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Since $\mathbb{C}$ is simply connected, writefor an entire function $f$. If $u$ were bounded above, thenwould be bounded, so the Liouville theorem would make $e^f$ constant. Its derivative $f^{\prime }{e}^f$ would then vanish, whence $f$ and $u$ would be constant. Applying the same argument to $e^{-f}$ rules out a lower bound. Thus a nonconstant $u$ is unbounded both above and below.

Given any $c\in \mathbb{R}$, choose $z_{-},z_{+}\in \mathbb{C}$ withThe restriction of $u$ to the line segment from $z_{-}$ to $z_{+}$ is continuous, so the intermediate value theorem supplies a point on that segment where $u=c$. ThereforeThis proves the surjectivity of a nonconstant entire harmonic function.

Lift $u$ through the complex exponential by settingThe conformal invariance of harmonicity makes $v$ harmonic. Since $e^{w+2\pi i}=e^w$ and $u(2z)=u(z)$,Thus $v$ is bounded on the compact rectangleand its two periods make it bounded on the whole plane. The harmonic Liouville theorem makes $v$ constant. Since the exponential map is surjective onto ${\mathbb{C}}^{\ast }$, $u$ is constant.

For the requested counterexample after deleting a countable set, takeand let ${\wp }_{\Lambda }$ be the Weierstrass elliptic function for the latticeDefineChanging a branch of the logarithm adds $2\pi i$, a period of ${\wp }_{\Lambda }$, so $u$ is well defined. Its poles project precisely to $S$, and the conformal invariance of harmonicity makes $u$ harmonic away from them. The other period givesMoreover, $z\notin S$ implies $2z\notin S$. Finally, $u$ is nonconstant because as positive real $z\to 1$,which is real and unbounded. This is a scale-periodic harmonic function from an elliptic function.

Take the punctured complex planeandIn polar coordinates, $u(r,\theta )=\log r$, soThus $u$ is harmonic on $D$.

If $u=\mathrm{Re}f$ for a holomorphic function $f$, the calculation from part (a) would giveThe integral of a derivative around a closed curve is zero, whereas the residue theorem givesThis contradiction proves that no such $f$ exists. It is precisely the log modulus has no global harmonic conjugate on the punctured plane obstruction.

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 $D$.

Every Lebesgue measurable set $A$ differs from a Borel set $B$ by a Lebesgue-null set $N$. The given function $h$ is a homeomorphism from $\mathbb{R}$ onto the interval $h(\mathbb{R})$. Hence $h(B)$ is Borel relative to $h(\mathbb{R})$ and therefore Lebesgue measurable in $\mathbb{R}$. Moreover,and $h(N)$ is null by Lusin condition N. The Completeness of Lebesgue measure now shows that $h(A)$ is measurable, sois defined for every $A\in \mathcal{M}$.

Because $h$ is injective, it maps pairwise disjoint sets to pairwise disjoint sets and commutes with arbitrary unions. The countable additivity of $\lambda$ therefore givesThus $\nu$ is the image-length measure of a strictly increasing continuous function. If $\lambda (A)=0$, the null-set hypothesis gives $\nu (A)=\lambda (h(A))=0$, so

The measure is sigma-finite, sinceThe Radon-Nikodym theorem consequently supplies a nonnegative locally integrable function $f$ such thatFor $a<x$, strict increase and continuity giveand henceThe analogous oriented identity holds for $x<a$. By differentiation of an indefinite Lebesgue integral,

Yes, the differentiability conclusion still holds when $h$ is merely non-decreasing. In fact, the Lebesgue theorem on differentiability of monotone functions says that every monotone function is differentiable almost everywhere; neither continuity nor the null-set hypothesis is needed for that conclusion. Flat intervals mean that the preceding image-set argument no longer gives disjoint images, so its natural replacement is the Lebesgue-Stieltjes measure determined by

For an integrable function $f:{\mathbb{R}}^n\to \mathbb{C}$, the Lebesgue differentiation theorem states thatfor almost everywhere $x\in {\mathbb{R}}^n$. In particular,at every Lebesgue point of $f$.

Almost every $x\in \mathbb{R}$ is a Lebesgue point of $g$. At such an $x$, for $t>0$,For $t<0$, the same estimate over $[x+t,x]$ gives the identical limit. Thereforeat every Lebesgue point of $g$, so $G$ is differentiable $\lambda$-almost everywhere. This is the differentiation of an indefinite Lebesgue integral.

