Past exam of the metric spaces and complex analysis course of the University of Oxford / 2017 / 4

The identity theorem says that two holomorphic functions on a connected domain which agree on a set with an accumulation point inside that domain agree everywhere there.

No such entire function exists. On the subsequence $1/(2m)\to 0$ indexed by even numbers, the proposed function equals $\sin z$. The identity theorem would force $f(z)=\sin z$ throughout $\mathbb{C}$. At the points indexed by odd numbers it is instead required to equal $-\sin z$, a contradiction since $\sin (1/n)>0$ for positive $n$. The accumulation point zero lies inside the assumed domain of holomorphy, which is essential to this argument.

Yes: take the unnormalized sinc function with its removable value at zero,Its power series is ${\sum }_{m=0}^{\infty }(-1{)}^m{z}^{2m}/(2m+1)!$, with infinite radius of convergence, so it is an entire function. At $z=1/n$ it gives $n\sin (1/n)$ as required. The identity theorem also proves this entire solution is unique.

No such entire function exists. If it did, the entire function $h(z{)}^2-z$ would vanish at the distinct points $(n/(n+1){)}^2$, whose accumulation point is one. The identity theorem would give $h(z{)}^2=z$ on all of $\mathbb{C}$.

At zero this forces $h(0)=0$. Differentiating the identity gives $2h(z)h^{\prime }(z)=1$, impossible at zero. Equivalently, a squared holomorphic function has an even-order zero, whereas $z$ has a simple zero. This is the obstruction expressed by no entire square root of the identity function; a square-root branch near one cannot extend holomorphically through zero.

Morera's theorem states that a continuous function on a domain is holomorphic if its contour integral around every triangle whose filled-in interior lies in the domain is zero.

The uniform limit $f$ is a continuous function, by the uniform limit theorem. For any such triangle $T$, the estimation lemma givesEach integral of $f_n$ vanishes by the Cauchy integral theorem. Hence the integral of $f$ vanishes too, and Morera's theorem proves $f$ is holomorphic on $D$. This is the basic locally uniform convergence of holomorphic functions argument; the given uniform convergence on all of $D$ is more than needed.

For $|z_0|<1$ and $k\ge 0$, the Cauchy integral formula for derivatives isThe formula applies because the closed unit disc lies inside the domain of holomorphy. For $k=0$ this is the ordinary value formula, with $0!=1$.

Fix $0<r<1$ and $k\ge 1$. On $|z|=1$ and $|z_0|<r$, the reverse triangle inequality gives $|z-z_0|\ge 1-r$. Subtract the two Cauchy integral formula expressions and apply the estimation lemma, using circle length $2\pi$:Thus every fixed-order derivative converges uniformly on each smaller disc. The right-hand factor tends to zero by the assumed uniform convergence on $D$. Keeping a positive gap $1-r$ from the integration contour is the essential feature of this Cauchy derivative convergence bound. If $r\le 0$, the specified open disc is empty and the claim is vacuous.

Take $g_n(z)=z^n/n$ and $g(z)=0$, for $n\ge 1$. These functions are entire andso convergence is uniform on the closed unit disc. But $g_n^{\prime }(z)=z^{n-1}$, givingfor every $n$. In particular $g_n^{\prime }(1)=1$ never approaches zero. On every strictly smaller disc the derivative does converge uniformly, consistent with part (iii). This is uniform holomorphic convergence without boundary derivative convergence.