed complete valued field, for example.

Let $$f(z)=\sum_{n=0}^\infty a_nz^n$$ be a bounded entire function; we wish to show that $a_n=0$ for $n>0$. We use the fact that $$\sum_{\zeta \text{ a k-th rooth of unity}} \zeta^n$$ equals $0$ if $k$ divides $n$ and equals $k$ otherwise. So in particular setting $g_n(z)=f(z)/z^n$ we have

$$\frac{1}{k}\sum_{\zeta \text{ a k-th rooth of unity}} g_n(r\zeta)=a_n+\sum_{m=1}^\infty a_{km+n}r^{km}.$$ The sum on the right tends to zero with $k$, as the $a_{km+n}$ decrease very rapidly by the entireness of $f$. So we have $$a_n=\lim_{k\to\infty}\frac{1}{k}\sum_{\zeta \text{ a k-th rooth of unity}} g_n(r\zeta).$$ But taking the limit as $|r|\to \infty$ and using the boundedness of $f$, this is zero for $n>0$.

I say that this is a recasting of the Cauchy integral proof because the character sum we use is essentially a Riemann sum for the integral of $g_n$ around a circular contour. Indeed, may prove the Cauchy integral formula for circular contours this way.