Solution. Write the Taylor expansions about \(a\). Since \(f\) has a zero of multiplicity \(\ell \) and \(g\) one of multiplicity \(m\), \[f(z) = (z-a)^{\ell }F(z),\qquad g(z) = (z-a)^{m}G(z),\] with \(F, G\) analytic near \(a\) and \(F(a) \neq 0 \neq G(a)\). Then for \(z \neq a\) \[\frac {f(z)}{g(z)} = (z-a)^{\ell -m}\,\frac {F(z)}{G(z)}, \qquad \frac {F(z)}{G(z)} \longrightarrow \frac {F(a)}{G(a)} \neq 0 .\] Three cases follow immediately from the exponent \(\ell - m\):

\(\ell > m\). Then \((z-a)^{\ell -m} \rightarrow 0\) and the limit is \(0\).

\(\ell < m\). Then \((z-a)^{\ell -m} \rightarrow \infty \) in modulus, and since \(F/G\) tends to a non-zero limit, \(\left |f/g\right | \rightarrow \infty \).

\(\ell = m\). The power disappears and the limit is \(F(a)/G(a)\). To see this is \(f^{(m)}(a)/g^{(m)}(a)\), differentiate \(f = (z-a)^{m}F\) exactly \(m\) times and evaluate at \(a\): every term carrying a surviving factor of \((z-a)\) dies, leaving \(f^{(m)}(a) = m!\,F(a)\), and likewise \(g^{(m)}(a) = m!\,G(a)\). The factorials cancel: \[\lim _{z\to a}\frac {f(z)}{g(z)} = \frac {F(a)}{G(a)} = \frac {f^{(m)}(a)}{g^{(m)}(a)} . \qquad \blacksquare \]

The complex proof is cleaner than the real one because analyticity hands us the factorisation outright; no mean value theorem is needed.

Solution. Suppose \(f \not \equiv 0\). Then there is \(z_0 \in G\) with \(f(z_0) \neq 0\), and by continuity \(f \neq 0\) on some disc \(D\) about \(z_0\). Since \(fg = 0\) throughout \(G\), we must have \(g \equiv 0\) on \(D\).

The zero set of \(g\) therefore contains a disc, which certainly has a limit point in \(G\). As \(G\) is a region — open and connected — the Identity Theorem gives \(g \equiv 0\) on all of \(G\). \(\blacksquare \)

Connectedness is essential. On \(G = D_1 \cup D_2\), two disjoint discs, put \(f = 0\) on \(D_1\) and \(f = 1\) on \(D_2\), and \(g\) the other way round: then \(fg \equiv 0\) with neither factor identically zero. The statement is really about \(\mathcal {O}(G)\) being an integral domain, which holds precisely when \(G\) is connected.

Solution. Write \(f = u + iv\). That \(f\) is real-valued means \(v \equiv 0\) on \(D\), so all partial derivatives of \(v\) vanish. The Cauchy–Riemann equations \[u_x = v_y,\qquad u_y = -v_x\] then give \(u_x = 0\) and \(u_y = 0\) throughout \(D\).

A real function on a region with both partials identically zero is constant: any two points of \(D\) are joined by a polygonal path in \(D\), and \(u\) is constant along each segment because its directional derivative vanishes.

Hence \(u\) is constant, \(v \equiv 0\), and \(f\) is constant. \(\blacksquare \)

Alternatively in one line: \(f(D)\) lies in \(\mathbb {R}\), which contains no open set of \(\mathbb {C}\), so \(f\) is not an open map — and by the Open Mapping Theorem a non-constant analytic function on a region is open. The same argument shows an analytic function with constant modulus, or constant argument, is constant.

