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Use the convention that a weak modular form is a holomorphic function $f$ on the complex upper half-plane satisfyingwith no condition at infinity. In particular $f(z+1)=f(z)$, so it has a Laurent series $f(z)={\sum }_{n\in \mathbb{Z}}{a}_n{q}^n$ on $0<|q|<1$, where $q=e^{2\pi iz}$. It is a modular form when it is holomorphic at a cusp, here the unique level-one cusp at infinity, equivalently $a_n=0$ for $n<0$. It is a cusp form when additionally $a_0=0$. If one uses the narrower weakly holomorphic convention, one also requires finitely many negative terms; the derivative argument below holds in that subspace as well. The matrix $-I$ forces odd-weight forms to be zero.

Put $(a{)}_r=a(a+1)\cdots (a+r-1)$, with $(a{)}_0=1$, andWe prove the derivative transformation of a weak modular form in the uniform formFor $\ell =0$ this is the weight-$k$ transformation under $S$. Differentiating the formula and multiplying by $z^2$, since $d(-1/z)/dz=z^{-2}$, gives the recurrencewhere out-of-range coefficients are zero. The displayed explicit coefficients satisfy it. Indeed, for interior $j$, extract the common polynomial $(k+j{)}_{\ell -j}$; the remaining identity isThis follows from Pascal's identity and $j\left(\genfrac{}{}{0ex}{}{\ell }{j}\right)=(\ell -j+1)\left(\genfrac{}{}{0ex}{}{\ell }{j-1}\right)$. The endpoint coefficients satisfy the recurrence directly. This polynomial proof also applies when a factor is zero, avoiding division by that factor. As $C_{\ell ,\ell }=1$, the result isThe letter $l$ in the printed exponent is the same derivative order $\ell$.

For integer $k<0$, take $\ell =1-k$. If $j<\ell$, the product runs from $k+j\le 0$ through $k+\ell -1=0$, and therefore vanishes. Consequently $f^{(1-k)}$ has weight $k+2(1-k)=2-k$ under $S$. It is also invariant under $T$, by differentiating $f(z+1)=f(z)$, and $S,T$ generate the modular group. Its holomorphy is preserved by differentiation. With $D=(2\pi i{)}^{-1}d/dz=qd/dq$, termwise differentiation of the locally convergent Laurent series gives the Bol identity for modular forms:For negative indices the exponent $1-k$ is a positive integer, and the constant term is killed, so every term is unambiguous.

A modular form of integer weight $k$ for a finite-index subgroup $\Gamma \le S{L}_2(\mathbb{Z})$ is a holomorphic function on the complex upper half-plane satisfying $f{|}_k\gamma =f$ for $\gamma \in \Gamma$ and holomorphic at every cusp of a modular group. Explicitly, for each $\sigma \in S{L}_2(\mathbb{Z})$ choose a positive integer $h_{\sigma }$ with $\sigma T^{h_{\sigma }}{\sigma }^{-1}\in \Gamma$. Then $f{|}_k\sigma$ is periodic with period $h_{\sigma }$, and its expansion in $q_{h_{\sigma }}=e^{2\pi iz/h_{\sigma }}$ must have no negative powers. Equivalently it must be bounded as $\mathrm{Im}z\to \infty$, uniformly in one period. This definition applies to noncongruence subgroups too. If $-I\in \Gamma$, nonzero forms require even weight.

Put $d=[S{L}_2(\mathbb{Z}):\Gamma ]$ and let $f\ne 0$. Form the product over left cosetsChanging a representative does not affect a factor. Right multiplication permutes the cosets, so the product is a nonzero level-one modular form of weight $kd$. It is holomorphic on the upper half-plane and at infinity, since every factor is bounded there by cusp holomorphy.

Let $h$ be the least positive integer with $T^h\in \Gamma$, and write $f={\sum }_{n\ge 0}{b}_n{q}_h^n$. If $r$ is its first nonzero index, the $h$ distinct cosets $\Gamma T^j$, $0\le j<h$, give the factors $f(z+j)$. Their product has leading term a nonzero constant times $e^{2\pi irz}=q^r$. All remaining factors are bounded at infinity. Hence $F(z)=O(e^{-2\pi r\mathrm{Im}z})$, so its level-one order satisfies $v_{\infty }(F)\ge r$. Notice that using just the single factor $f$ would only give $r/h$; the entire translation orbit is needed.

