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Let $q$ be the Holder conjugate exponent of $p$. On ${\mathbb{R}}^n$ with Lebesgue measure, the duality of Lp spaces isunder the pairing ${\Lambda }_g(f)=\int fg$. Thus $(L^1{)}^{\ast }=L^{\infty }$, while for $1<p<\infty$ the dual exponent satisfies $1/p+1/q=1$.

A Banach space is reflexive when its canonical map into its bidual is surjective. The spaces $L^p({\mathbb{R}}^n)$ are reflexive exactly forFor these exponents, applying $L^p$ duality twice identifies the bidual with $L^p$. The endpoint spaces $L^1$ and $L^{\infty }$ are not reflexive; for example, disjoint unit cubes give closed copies of ${\ell }^1$ and ${\ell }^{\infty }$, respectively.

Integration gives an isometric embedding $L^1\hookrightarrow (L^{\infty }{)}^{\ast }$, but it is not onto. On the subspacedefine $\Lambda (c+h)=c$, the limit at infinity. Since $|\Lambda (v)|\le \Vert v{\Vert }_{\infty }$, the Hahn-Banach theorem extends $\Lambda$ to all of $L^{\infty }$. If this extension were integration against some $g\in L^1$, its vanishing on every compactly supported bounded function would force $g=0$ almost everywhere, but $\Lambda (1)=1$. ThereforeThis is the Dual of L infinity is larger than L1.

For $1\le p<\infty$, the Lebesgue space $L^p(X)$ consists of equivalence classes modulo equality almost everywhere of measurable functions satisfyingThe space $L^{\infty }(X)$ consists of essentially bounded measurable functions modulo equality almost everywhere, with norm

Let $1\le p<q<\infty$ and put $r=q/p>1$. By Hölder's inequality,Taking $p$th roots givesFor $q=\infty$, directly integrating $|f{|}^p\le \Vert f{\Vert }_{\infty }^p$ gives the same endpoint estimate. Hence $L^q(X)\subseteq L^p(X)$ when $\mu (X)<\infty$, as in Lp inclusion on a finite measure space.

For every finite scalar sequence $a=(a_n)$ with $|a_n|\le 1$, define a bounded linear functional on the continuous dual space $X^{\ast }$ byFor each fixed $f\in X^{\ast }$, the hypothesis giveswhere the supremum is over all finite choices and the phases $a_n$ may be chosen to align the complex numbers $f(x_n)$.

The dual $X^{\ast }$ is a Banach space even when $X$ is incomplete. The Uniform boundedness principle therefore gives a constant $C$ such thatFor each $N$, choose $a_n$ for $n\le N$ so that $a_{n}f(x_n)=|f(x_n)|$. ThenLetting $N\to \infty$ provesThis is the uniform bound from pointwise absolute summability of dual evaluations.

For $x\in X$, choose $x_m\in D$ with $x_m\to x$ and defineThe sequence on the right is Cauchy becauseand it converges because $Y$ is complete. The same estimate shows that the limit is independent of the approximating sequence, is linear, and satisfiesIts norm is at most $\Vert T\Vert$, while restriction to $D$ gives the reverse inequality, soAny two continuous extensions agree on the dense set $D$ and hence everywhere, proving uniqueness. This is the extension of a bounded linear operator from a dense subspace.

Completeness of $X$ was never used, so the result remains true when $X$ is incomplete. Completeness of $Y$ is essential. For example, with $D=c_{00}\subset X={\ell }^2$, let the codomain be the incomplete normed space $Y=c_{00}$ with the ${\ell }^2$ norm and let $T$ be the identity. An extension would have to send every limit in ${\ell }^2$ to itself and therefore could not take values in $c_{00}$. This is the failure of dense-subspace extension into an incomplete codomain.

Let $(T_m)$ be Cauchy in the operator norm on $\mathcal{B}(X,Y)$. For each $x\in X$,so $(T_{m}x)$ is Cauchy in the Banach space $Y$. DefinePointwise passage to the limit shows that $T$ is linear. A norm-Cauchy sequence is bounded, say $\Vert T_m\Vert \le M$, and henceThus $T$ is bounded. Finally, for fixed $m$,The right-hand coefficient tends to zero with $m$, so $T_m\to T$ in operator norm. ThereforeThis is the Banach space of bounded linear operators.

Let $V_4=⟨u,v\mid u^2=v^2=[u,v]=1⟩$. Define a surjectionIts kernel determines a connected regular four-sheeted cover of a wedge of circles and a real projective plane, say ${\tilde{X}}_n\to X_n$.

Use the CW structure on $X_n$ with one vertex, one edge for every $a_i$ and for $b$, and one two-cell attached along $b^2$. In the regular cover, the degree-two cellular chain group isThe cellular boundary formula gives the lifted attaching word $b^2$ the boundaryin the $b$-edge summand of $C_1$. ThereforeThere are no three-cells, so

If $X_n$ were homotopy equivalent to a compact connected surface $S$, the subgroup corresponding to this cover would produce a connected four-sheeted cover $\tilde{S}$ homotopy equivalent to ${\tilde{X}}_n$. A finite cover of a compact surface is again a compact connected surface. But the integral top homology of a compact connected surface has rank at most one, contradicting the displayed rank two. Hence $X_n$ is not homotopy equivalent to a compact connected surface for any $n\ge 1$.