Solution. The hypothesis says \(\frac {f'}{f}\) and \(\frac {g'}{g}\) agree at every \(a_n\). Both are analytic on \(D\), since \(f\) and \(g\) are non-vanishing there. The set \(\{a_n\}\) has the limit point \(a \in D\), so by the Identity Theorem \[\frac {f'(z)}{f(z)} = \frac {g'(z)}{g(z)} \qquad \text {for all } z \in D .\]

Now consider \(h = f/g\), analytic and non-vanishing on \(D\). By the quotient rule, \[h' = \frac {f'g - fg'}{g^{2}} = \frac {fg}{g^{2}}\left (\frac {f'}{f} - \frac {g'}{g}\right ) = 0\] throughout \(D\). A function with vanishing derivative on a region is constant, so \(h \equiv c\) and \[\boxed {f(z) = c\,g(z) \qquad (z \in D)}\] with \(c \neq 0\) since \(f\) does not vanish. \(\blacksquare \)

The quantity \(f'/f\) is the logarithmic derivative, and the result says it determines \(f\) up to a multiplicative constant — the exact analogue of an antiderivative being determined up to an additive one.

Solution. (a). Write \(f = u+iv\) and \(g(z) = \overline {f(\overline {z})}\). If \(z = x+iy\) then \(\overline {z} = x - iy\) and \[g(x+iy) = u(x,-y) - i\,v(x,-y),\] so \(g\) has real part \(U(x,y) = u(x,-y)\) and imaginary part \(V(x,y) = -v(x,-y)\). Differentiating, and writing the arguments of \(u,v\) as \((x,-y)\) throughout, \[U_x = u_x,\qquad U_y = -u_y,\qquad V_x = -v_x,\qquad V_y = v_y .\] The Cauchy–Riemann equations for \(g\) are \(U_x = V_y\) and \(U_y = -V_x\), that is \[u_x = v_y \qquad \text {and}\qquad -u_y = v_x ,\] which are exactly the Cauchy–Riemann equations for \(f\) at the point \(\overline {z}\), and hold by hypothesis. The partials are continuous, so \(g\) is holomorphic on \(D\) (which is symmetric, so \(\overline {z} \in D\) whenever \(z \in D\)). \(\blacksquare \)

(b). Suppose \(f(r) \in \mathbb {R}\) for every \(r \in D \cap \mathbb {R}\). Then for such \(r\), \[g(r) = \overline {f(\overline {r})} = \overline {f(r)} = f(r).\] So \(f\) and \(g\) are two holomorphic functions on the region \(D\) agreeing on \(D \cap \mathbb {R}\), a set with a limit point in \(D\) (it is a non-empty open subset of a line, as \(D\) is open and symmetric). By the Identity Theorem \(f \equiv g\), that is \[f(z) = \overline {f\!\left (\overline {z}\right )} \qquad \text {for all } z \in D . \qquad \blacksquare \]

This is the Schwarz reflection principle in the form used everywhere: a function real on the reals commutes with conjugation.

Solution. Two symmetries of \(\zeta \) are needed, and the problem is really a test of knowing which.

Conjugation. \(\zeta \) is real on the real axis where it is defined — its Dirichlet series \(\sum n^{-s}\) has real coefficients — so by the reflection principle of the previous problem, applied to the region of analyticity, \[\zeta \!\left (\overline {s}\right ) = \overline {\zeta (s)} .\] Hence if \(\zeta (a) = 0\) then \(\zeta (\overline {a}) = \overline {0} = 0\), giving \(\overline {a}\).

The functional equation. \(\zeta \) satisfies \[\zeta (s) = 2^{s}\pi ^{s-1}\sin \!\left (\frac {\pi s}{2}\right )\Gamma (1-s)\,\zeta (1-s).\] Suppose \(\zeta (a) = 0\) with \(0 \leq \operatorname {Re}a \leq 1\). Reading the equation at \(s = 1-a\), \[\zeta (1-a) = 2^{1-a}\pi ^{-a}\sin \!\left (\frac {\pi (1-a)}{2}\right )\Gamma (a)\,\zeta (a) = 0,\] because the last factor vanishes and none of the others has a pole in the strip that could cancel it: \(\Gamma \) has poles only at \(0,-1,-2,\ldots \), and the only critical-strip point at risk is handled by \(\zeta \) having no zero at \(s = 0\) or \(s=1\) inside the strip’s interior. So \(1-a\) is a zero.

Combining the two, \(\overline {1-a} = 1-\overline {a}\) is also a zero.

Hence \(a\), \(\overline {a}\), \(1-a\) and \(1-\overline {a}\) are all zeros. \(\blacksquare \)

These four points form a rectangle symmetric about both the real axis and the line \(\operatorname {Re}s = \tfrac 12\). The Riemann Hypothesis is the assertion that the rectangle always degenerates to a segment — that is, \(a = 1-\overline {a}\).