The permitted valence formula for the modular group bounds $v_{\infty }(F)\le kd/12$, because all its other weighted zero orders are nonnegative. Thus $r\le kd/12$. With $R=\lfloor kd/12\rfloor$, the map $f\mapsto (b_0,\dots ,b_R)$ is injective: a nonzero form in its kernel would have $r\ge R+1$. This proves the dimension bound for modular forms on a finite-index subgroup:

For the dilation assertion, the original PDF states $N_{1}D\mid N$, a product-divisibility condition, rather than the TeX's chain $N_1\mid D\mid N$. Use the PDF condition. Put $A=\mathrm{diag}(D,1)$. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N)$, thensince $N_{1}D\mid c$. Therefore, for $g(z)=f(Dz)$, either direct substitution or the slash operator for modular forms gives $g{|}_k\gamma =g$. It is holomorphic on the upper half-plane.

To verify every cusp, take $\sigma \in S{L}_2(\mathbb{Z})$ and choose $\tau \in S{L}_2(\mathbb{Z})$ taking infinity to the rational cusp $A\sigma \infty$. Then $B={\tau }^{-1}A\sigma$ is rational upper triangular, with positive ratio of its two diagonal entries. Consequentlyis a nonzero constant factor times $f{|}_k\tau$ evaluated at an affine map $z\mapsto uz+v$ with $u>0$. The imaginary part tends to infinity, where $f{|}_k\tau$ is bounded. Thus $g{|}_k\sigma$ is bounded too. Together with its periodicity this proves cusp holomorphy by the Riemann removable singularity theorem. Hence the oldform by argument dilation satisfies

Finally ${\Gamma }_0(3)$ has index four: reduction modulo three gives a transitive $S{L}_2({\mathbb{F}}_3)$ action on the four points of ${\mathbb{P}}^1({\mathbb{F}}_3)$, and ${\Gamma }_0(3)$ is the inverse image of the stabilizer of infinity. Reduction is surjective, since elementary determinant-one matrices generate $S{L}_2({\mathbb{F}}_3)$ and lift integrally. The dimension bound gives $\dim M_4({\Gamma }_0(3))\le 2$.

The normalized Eisenstein series $E_4(z)$ and the oldform $E_4(3z)$ both belong to this space. Their Fourier expansion of a normalized Eisenstein series starts asA linear relation between them has zero constant term only if its two scalars sum to zero; its $q$ coefficient then forces both scalars to vanish. They are independent, proving the weight-four modular forms on Gamma 0 3 description

Use the canonical Kähler potential and normalize superspace integration by $\int d^2\theta {\theta }^2=1$. The superspace Lagrangian density isThe full superspace integration produces a D-term, while the chiral superfield integral produces an F-term. Work in chiral coordinates for the chiral-superfield component expansion, so the omitted spacetime-derivative terms do not enter the extraction of $F$.

Set $\delta \Phi =\sqrt{2}\theta \psi +{\theta }^{2}F$. A Taylor expansion of the superpotential givesHigher terms vanish because there are only two Grassmann variables ${\theta }^{\alpha }$. With $\theta \psi ={\theta }^{\alpha }{\psi }_{\alpha }$ and $\psi \psi ={\psi }^{\alpha }{\psi }_{\alpha }$, their anticommutation gives $(\theta \psi {)}^2=-\frac{1}{2}{\theta }^2\psi \psi$. ThusThe canonical Kähler potential contributes $F^{\ast }F$. Consequently the part of the Lagrangian density containing the auxiliary field isThe fermion terms coming from the superpotential areThe Weyl spinor kinetic term and the complex scalar field kinetic term are canonical because $K={\Phi }^{†}\Phi$.

Treat $F$ and $F^{\ast }$ as independent variables in their algebraic Euler-Lagrange equations. They giveEquivalently,After eliminating the auxiliary field, the Lagrangian density contains $-V$, so the F-term scalar potential isIt is nonnegative for arbitrary complex $m,g,\phi$, since it is a modulus squared. A supersymmetric vacuum satisfies F-flatness, so its zeros are:If $m=0$ and $g\ne 0$ these coincide; if $g=0$ and $m\ne 0$ only the origin remains; if $m=g=0$ every constant $\phi$ is a supersymmetric vacuum.

