Compactness makes the continuous function $|f|$ attain a maximum. Apply the preceding argument in a chart at a maximizing point to make $f$ constant on a neighborhood. The permitted identity theorem and connectedness then make it constant on all of $S$. Hence every holomorphic function from a compact connected Riemann surface to $\mathbb{C}$ is constant.

Choose a disk on which $|f(z)|\le M=|f(a)|$. If $M=0$ the assertion is immediate. Otherwise multiply $f$ by a fixed unit scalar so that its value at $a$ is the positive real number $M$. The mean-value identity shows the average of $\mathrm{Re}f$ on every sufficiently small circle is $M$. Since $\mathrm{Re}f\le |f|\le M$ there, continuity forces equality at every point on each circle. Equality of real part to $M$ together with modulus at most $M$ forces the imaginary part to vanish. Varying the radius covers a neighborhood of $a$. Thus both $|f|$ and $f$ are constant near $a$, proving the local maximum-modulus assertion.

Choose $r>0$ so the closed disk centered at $a$ of radius $r$ lies in $U$. The Cauchy integral formula givesParametrize $z=a+r{e}^{i\theta }$; then $dz=ir{e}^{i\theta }d\theta$ and the denominator cancels. Thus

A Riemann surface is a connected Hausdorff second-countable topological surface equipped with coordinate charts to open subsets of $\mathbb{C}$ whose transition maps are holomorphic. This is a complex manifold of complex dimension one. The atlas makes holomorphic functions, meromorphic functions and holomorphic differentials coordinate-independent notions.

For degree two every ramification point has index two, so contributes exactly one. The formula gives $2g+2$ ramification points. No two can share an image: that would give a fiber of total local degree at least four. Hence there are also $2g+2$ distinct branch points on the sphere.

For a degree-one map, the sum of local degrees in every fiber is one. Therefore each fiber contains exactly one point and its local degree is one. There is no ramification, and the Riemann-Hurwitz formula gives $2g-2=-2$, so $g=0$. The map is a continuous bijection from compact $C$ onto the Hausdorff sphere and hence a homeomorphism. In fact its local degree-one coordinate inverses are holomorphic, so it is biholomorphic as well.

The ramification sum equals $2g-2+2d$. Since $g\ge 0$ and $d>1$, it is at least $2d-2>0$. Thus some point has analytic ramification index greater than one.

For a smooth complex projective plane curve of degree $d$,Project generically from a point outside the curve to a projective line. Its degree is $d$, by intersection with a generic line through the center. Ramification occurs when such a line is tangent. In coordinates for projection $[x:y:z]\mapsto [x:z]$, these points satisfy $F=F_y=0$. The polar has degree $d-1$, and Bézout gives total ramification $d(d-1)$; generic projection makes the ramification simple, or the intersection multiplicities can be read as the corresponding ramification orders. The Riemann-Hurwitz formula yields $2g-2=-2d+d(d-1)$ and hence the formula. The smoothness qualification is essential for geometric genus: singular plane curves have the same arithmetic degree-genus expression, while the genus of their normalization can be smaller.

For $d>1$, the map $z\mapsto z^d$ has local index $d$ at zero and at infinity; at infinity use coordinates $w=1/z$ in source and target, giving again $w\mapsto w^d$. At every nonzero finite point its derivative is nonzero, so there are no further ramification points. Thus the ramification sum is $2(d-1)$ and the formula gives $-2=-2d+2(d-1)$, as required for the sphere. For $d=1$ the map is the identity and has no ramification, giving the same equality separately.

With local indices $e_p$, the required specialization isOnly the finitely many points of index greater than one contribute. Their images are the branch points on the target.

For $f=y^2-x^3-17$, the partials are $f_x=-3x^2$ and $f_y=2y$. If $p\notin \{2,3\}$ their simultaneous vanishing forces $x=y=0$, and that point lies on the curve exactly when $p=17$. In characteristic two, the equations force $x=0$ and then $y^2=1$, so $y=1$. In characteristic three they force $y=0$ and then $x^3=1$, so $x=1$. Consequentlyand there are no singular points for any other prime.