Solution. The statement needs its hypothesis read carefully: “differentiable” must mean complex-differentiable at every point of an open set, not at a single point. With that reading:

Let \(f\) be complex-differentiable at every point of an open set \(\Omega \). Fix a disc \(D \subseteq \Omega \) and let \(\Delta \) be any triangle in \(D\). Goursat’s theorem — proved by repeated bisection of \(\Delta \), using only the existence of \(f'\) and no continuity assumption on it — gives \[\int _{\partial \Delta } f(z)\,dz = 0 .\] Since \(f\) is differentiable it is continuous, so Morera’s theorem applies and \(f\) is analytic on \(D\); analyticity being local, \(f\) is analytic on \(\Omega \). \(\blacksquare \)

Two warnings. Differentiability at a point is not enough: \(f(z) = \left |z\right |^{2}\) is complex-differentiable at \(0\) and nowhere else, so it is analytic nowhere. And the result has no real analogue — a once-differentiable real function need not be twice differentiable, whereas here one complex derivative on an open set delivers infinitely many.

Goursat’s contribution is precisely the removal of the continuity assumption on \(f'\); without it, the implication would be circular.

Solution. Fix \(z_0 \in G\) and let \(D\) be a disc about \(z_0\) with \(\overline {D} \subseteq G\). On a disc every harmonic function has a harmonic conjugate, so there is an analytic \(f = u + iv\) on \(D\).

Since \(u\) is non-constant on \(G\) and \(G\) is connected, \(u\) is non-constant on \(D\) — otherwise \(u_x = u_y = 0\) on \(D\), hence \(f' = u_x - iu_y \equiv 0\) there, and the Identity Theorem applied to \(f'\) on a suitable chain of discs would make \(u\) constant on all of \(G\).

Therefore \(f\) is non-constant and analytic on \(D\), so by the Open Mapping Theorem \(f(D)\) is an open subset of \(\mathbb {C}\). Now \[u(D) = \operatorname {Re}\big (f(D)\big ),\] and the projection \(\operatorname {Re} : \mathbb {C} \rightarrow \mathbb {R}\) is an open map (the image of an open disc is an open interval). The composition of open maps is open, so \(u(D)\) is open in \(\mathbb {R}\).

Every point of \(G\) has such a neighbourhood, and a map that is open on a neighbourhood of each point is open. \(\blacksquare \)

Consequence: a non-constant harmonic function has no interior maximum or minimum, since an open subset of \(\mathbb {R}\) contains no largest or smallest element — which is the maximum principle for harmonic functions.

Solution. Suppose such a \(u\) existed. Consider, for \(\varepsilon > 0\), \[u_{\varepsilon }(z) = u(z) + \varepsilon \log \left |z\right | .\] Now \(\log \left |z\right |\) is harmonic on the punctured disc, so \(u_\varepsilon \) is harmonic on \(\Delta ^{*}\).

Examine its boundary behaviour on the annulus \(\delta \leq \left |z\right | \leq 1\):

-

on \(\left |z\right | = 1\): \(u_\varepsilon = 0 + \varepsilon \log 1 = 0\);

-

on \(\left |z\right | = \delta \): \(u_\varepsilon \leq M + \varepsilon \log \delta \), where \(M = \max _{\overline {\Delta ^{*}}}\left |u\right |\), finite by continuity on a compact set. Since \(\log \delta \rightarrow -\infty \), for each fixed \(\varepsilon \) we may choose \(\delta \) so small that this is \(\leq 0\).

By the maximum principle on the annulus, \(u_\varepsilon \leq 0\) throughout it. Fix \(z\) with \(0 < \left |z\right | < 1\) and let \(\delta \rightarrow 0\): we get \[u(z) + \varepsilon \log \left |z\right | \leq 0 .\] Now let \(\varepsilon \rightarrow 0\), giving \(u(z) \leq 0\) for every \(z\) in the punctured disc. By continuity, \(u(0) \leq 0\) — contradicting \(u(0) = 1\). \(\blacksquare \)

The point is that a puncture is invisible to harmonic functions with bounded behaviour: an isolated boundary point cannot carry boundary data. The logarithm is the standard barrier used to say so, and this is why the Dirichlet problem on the punctured disc is not solvable — the same fact the Perron problem exploits.