To compare masses, expand around a supersymmetric vacuum $v$ satisfying $W^{\prime }(v)=0$, writing $\phi =v+\eta$. With $M=W^{\prime \prime }(v)=m+2gv$,Writing $\eta =(A+iB)/\sqrt{2}$ gives the scalar field mass term $-\frac{1}{2}|M{|}^2(A^2+B^2)$; the Weyl spinor has physical mass $|M|$. Therefore the supersymmetric mass degeneracy isIn particular, at $v=0$ the boson and fermion masses are both $|m|$. At the second supersymmetric vacuum $v=-m/g$ the sign of $M$ reverses, but the physical mass is again $|m|$. Equality here concerns fluctuations about a supersymmetric vacuum, not arbitrary backgrounds with $W^{\prime }(v)\ne 0$.

In the convention $V\supset \lambda |\phi {|}^4$ and ${\mathcal{L}}_{\mathrm{Yukawa}}=-y\phi \psi \psi +h.c.$, the same superpotential coefficient gives $\lambda =|g{|}^2$ and $y=g$. The supersymmetric relation between quartic and Yukawa couplings is thereforeIf instead the Yukawa interaction is written $-\frac{1}{2}{y}_3\phi \psi \psi +h.c.$, then $y_3=2g$ and the same physical relation is $\lambda =|y_3{|}^2/4$. The numerical factor depends on the definition of the Yukawa coupling.

The perturbative non-renormalization theorem applies to the local holomorphic superpotential in the Wilsonian effective action, using regularization in quantum field theory that preserves supersymmetry. There are two complementary ways to see why it applies here.

First, in supergraph calculations in perturbative quantum field theory, the algebra of supersymmetric covariant derivatives puts any candidate local correction to the superpotential into a full $d^4\theta$ superspace integration. It therefore corrects a D-term, such as the Kähler potential, rather than a local chiral $d^2\theta$ F-term. Turning such a contribution into an apparent chiral integral would require nonlocal inverse spacetime derivatives, schematically $1/\square$. These are excluded from the local Wilsonian effective action with a fixed nonzero momentum cutoff. This gives the diagrammatic non-renormalization theorem.

Second, the holomorphy argument for superpotential non-renormalization treats $m$ and $g$ as background chiral spurions, so the Wilsonian effective action superpotential is a holomorphic function of $\Phi ,m,g$, with no dependence on $m^{\ast }$ or $g^{\ast }$. Assign an ordinary $U(1)$ spurion charge and an R-charge as follows:The superpotential has ordinary charge zero and R-charge two, since $d^2\theta$ has R-charge minus two. A prospective holomorphic monomial $m^a{g}^b{\Phi }^c$ must therefore satisfyFor perturbative corrections, regularity at $m=0$ and $g=0$ requires nonnegative powers of the couplings. Regularity in $m$ is justified by keeping a nonzero Wilsonian momentum cutoff and retaining $\Phi$ as a light field. Hence only $c=2$ and $c=3$ survive: the existing $m{\Phi }^2$ and $g{\Phi }^3$ structures. The quadratic coefficient is fixed by the free theory at $g=0$, where there are no interaction loops. The cubic coefficient is fixed at first order in $g$, where a connected three-field interaction has only its tree-level Feynman diagram; loop corrections would require further powers of the interaction couplings, which the holomorphy argument for superpotential non-renormalization charge constraints forbid. ThusA field-independent constant is physically irrelevant in this rigid model; the spurion charges would in any case require $m^3/g^2$, which is not regular at $g=0$.

The Kähler potential can still receive wave-function renormalization. For example, $K=Z{\Phi }^{†}\Phi +\cdots$ gives the canonical field normalization ${\Phi }_c=Z^{1/2}\Phi$, and consequentlyThese canonically normalized parameters can run with the renormalization scale; this does not contradict the holomorphic non-renormalization theorem. When massless modes are included in the full quantum effective action, nonlocal terms associated with infrared divergences can also imitate an F-term. Distinguishing that action from the local Wilsonian effective action is part of the theorem's convention.