Put $F=y^{2}z-x^3+3x{z}^2-2z^3$. Its partials areAt infinity $z=0$, the curve equation forces $x=0$, giving $[0:1:0]$, where $F_z\ne 0$. In the chart $z=1$, the derivative equations give $y=0$ and $x=\pm 1$. Only $x=1$ lies on the curve and also makes $F_z=0$. Thus the only singular point is $[1:0:1]$. As a check, the affine equation factors as $y^2=(x-1{)}^2(x+2)$, whose quadratic tangent cone there is $y^2-3(x-1{)}^2$, giving two distinct tangent lines.

Bézout's theorem says that two projective plane curves over $\mathbb{C}$ of degrees $d,e$ with no common irreducible component have total intersection multiplicity $de$, so at most $de$ distinct intersection points. Let the irreducible curve have reduced homogeneous equation $F$ of positive degree $d$. In characteristic zero some partial derivative is nonzero. For $d>1$ this derivative has degree $d-1$ and cannot share a component with the irreducible degree-$d$ polynomial $F$. Every singular point belongs to their intersection. Bézout bounds their number by $d(d-1)$, proving there are finitely many singular points. For $d=1$ a partial is a nonzero constant and there are none.

For a plane affine curve given by a reduced equation $f(x,y)=0$, a point is singular when $f=f_x=f_y=0$ there. For a projective curve given by a homogeneous equation $F(x,y,z)=0$, a point is singular when $F=F_x=F_y=F_z=0$ on a nonzero representative. These conditions are invariant under projective rescaling. In characteristic zero, Euler's identity $dF=x{F}_x+y{F}_y+z{F}_z$ makes the equation $F=0$ follow from vanishing of the partials when $d\ne 0$ in the field. Over an arbitrary field one should retain the equation itself, particularly if the characteristic divides the degree.

Write $P_5=[u:v:w]$ in the same normalized coordinates. Its coordinates are all nonzero, because the coordinate lines join pairs of the first three points. Evaluation at $P_5$ gives the additional linear equation $avw+bwu+cuv=0$ on the two-dimensional coefficient plane $a+b+c=0$. It is not identically zero there: that would imply $vw=wu=uv$, hence $u=v=w$ and $P_5=P_4$, contradicting the hypothesis. Its kernel is therefore one-dimensional. Thus there is exactly one projective conic through all five points. If it were reducible, $P_5$ would lie on a line joining two of the other points, contrary to no three being collinear. Hence the unique conic is irreducible.

By the preceding frame result take the four points to be $[1:0:0],[0:1:0],[0:0:1],[1:1:1]$. The conics through them have equationsCoefficient vectors differing by a common nonzero scalar define the same conic. Their parameter space is the projectivization of a two-dimensional vector space, hence a projective line.

Each partition of the four points into two unordered pairs gives a reducible conic formed by the two joining lines. There are three pairings. Conversely a reducible conic is a union of two lines, possibly repeated. General position prevents any one line from carrying three of the points, so the lines must be distinct and carry exactly one of these pairings. Thus there are exactly three reducible members. In these coordinates they are the parameters $[0:1:-1]$, $[1:0:-1]$, $[1:-1:0]$, giving respectively $x(z-y)=0$, $y(z-x)=0$, and $z(y-x)=0$.

Choose representative vectors $x_1,\dots ,x_{n+1}$ of the first $n+1$ points; they form a basis. Write a representative of the last point as ${\sum }_i{a}_i{x}_i$. General position forces each $a_i\ne 0$, since otherwise $n+1$ of the points would lie in a hyperplane. Rescale the basis vectors to $a_i{x}_i$. The tuple is then projectively equivalent to the standard frameApply the same normalization to the second tuple and compose the two transformations to obtain existence. An automorphism fixing the coordinate points is represented by a diagonal matrix. Fixing their sum forces all its diagonal entries to agree, so its projective action is the identity. This proves existence and uniqueness. At $n=1$ it says the Möbius group acts sharply triply transitively on the projective line: any ordered triple of distinct points can be sent to any other in exactly one way.