The wedge intersection is simply connected after taking suitable neighbourhoods, so repeated application of the Seifert-van Kampen theorem gives the free product of $n$ copies of $\mathbb{Z}$ withThus, for any basepoint,This is the fundamental group of a wedge of circles and a real projective plane.

Let $X=Y\cup Z$, where $Y$, $Z$, and $Y\cap Z$ are path-connected subcomplexes containing the basepoint; small open neighbourhoods may be used to reduce to the usual open-cover statement. The Seifert-van Kampen theorem says thatMore explicitly, this is the quotient of the free product ${\pi }_1(Y)\ast {\pi }_1(Z)$ by the normal closure ofEquivalently, ${\pi }_1(X)$ is the pushout of the two inclusion-induced homomorphisms.

Let $|G|=pq$ with primes $p>q$ and $G$ nonabelian. The Sylow theorems give $n_p\mid q$ and $n_p\equiv 1(\mathrm{mod}p)$, so $n_p=1$ and the Sylow $p$-subgroup $P$ is normal. Also $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. If $n_q=1$, both Sylow subgroups are normal and $G$ is their abelian direct product, a contradiction. Hence $n_q=p$, and

Because $G/P$ is abelian, the commutator subgroup $G^{\prime }$ lies in $P$. It is nontrivial because $G$ is nonabelian, and $|P|=p$ is prime, so $G^{\prime }=P$. Thereforeand $G$ has exactly $q$ linear characters.

Every nonlinear irreducible degree divides $pq$. Its square is at most $pq$, so the only possibility is degree $q$. If there are $b$ such characters, the sum of squares of irreducible degrees givesThe number of irreducible characters equals the number of conjugacy classes, soThese are the irreducible characters of a nonabelian group of order p q.

Suppose a finite simple group had an irreducible representation $\rho$ of degree two. Its kernel is normal and cannot be the whole group, so $\rho$ is faithful. A nonabelian simple group has no nontrivial linear character; consequently $\det \rho$ is trivial and $\rho (G)\subseteq {\mathrm{SL}}_2(\mathbb{C})$.

Since $2\mid |G|$ by degree divisibility, Cauchy's theorem for finite groups gives an involution $x\in G$. A nonidentity matrix of order two in ${\mathrm{SL}}_2(\mathbb{C})$ has both eigenvalues $-1$, hence equals $-I$. Thus $\rho (x)=-I$ is central in $\rho (G)$. Faithfulness makes $x$ central in $G$, contradicting the trivial centre of a nonabelian simple group. Therefore no finite simple group has an irreducible character of degree two.

For $p=2$, every group of order four is abelian. Now let $p$ be odd. Degree divisibility and the sum-of-squares formula show that every nonlinear irreducible character has degree two: the other divisors $p$ and $2p$ have squares larger than $2p$.

Let $a$ be the number of linear characters and $b$ the number of degree-two characters. The linear characters are precisely the characters of the abelianization, soThe degree-sum formula givesso $a\equiv 2(\mathrm{mod}4)$. Among the divisors of $2p$, this forces $a=2$ or $a=2p$. In the latter case $b=0$ and all degrees are one. In the former,Thus the degrees areunless they are all one.

Let $|G|=p^2$. By irreducible character degree divides the group order, each irreducible degree divides $p^2$. A degree is at most $\sqrt{|G|}=p$ by the sum of squares of irreducible degrees. If a degree $p$ occurred, its square would already consume the entire sum $p^2$, leaving no room for the trivial character. Hence every irreducible degree is one. A finite group has only linear irreducible characters exactly when it is abelian, so every group of order $p^2$ is abelian.

Put $F=L^G$ and $n=|G|$. The automorphism-count bound for a finite field extension givesFor the reverse inequality, let $x_1,\dots ,x_m$ be linearly independent over $F$. We claim that the $n\times m$ matrixhas linearly independent columns over $L$. If not, choose a nonzero relationwith the fewest nonzero coefficients, and normalize one coefficient to $c_1=1$. Applying $h\in G$ and reindexing the rows gives another relation with coefficients $h(c_j)$. Subtracting eliminates the first term, so minimality forces $h(c_j)=c_j$ for every $h$ and $j$. Thus every $c_j\in F$. The row for the identity automorphism then contradicts the $F$-linear independence of the $x_j$.

Therefore $m\le n$. Taking an $F$-basis of $L$ gives $[L:F]\le n$, and henceThis proves the degree assertion in the Artin fixed-field theorem.

The class of well-orderings is not first-order axiomatizable. If $S$ axiomatized it, add constants $c_0,c_1,\dots$ and sentencesEvery finite subset has a finite well-order containing the required finite descending chain. By the compactness theorem, the whole set has a model; that model satisfies $S$ but contains an infinite descending chain, so it is not a well-order. This contradiction proves the claim, as summarized by well-orders are not first-order axiomatizable.

The class is axiomatizable by the partial order axioms together with totality:Explicitly, the remaining axioms state reflexivity, antisymmetry, and transitivity. This is the first-order axiomatization of total orders.

This class is the class of torsion-free groups and is axiomatizable. Take the group axioms and, for every integer $n\ge 2$, the sentenceThese sentences exclude every nonidentity element of finite order and give the first-order axiomatization of torsion-free groups.