A function element is a pair $(f,D)$ consisting of a domain $D\subseteq G$ and a holomorphic function $f:D\to \mathbb{C}$ belonging to the given complete analytic function. Its germ of a holomorphic function at $z\in D$ is the equivalence class $[f{]}_z$, where two elements are equivalent when they agree on a neighbourhood of $z$.

For every function element $(f,D)$ and sufficiently small open $U\subseteq D$, setThese sets form a basis for the space of germs of holomorphic functions. The projectionrestricts to a homeomorphism $U_f\to U$. The inverse projections $\pi {|}_{U_f}$ are compatible complex charts, making $\mathcal{G}$ a Riemann surface and $\pi$ a covering map and local biholomorphism. In such a chart, the evaluation map on a space of germs has coordinate expressionIt is therefore holomorphic on every chart, hence analytic on each component of $\mathcal{G}$.

Now let $f:R\to S$ be a nonconstant analytic map of compact connected Riemann surfaces, and let $B$ be its finite branch-value set. Each point of $B$ is a branch value of a holomorphic map. At every point of $R\setminus f^{-1}(B)$ the local degree is one, so the holomorphic inverse-function theorem supplies a neighbourhood on which $f$ is biholomorphic. For $y\in S\setminus B$, the fibre is finite. Choose disjoint inverse neighbourhoods around all its points and shrink their common image; compactness, or equivalently properness of $f$, ensures that no additional preimages enter. Thusis an ordinary unramified covering map.

Fix $P\in S\setminus B$. Given a loop $\gamma$ based at $P$, lift it from every $Q\in f^{-1}(P)$ and send $Q$ to the endpoint of its lift. The path lifting theorem makes this a permutation; concatenation of loops composes the permutations and reversal gives inverses. Its image is the monodromy group of a covering. The punctured surface $R\setminus f^{-1}(B)$ is path connected. Given two points of the fibre, join them there by a path; its projection is a loop whose lift joins those points. Hence the monodromy action is transitive.

Finally considerIt has degree four. Its finite critical points satisfywith branch values $2$ and $-2$; the double poles at zero and infinity give the branch value infinity. For a regular value, the fibre isThe two commuting involutions $z\mapsto -z$ and $z\mapsto z^{-1}$ preserve $F$ and act transitively on this fibre. They generate all four deck transformations, so the covering is normal and its monodromy group isWith the displayed ordering of the fibre, its elements areThis is the monodromy group of z squared plus z to the minus two.

One form of the Baire category theorem says that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets.

For each nonnegative integer $n$, putBecause $f$ is an entire function, every derivative $f^{(n)}$ is continuous and $Z_n$ is closed. No $f^{(n)}$ is identically zero: if $f^{(n)}\equiv 0$ for some $n$, repeated integration shows that $f$ is a polynomial. By the identity theorem, the zero set of the nonzero holomorphic function $f^{(n)}$ has empty interior. Thus each $Z_n$ is nowhere dense.

The complex plane is complete, so the Baire category theorem givesChoose $z_0$ outside this union. Then $f^{(n)}(z_0)\ne 0$ for every $n$, and hence every coefficient $f^{(n)}(z_0)/n!$ in the Taylor series about $z_0$ is nonzero. Such a point $z_0$ therefore exists.

The set $K$ is compact, and its complement is connected: two disjoint closed disks do not enclose any bounded complementary component. Since $f$ is holomorphic on the open neighbourhood $\Omega$ of $K$, the polynomial Runge theorem gives, for each $n$, a polynomial $p_n$ such thatHence

Define a function on the disconnected open set $\Omega$ byIt is holomorphic on $\Omega$. Applying the first part produces polynomials $P_n$ with

For an obstruction, takeThese are disjoint nonempty bounded closed sets. If polynomials $Q_n$ converged uniformly to zero on $K_1$, the maximum modulus principle on a bounded domain would giveThey therefore cannot also satisfy $Q_n(0)\to 1$. Thus this pair $K_1,K_2$ has the requested non-approximation property.

The Cauchy integral formula says that if $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D(z_0;R)}$, then for $z\in D(z_0;R)$,In particular, at the centre,for every $0<r<R$.