Let $1<p<\infty$ and $q=p/(p-1)$. The dual of ${\ell }^p$ is ${\ell }^q$. For every $a\in {\ell }^q$, $⟨a,e_n⟩=a_n\to 0$, so $e_n\rightharpoonup 0$ weakly. A compact operator sends a weakly convergent sequence to a norm-convergent sequence with the corresponding limit. Indeed, any failure of $\Vert T{e}_n\Vert \to 0$ would give a subsequence bounded away from zero, with a further norm-convergent subsequence by compactness; weak continuity of $T$ forces that norm limit to be zero, a contradiction. The condition is therefore necessary.

For failure of sufficiency, partition the positive integers into consecutive blocksand define the block averaging compactness counterexampleBy Hölder's inequality, $|(Tx{)}_k{|}^p\le {\sum }_{j\in B_k}|x_j{|}^p$. Summing over disjoint blocks shows $\Vert Tx{\Vert }_p\le \Vert x{\Vert }_p$, so $T$ is bounded. If $j\in B_k$, then $T{e}_j=k^{-1/q}{e}_k$; as $j\to \infty$, also $k\to \infty$, so $\Vert T{e}_j{\Vert }_p\to 0$.

However, the vectors $x^{(k)}=k^{-1/p}{1}_{B_k}$ have norm one and satisfy $T{x}^{(k)}=e_k$. Distinct output vectors have distance $2^{1/p}$, so this bounded sequence of inputs has no norm-convergent image subsequence. Thus $T$ is not compact.

To connect with the hint, define $(Uy{)}_j=k^{-1/q}{y}_k$ for $j\in B_k$. Then $U:{\ell }^q\to {\ell }^q$ is an isometry, and the dual pairing calculation gives $T=U^{\ast }$. The vectors $U{e}_k$ have disjoint supports and norm one. The direct estimates above establish the counterexample without relying only on the hint.

Let $Q_N$ be truncation to the first $N$ coordinates in ${\ell }^1$. For $x\in {\ell }^1$, boundedness of $T$ and absolute convergence giveApplying the difference to each $e_n$ with $n>N$ gives the reverse bound on the operator norm. HenceThe stated condition makes this tend to zero. Each $T{Q}_N$ has finite rank, so part (a)(ii) proves compactness. Thus the condition is sufficient on ${\ell }^1$.

For failure of necessity, set $Tx=({\sum }_{n\ge 1}{x}_n)e_1$. The sum defines a bounded linear functional on ${\ell }^1$, so $T$ is a bounded rank-one operator and therefore compact. But $T{e}_n=e_1$ for every $n$, so $\Vert T{e}_n{\Vert }_1=1$. Thus compactness need not make the basis images tend to zero on ${\ell }^1$.

Let $A_n$ be finite-rank bounded operators with $\Vert T-A_n\Vert \to 0$ in operator norm. Fix $ϵ>0$ and choose $n$ with $\Vert T-A_n\Vert <ϵ/3$. The set $A_n(B_X)$ is bounded in a finite-dimensional subspace, hence has a finite $ϵ/3$-net $y_1,\dots ,y_m$.

For each $x\in B_X$, some $j$ satisfiesThus $T(B_X)$ is totally bounded. Its closure is totally bounded and, as a closed subset of the Banach space $Y$, complete. A complete totally bounded metric space is compact. Consequently $T$ is compact.

The operator-norm limit is bounded, since $\Vert T\Vert \le \Vert A_n\Vert +\Vert T-A_n\Vert$ for any sufficiently good approximation. This proves directly that an approximable operator into a Banach space is compact; no assertion that every compact operator on an arbitrary Banach space is approximable is needed.