Solution. Statement. Let \(u\) be harmonic and non-negative on \(B(a,R)\) and continuous on the closure. Then for \(0 \leq r < R\) and all \(\theta \), \[\frac {R-r}{R+r}\,u(a) \ \leq \ u\!\left (a + re^{i\theta }\right ) \ \leq \ \frac {R+r}{R-r}\,u(a).\]

Proof. By the Poisson integral formula, for \(z = a + re^{i\theta }\), \[u(z) = \frac {1}{2\pi }\int ^{2\pi }_{0} \frac {R^{2}-r^{2}}{R^{2} - 2Rr\cos (\theta -t) + r^{2}}\; u\!\left (a + Re^{it}\right )dt .\] The kernel is bounded above and below using \((R-r)^{2} \leq R^{2}-2Rr\cos \phi + r^{2} \leq (R+r)^{2}\), which is just \(-1 \leq \cos \phi \leq 1\). Hence \[\frac {R^{2}-r^{2}}{(R+r)^{2}} \ \leq \ \text {kernel} \ \leq \ \frac {R^{2}-r^{2}}{(R-r)^{2}},\] that is \[\frac {R-r}{R+r} \ \leq \ \text {kernel} \ \leq \ \frac {R+r}{R-r}.\] Because \(u \geq 0\) on the boundary circle, multiplying by \(u(a+Re^{it}) \geq 0\) preserves the inequalities, and integrating gives \[\frac {R-r}{R+r}\cdot \frac {1}{2\pi }\int ^{2\pi }_{0}u\,dt \ \leq \ u(z)\ \leq \ \frac {R+r}{R-r}\cdot \frac {1}{2\pi }\int ^{2\pi }_{0}u\,dt .\] The mean value property identifies the average with \(u(a)\), completing the proof. \(\blacksquare \)

Non-negativity is what allows the kernel bounds to be multiplied through; without it the inequalities reverse on the negative part and the result is false. The consequence — Harnack’s principle — is that a non-negative harmonic function cannot be large somewhere and tiny nearby: the two are pinned within a factor depending only on the geometry.

Solution. (a). Perron’s method, described. Call a continuous \(\varphi \) on \(\overline {\Omega }\) subharmonic if it satisfies the sub-mean-value inequality on every small circle. Given boundary data \(f\), form the Perron family \[\mathcal {P} = \left \{\varphi \text { subharmonic on } \Omega \ :\ \limsup _{z \to \xi }\varphi (z) \leq f(\xi ) \text { for all } \xi \in \partial \Omega \right \},\] and define \[u(z) = \sup _{\varphi \in \mathcal {P}} \varphi (z).\] This upper envelope is always harmonic on \(\Omega \). Whether it attains the boundary values — that is, whether \(u(z) \to f(\xi )\) as \(z \to \xi \) — depends on the boundary point: it holds at \(\xi \) exactly when \(\xi \) admits a barrier, a local subharmonic function that is negative on \(\Omega \) and tends to \(0\) at \(\xi \) alone. So Perron always produces a candidate; the geometry of \(\partial \Omega \) decides whether the candidate solves the problem.

(b). A boundary point with no barrier. Take \(\Omega = \{z : 0 < \left |z\right | < 1\}\), the punctured disc, whose boundary is the unit circle together with the isolated point \(0\). Prescribe \[f \equiv 0 \text { on } \left |z\right | = 1,\qquad f(0) = 1 .\] This is continuous on \(\partial \Omega \), the two pieces being disjoint compact sets.

The Dirichlet problem is not solvable, and that is exactly the previous problem: no harmonic \(u\) on the punctured disc, continuous up to the closure, can vanish on the unit circle and equal \(1\) at the origin. The proof there — comparing with \(u + \varepsilon \log \left |z\right |\) and letting \(\varepsilon \to 0\) — shows \(u \leq 0\) everywhere, contradicting \(u(0) = 1\). \(\blacksquare \)

The isolated boundary point is invisible to harmonic functions; it admits no barrier, and Perron’s envelope simply ignores the datum placed there.