If $f(z_0)=0$, the assumed bound gives $|f(z)|\le 0$, so $f=0$. Otherwise define $g=f/f(z_0)$. Then $g(z_0)=1$ and $|g|\le 1$. Applying the centre formula and taking real parts givesThe continuous nonnegative function $1-\mathrm{Re}g$ has integral zero, so it vanishes around the circle. Equality $\mathrm{Re}g=1$ together with $|g|\le 1$ forces $g=1$. This holds for every $r<R$, hence $f(z)=f(z_0)$ throughout the disc. This is the equality case underlying the maximum modulus principle.

In a holomorphic coordinate chart, $T^{1,0}X$ has frame $\partial /\partial z^1,\dots ,\partial /\partial z^n$. Under a holomorphic coordinate change $w=w(z)$, the chain rule givesThe Jacobian is an invertible matrix of holomorphic functions. These frames therefore make $T^{1,0}X$ the holomorphic tangent bundle.

For $\sigma >1$, summation over the intervals $[n,n+1)$ givesSince $\lfloor u\rfloor =u-\{u\}$,The integral converges absolutely and defines a holomorphic function for $\sigma >0$. This formula therefore gives the Meromorphic continuation of the Riemann zeta function to the right half-plane, with only a simple pole at $s=1$ and residue one.

The polynomial Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $p$ such that ${\sup }_{z\in K}|p(z)-f(z)|<\epsilon$.

For an explicit approximation on the left semicircle $K=\{z:|z|=1,\mathrm{Re}z\le 0\}$, defineThe two power-series expansionsproduce this formula. The first is a geometric series whose ratio has modulus at most $4/\sqrt{17}<1$ on $K$. The second converges uniformly on the whole unit circle; the elementary bound $\left(\genfrac{}{}{0ex}{}{j+k}{k}\right)\le 2^{j+k}$ shows that truncating it at $k=n^2$ gives a total error tending to zero even after summing over $0\le j\le n$. Consequently $p_n\to 1/z$ uniformly on $K$. This is an explicit polynomial approximation of the reciprocal on the left semicircle.

No such sequence exists on the punctured unit circle $S=\{z:|z|=1,z\ne 1\}$. If polynomials $p_n$ converged uniformly there, they would be Uniformly Cauchy. Continuity givesso they would converge uniformly on the entire unit circle. The uniform limit theorem would force the value at the missing point to be $1$, and hence the limit would be $1/z$ on the full circle. But the Cauchy integral theorem gives $\oint p_n(z)dz=0$ for every $n$, while uniform convergence and contour integration would implya contradiction. This is the polynomial approximation obstruction on a punctured circle.

The Gamma 1 congruence subgroup isThe space $M_k({\Gamma }_1(N))$ consists of the holomorphic functions $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma \in {\Gamma }_1(N)$ and which are holomorphic at a cusp for every cusp. Its subspace $S_k({\Gamma }_1(N))$ consists of the forms that vanish at every cusp.

The identity theorem states that if two holomorphic functions on a connected open set $\mathcal{D}$ agree on a subset having an accumulation point in $\mathcal{D}$, then they agree throughout $\mathcal{D}$. In particular, agreement on any nonempty open subset is sufficient.

Suppose first that $u$ is harmonic. Sincethe Cauchy-Riemann equations show thatis holomorphic on $D$. An open disc is a simply connected domain, so $g$ has a holomorphic primitive $f$. If $f=p+iq$, thenThus $p-u$ has zero gradient and is constant on the connected disc. Subtracting that real constant from $f$ gives

Conversely, if $u=\mathrm{Re}f$ for a holomorphic $f=u+iv$, differentiating the Cauchy-Riemann equations givesHence $u$ is harmonic. This proves the harmonic function as the real part of a holomorphic function criterion on $D$.

The point $z_0$ is an ordinary point when both $p$ and $q$ are holomorphic at $z_0$.

The hypothesis makes $\Omega$ star-shaped with centre $z_0$. For small $h$ with the triangle having vertices $z_0,w,w+h$ contained in $\Omega$, the vanishing triangular integral givesParametrizing the final segment,by continuity. Thus $g$ is holomorphic andThis is the triangle-integral construction behind Morera's theorem.