Let $\Vert x_n\Vert \le C$. For each $i$, Bessel inequality givesso the $i$th coordinate sequence is bounded. By the Bolzano-Weierstrass theorem, choose a subsequence on which the first coordinate converges. From it choose a further subsequence on which the second coordinate converges, and continue.

Take the diagonal subsequence $(x_{\varphi (n)})$. For every fixed $i$, its $i$th coordinate converges, and it remains bounded by $C$. The coordinate criterion for weak convergence in a separable Hilbert space supplies an $x_{\infty }\in H$ such thatThis proves the theorem on a weak subsequence of a bounded Hilbert-space sequence.

Choose a contractionApply the homotopy extension property to the initial map ${\mathrm{id}}_X:X\to X$ and to $h$, regarded as a homotopy into $X$. It suppliessuch that

At $t=1$, the map $H_1$ sends every point of $A$ to $a_0$. It is therefore constant on the equivalence classes of the quotient topology $X/A$. By the universal property of the quotient topology, there is a continuous mapsuch thatwhere $q:X\to X/A$ is the quotient map. The homotopy $H$ immediately gives

For every $t$, the map $q\circ H_t$ is constant on $A$, because $H(a,t)=h(a,t)\in A$. It therefore descends toThis descended map is continuous: the product $q\times {\mathrm{id}}_I$ is a quotient map because $I$ is compact Hausdorff, and $q\circ H$ is constant on its fibres.
At the endpoints,ThusThe maps $q$ and $g$ are homotopy inverses, soThis proves the theorem on collapsing a contractible cofibration.

A subspace $Y\subseteq X$ is a retract of $X$ if there is a continuous retractionsuch that $r(y)=y$ for every $y\in Y$. A space is a contractible space if its identity map is homotopic to a constant map.

Let $i:Y\hookrightarrow X$ be the inclusion and letbe a contraction, with $H(x,0)=x$ and $H(x,1)=x_0$. Thenis continuous and satisfiesThus $K$ contracts $Y$, proving that every retract of a contractible space is contractible.

Spaces $X$ and $Y$ are homotopy equivalent if there are continuous mapssuch thatThe maps $f$ and $g$ are called homotopy inverses.

The minimal polynomial of $\beta$ isand part (b) showed that ${\mathcal{O}}_K=\mathbb{Z}[\beta ]$. Modulo $3$,The quadratic has no root in ${\mathbb{F}}_3$, so it is irreducible. The Dedekind factorization theorem therefore givesThe two factors are distinct prime ideal, with norms $3$ and $3^2=9$, respectively.

Modulo $2$,and the quadratic is irreducible over ${\mathbb{F}}_2$. HenceThis is a product of two distinct proper prime ideals, so $(2)$ is not prime. These decompositions are collected in prime ideals above two and three in the cubic field of discriminant minus 307.

PutUsing ${\alpha }^3=5\alpha -8$, multiplication by $\beta$ on the ordered basis $(1,\alpha ,\beta )$ is determined byIts matrix, with these coordinate vectors as columns, isThe characteristic polynomial isBy the Cayley-Hamilton theorem, this monic integer polynomial annihilates $\beta$, so $\beta$ is an algebraic integer.

LetThe element $\alpha$ is also integral because it satisfies the monic polynomial $f$. Hence $A\subseteq {\mathcal{O}}_K$. Sincethe lattice $\mathbb{Z}[\alpha ]$ has index two in $A$. The given power-basis discriminant isso the discriminant-index formula for an integral lattice gives

If $m=[{\mathcal{O}}_K:A]$, thenBecause $307$ is prime, the square $m^2$ can divide $307$ only when $m=1$. Thusso $(1,\alpha ,\beta )$ is an integral basis of the cubic field of discriminant minus 307.

Finally, the relation ${\beta }^2=-4-\alpha +3\beta$ givesConsequently ${\mathcal{O}}_K=\mathbb{Z}[\beta ]$, which will allow direct use of Dedekind's theorem.

The element $\alpha$ is a root ofBy the rational root theorem, any rational root of this monic polynomial must be one ofDirect substitution shows that none is a root. A reducible cubic over a field has a linear factor, so $f$ is an irreducible polynomial over $\mathbb{Q}$. It is therefore the minimal polynomial of $\alpha$, and