Solution. Let \(u\) be harmonic on a domain \(\Omega \) and fix \(z_0 \in \Omega \). Choose a disc \(D \subseteq \Omega \) about \(z_0\). On a disc — simply connected — \(u\) has a harmonic conjugate \(v\): define \[v(x,y) = \int _{(x_0,y_0)}^{(x,y)} \left (-u_y\,dx + u_x\,dy\right ),\] the integral being path-independent because \(\partial _y(-u_y) = -u_{yy} = u_{xx} = \partial _x(u_x)\), which is precisely \(\Delta u = 0\).

Then \(f = u + iv\) satisfies the Cauchy–Riemann equations by construction and has continuous partials, so \(f\) is analytic on \(D\).

An analytic function is infinitely complex-differentiable, hence its real and imaginary parts have partial derivatives of every order. Therefore \(u = \operatorname {Re}f\) is \(C^{\infty }\) on \(D\), and as \(z_0\) was arbitrary, on \(\Omega \). \(\blacksquare \)

The hypothesis was merely \(C^{2}\), and the conclusion is \(C^{\infty }\): solutions of Laplace’s equation are automatically as smooth as one could want, and in fact real-analytic. This is elliptic regularity in its simplest instance, and there is nothing like it for, say, the wave equation.

Solution. Fix \(z_0 \in G\) and a closed disc \(\overline {D} = \overline {B}(z_0,R) \subseteq G\). Let \(h\) be the solution of the Dirichlet problem on \(D\) with boundary values equal to \(u\) on \(\partial D\) — available explicitly by the Poisson integral, since \(u\) is continuous on the circle. So \(h\) is harmonic on \(D\), continuous on \(\overline {D}\), and \(h = u\) on \(\partial D\).

Put \(w = u - h\) on \(\overline {D}\). Then \(w\) is continuous, vanishes on \(\partial D\), and satisfies the mean value property on \(D\): \(u\) does by hypothesis, \(h\) does because it is harmonic, and the property is preserved by subtraction.

A continuous function with the mean value property obeys the maximum principle — the standard argument applies verbatim, since it uses only the mean value property and not harmonicity. So \(w\) attains its maximum and minimum on \(\partial D\), where both are \(0\). Hence \(w \equiv 0\) and \[u = h \quad \text {on } D,\] so \(u\) is harmonic on \(D\), in particular at \(z_0\). As \(z_0\) was arbitrary, \(u\) is harmonic on \(G\) and \(\Delta u = 0\). \(\blacksquare \)

The device — subtract off the Poisson solution and use the maximum principle to kill the difference — is the standard way to promote a mean-value hypothesis to harmonicity, and it is worth learning as a pattern.

Solution. Consider \(w = u - v\). It is harmonic on all of \(\mathbb {R}^{2}\) and, by hypothesis, \(w \geq 0\) everywhere.

Identify \(\mathbb {R}^{2}\) with \(\mathbb {C}\). Since \(\mathbb {C}\) is simply connected, \(w\) has a harmonic conjugate \(\widetilde {w}\) on the whole plane, so \[F = w + i\widetilde {w}\] is entire. Now consider \[g = e^{-F}.\] This is entire, and \[\left |g(z)\right | = e^{-\operatorname {Re}F(z)} = e^{-w(z)} \leq e^{0} = 1,\] using \(w \geq 0\). So \(g\) is a bounded entire function, and Liouville’s theorem makes it constant.

Then \(\left |g\right | = e^{-w}\) is constant, so \(w\) is constant, say \(w \equiv c\) with \(c \geq 0\). That is \[\boxed {u(x,y) = v(x,y) + c \qquad \text {for all } (x,y) \in \mathbb {R}^{2}.}\] \(\blacksquare \)

The trick worth remembering is the passage from a one-sided bound on a harmonic function to a two-sided bound on an analytic one: \(w \geq 0\) says nothing directly about \(\left |F\right |\), but it bounds \(\left |e^{-F}\right |\) perfectly. This is Liouville’s theorem for harmonic functions, and the same device proves that a harmonic function on \(\mathbb {R}^2\) bounded above is constant.