The Taylor series of the complex sine isTherefore, for $z\ne 0$,and the right side defines a holomorphic function at $z=0$ with value $1$. Thus the apparent singularity is a removable singularity.

The resulting power series isThe ratio test shows convergence for every $z\in \mathbb{C}$, so its radius of convergence is

For an integer $k$ and a congruence subgroup $\Gamma \le S{L}_2(\mathbb{Z})$, the space $M_k(\Gamma )$ consists of holomorphic functions $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma$, and which are holomorphic at every cusp. This is the space of modular forms of weight $k$ and level $\Gamma$.

Choose paths from $0$ to $z$ in the cut plane and deform them continuously with the endpoint, never crossing the specified cut. With $0\le \arg z<2\pi$, use the sheet reached from the positive real axis; when an endpoint is moved into the left half-plane, the path passes around the upper endpoint $i$ of the cut. The cut makes all such admissible paths homotopic with fixed endpoints, so the integral is single valued and its endpoint derivative is $(1+z^2{)}^{-1/2}$; hence it is analytic.

For $z=-\sinh u$, continue the endpoint counterclockwise from the positive real axis through the upper half-plane. Inthe continued square root equals $-\cosh u$ at $z=-\sinh u$. The logarithm is reached with argument $\pi$, and therefore

The identitiesshow directly why continuation without an argument restriction is multivalued. The complete set of values of the multivalued inverse hyperbolic sine at $\sinh u$ is

Rouché's theorem states that if $g$ and $h$ are holomorphic functions on a neighbourhood of the closure of a bounded domain andon its positively oriented boundary, then $g$ and $g+h$ have the same number of zeros in the domain, counted with multiplicity.

The Open mapping theorem states that a nonconstant holomorphic function on a domain maps every open subset to an open subset. To prove it, fix $z_0$ and put $w_0=f(z_0)$. By the identity theorem, the zeros of $f-w_0$ are isolated. Choose $r>0$ so that the closed disc $\overline{D(z_0,r)}$ lies in the domain and $z_0$ is the only zero of $f-w_0$ in that disc. ThenWhenever $|w-w_0|<\delta$, the constant $w_0-w$ has modulus smaller than $f-w_0$ on the circle. The Rouche theorem therefore says thathas the same positive number of zeros in the disc as $f-w_0$. Thus every $w\in D(w_0,\delta )$ lies in the image of $f$, proving that the image is open.

If $|f|$ had a local maximum at $z_0$, take a small open disc $D$ on which $|f(z)|\le |f(z_0)|$. The restriction of $f$ to $D$ cannot be constant, since the identity theorem would then make $f$ constant on the whole domain. The complex open mapping theorem says that $f(D)$ is an open neighbourhood of $f(z_0)$, and such a neighbourhood contains points of modulus greater than $|f(z_0)|$, a contradiction. This proves the maximum modulus principle from the complex open mapping theorem.

Now let $\Omega$ be bounded. Its closure is compact, so the continuous function $|f|$ attains a maximum there. If $f$ is nonconstant, the maximum modulus principle excludes an interior maximum; if $f$ is constant, the claimed bound is immediate. Hencewhich is the maximum modulus principle on a bounded domain.

Finally let $\Omega =\{z:\mathrm{Re}z>1\}$ and suppose $|f|\le K$ throughout $\overline{\Omega }$. Fix $z_0\in \Omega$. The principal complex logarithm is holomorphic on the right half-plane, so for each positive integer $n$ the functionis holomorphic on $\Omega$ and continuous on its closure. Given $\epsilon >0$, choose $R>|z_0|$ so large thatApply the bounded-domain result to $g_n$ on $\Omega \cap D(0,R)$. On the vertical part of the boundary, $|z|\ge 1$ and hence $|g_n(z)|\le M$. On the circular part,It follows thatLetting $n\to \infty$ and then $\epsilon \to 0$ gives $|f(z_0)|\le M$. Since $z_0$ was arbitrary, this proves the bounded half-plane maximum principle.

The boundedness assumption is necessary. The functionis holomorphic on $\Omega$ and continuous on its closure, and $|f(1+iy)|=e$ on the boundary, but $|f(x)|=e^x$ is unbounded for real $x>1$. Thus the boundary estimate does not control an unbounded holomorphic function on this unbounded domain.