Solution. This is the statement proved in Problem 10 above; the argument is repeated here in the form the question asks for.

By the Poisson integral formula on \(B(a,R)\), writing \(z = a + re^{i\theta }\), \[u(z) = \frac {1}{2\pi }\int ^{2\pi }_{0} P_r(\theta - t)\,u\!\left (a+Re^{it}\right )dt, \qquad P_r(\phi ) = \frac {R^{2}-r^{2}}{R^{2}-2Rr\cos \phi + r^{2}} .\] Since \(-1 \leq \cos \phi \leq 1\), \[(R-r)^{2} \leq R^{2}-2Rr\cos \phi + r^{2} \leq (R+r)^{2},\] and therefore \[\frac {R-r}{R+r} = \frac {R^{2}-r^{2}}{(R+r)^{2}} \ \leq \ P_r(\phi )\ \leq \ \frac {R^{2}-r^{2}}{(R-r)^{2}} = \frac {R+r}{R-r}.\] The hypothesis \(u \geq 0\) lets these bounds be multiplied through the integral without reversing, and the mean value property \(\frac {1}{2\pi }\int _0^{2\pi } u(a+Re^{it})dt = u(a)\) evaluates both ends: \[\boxed {\frac {R-r}{R+r}\,u(a)\ \leq \ u\!\left (a+re^{i\theta }\right )\ \leq \ \frac {R+r}{R-r}\,u(a).} \qquad \blacksquare \]

Both bounds tend to \(u(a)\) as \(r \rightarrow 0\) and degenerate as \(r \rightarrow R\), which is right: near the centre the value is pinned, near the boundary it is free.

Solution. (a). The Poisson integral. For \(\left |z\right | < 1\), writing \(z = re^{i\theta }\), \[u\!\left (re^{i\theta }\right ) = \frac {1}{2\pi }\int ^{2\pi }_{0} \frac {1-r^{2}}{1 - 2r\cos (\theta -t) + r^{2}}\; f\!\left (e^{it}\right )dt,\] with \(u = f\) on \(\left |z\right | = 1\). Equivalently \(u(z) = \frac {1}{2\pi }\int _0^{2\pi }\operatorname {Re}\!\left (\frac {e^{it}+z}{e^{it}-z}\right )f(e^{it})\,dt\), which exhibits \(u\) as the real part of an analytic function of \(z\) and so makes harmonicity evident.

(b). Take \(f \equiv 1\) in part (a). Then the solution is \(u \equiv 1\), and evaluating the formula at \(z = \tfrac 12\) (so \(r = \tfrac 12\), \(\theta = 0\)) gives \[1 = \frac {1}{2\pi }\int ^{2\pi }_{0}\frac {1-\frac 14}{1-\cos t+\frac 14}\,dt = \frac {1}{2\pi }\int ^{2\pi }_{0}\frac {3/4}{\left |e^{it}-\frac 12\right |^{2}}\,dt,\] using \(\left |e^{it}-\frac 12\right |^{2} = \frac 54 - \cos t\). Rearranging, \[\int _{\left |z\right |=1}\frac {d\theta }{\left |z-\frac 12\right |^{2}} = \frac {2\pi }{3/4} = \boxed {\frac {8\pi }{3}} .\] This is the arc-length integral computed directly in the Complex Integration chapter, obtained here for free as the total mass of the Poisson kernel.

(c). Proved in Problem 13 above: subtract the Poisson solution on a small disc and use the maximum principle — valid for any continuous function with the mean value property — to conclude that the difference vanishes.

Solution. Impose \(\Delta u = 0\) on \(u = ax^{3}+bx^{2}y+cxy^{2}+dy^{3}\): \[u_{xx} = 6ax + 2by,\qquad u_{yy} = 2cx + 6dy,\] \[\Delta u = (6a+2c)x + (2b+6d)y \equiv 0 \ \Longrightarrow \ c = -3a,\quad b = -3d .\] So the most general such harmonic polynomial is \[\boxed {u = a\left (x^{3}-3xy^{2}\right ) + d\left (y^{3}-3x^{2}y\right )}\] with \(a, d\) arbitrary reals — a two-parameter family, as expected: of the four coefficients, Laplace’s equation imposes two conditions.

Recognising it. Since \(z^{3} = (x^{3}-3xy^{2}) + i(3x^{2}y - y^{3})\), \[x^{3}-3xy^{2} = \operatorname {Re}z^{3}, \qquad y^{3}-3x^{2}y = \operatorname {Re}\!\left (iz^{3}\right ),\] so with \(A = a + id\), \[u = \operatorname {Re}\!\left (A z^{3}\right ).\]

Conjugate and analytic function. The harmonic conjugate is then the imaginary part, \[v = \operatorname {Im}\!\left (Az^{3}\right ) = a\left (3x^{2}y - y^{3}\right ) + d\left (x^{3}-3xy^{2}\right ),\] determined up to an additive real constant, and \[f(z) = u + iv = (a+id)\,z^{3} + iC, \qquad C \in \mathbb {R}.\]

Verified symbolically: expanding \(\operatorname {Re}\!\left ((a+id)(x+iy)^{3}\right )\) returns \(ax^{3} - 3ax y^{2} - 3d x^{2}y + dy^{3}\), exactly the \(u\) above.

Solution. (\(\Rightarrow \)) Harmonic implies the mean value property. Let \(u\) be harmonic on a domain and \(\overline {B}(z_0,r)\) contained in it. On a disc \(u\) has a harmonic conjugate, so \(u = \operatorname {Re}f\) with \(f\) analytic. Cauchy’s integral formula gives \[f(z_0) = \frac {1}{2\pi i}\oint _{\left |z-z_0\right |=r}\frac {f(z)}{z-z_0}\,dz = \frac {1}{2\pi }\int ^{2\pi }_{0} f\!\left (z_0+re^{it}\right )dt,\] on substituting \(z = z_0 + re^{it}\). Taking real parts, \[u(z_0) = \frac {1}{2\pi }\int ^{2\pi }_{0} u\!\left (z_0+re^{it}\right )dt .\]

(\(\Leftarrow \)) The mean value property implies harmonic. This is Problem 13: on a small disc, subtract the Poisson solution \(h\) with the same boundary values. The difference \(w = u - h\) is continuous, has the mean value property, and vanishes on the boundary circle; the maximum principle — which needs only the mean value property — forces \(w \equiv 0\), so \(u = h\) is harmonic. \(\blacksquare \)

The equivalence is worth stating as an equivalence because the two halves are used in opposite directions: the forward one to derive Gauss’s mean value theorem and the maximum principle, the reverse one to recognise harmonicity in functions defined by limits or integrals, where differentiating twice would be awkward.

Solution. Let \(M = \max _{\overline {\Omega }} f\), attained since \(f\) is continuous on the compact set \(\overline {\Omega }\), and suppose \(f(z_0) = M\) for some interior \(z_0 \in \Omega \). Put \[E = \left \{z \in \Omega : f(z) = M\right \},\] which is non-empty and, by continuity, closed in \(\Omega \).

\(E\) is open. Let \(z_1 \in E\) and choose \(\rho > 0\) with \(\overline {B(z_1,\rho )} \subseteq \Omega \). For any \(0 < s \leq \rho \) the mean value equality gives \[M = f(z_1) = \frac {1}{2\pi }\int ^{2\pi }_{0} f\!\left (z_1 + se^{i\theta }\right )d\theta .\] The integrand is \(\leq M\) throughout. If it were \(< M\) at a single point, then by continuity it would be \(< M - \varepsilon \) on an arc of positive length, making the average strictly less than \(M\) — a contradiction. Hence \(f \equiv M\) on every circle \(\left |z - z_1\right | = s \leq \rho \), that is on the whole disc \(B(z_1,\rho )\). So \(E\) is open.

Conclusion. \(E\) is a non-empty subset of the region \(\Omega \) that is both open and closed in \(\Omega \). Since \(\Omega \) is connected, \(E = \Omega \), so \(f \equiv M\) on \(\Omega \) and, by continuity, on \(\overline {\Omega }\). \(\blacksquare \)

The proof is direct as demanded: it uses only the mean value equality, continuity and connectedness, and never mentions analyticity, Poisson kernels or harmonic conjugates. That is what makes it available as a tool for proving functions harmonic, rather than a consequence of their being so.