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SetThe triangle inequality makes $f$ Lipschitz continuous and hence a continuous function. Since $\gamma$ is an isometric embedding,so $f(t)\to \infty$ as $|t|\to \infty$. The function is therefore coercive, and the extreme value theorem on a sufficiently large compact interval gives a minimizing parameter.

Suppose $t_1,t_2$ both minimize $f$, put $p=\gamma (t_1)$ and $q=\gamma (t_2)$, and letThe restriction of $\gamma$ between the two parameters is a geodesic from $p$ to $q$. Let $m$ be its midpoint. In the geodesic triangle with vertices $x,p,q$, the thin geodesic triangle condition gives a point $y$ on one of the other two sides with $d(m,y)\le \delta$. By symmetry suppose $y\in [x,p]$. ThensoBut $m$ lies on $\gamma (\mathbb{R})$ and $p$ is a closest point, so $R\le d(x,m)$. Therefore $L\le 4\delta$, which is stronger than the requiredThis is the closest point on a geodesic line in a hyperbolic metric space estimate.

For a continuous closed path $\gamma :[0,1]\to \mathbb{C}\setminus \{0\}$, choose a continuous argument liftIts winding number of a continuous closed path about zero isFor a piecewise smooth path this equals $(2\pi i{)}^{-1}{\int }_{\gamma }dz/z$.

If $|\gamma (t)|>|\varphi (t)|$, thennever vanishes, since $|H(s,t)|\ge |\gamma (t)|-s|\varphi (t)|>0$. Thus $H$ is a homotopy through closed paths avoiding zero. By homotopy invariance of winding number, or directly by the dominated-perturbation lemma,

More generally, ${\gamma }_0$ and ${\gamma }_1$ are homotopic by paths in $\Gamma$ when there is a continuous functionwith $H(0,t)={\gamma }_0(t)$, $H(1,t)={\gamma }_1(t)$, and $H(s,0)=H(s,1)$ for every $s$. The winding-number theorem states that such a homotopy implies

For the Fundamental theorem of algebra, let $P(z)=a_n{z}^n+\cdots +a_0$ with $n\ge 1$. For sufficiently large $R$,for every $t$. The dominated-perturbation lemma shows that $t\mapsto P(R{e}^{2\pi it})$ has the same winding number as $t\mapsto a_n{R}^n{e}^{2\pi int}$, namely $n$. If $P$ had no zero, however,would be a homotopy in $\Gamma$ from that loop to the constant loop $P(0)$, whose winding number is zero. This contradiction proves that $P$ has a complex root. This is the winding-number proof of the fundamental theorem of algebra.

Finally suppose that a continuous retraction $r:D^2\to S^1$ existed. The boundary loop $\eta (t)=e^{2\pi it}$ has winding number one, while $(1-s)\eta (t)$ contracts it to zero inside the disc. Composing this contraction with $r$ gives a homotopy through loops in $S^1$ from $r\circ \eta =\eta$ to the constant loop $r(0)$. Their winding numbers are respectively one and zero, contradicting homotopy invariance. Hence there is no such retraction, as in the winding-number proof of the no-retraction theorem.

Statement (iv) is true. Apply the Weierstrass approximation theorem separately to the real and imaginary parts of the continuous function $f$ on the compact interval $[0,1]$. Combining the two real polynomial sequences gives complex polynomials converging uniformly to $f$ on $I$.

For $\mathbf{a},\mathbf{b}\in {\mathbb{R}}^{n+1}$, the triangle inequality and the Cauchy-Schwarz inequality giveThus $S$ is a continuous function. The unit sphere in the finite-dimensional Euclidean normed vector space ${\mathbb{R}}^{n+1}$ is compact. Moreover, $S(\mathbf{a})=0$ implies that the polynomial $T\mathbf{a}$ vanishes identically, so every coefficient $a_r$ is zero. Hence $S$ is strictly positive on the unit sphere. By the extreme value theorem, it has a positive minimum

The supremum norm is homogeneous. For $\mathbf{a}\ne 0$,It follows directly that $\Vert T\mathbf{a}{\Vert }_{\infty }\to \infty$ whenever $\Vert \mathbf{a}{\Vert }_2\to \infty$.

Write each polynomial of degree at most $n$ as $T\mathbf{a}$ and defineThis is a continuous function, and the reverse triangle inequality together with part (a) givesConsequently $F$ is a coercive function. Choose $R$ so large that $F(\mathbf{a})>F(0)$ outside the closed Euclidean ball of radius $R$. That ball is compact, so the extreme value theorem supplies a minimizer ${\mathbf{a}}_{\ast }$ there. Then $P=T{\mathbf{a}}_{\ast }$ is a best uniform approximation to $f$ among polynomials of degree at most $n$.

As printed, this part contains an indexing error. The displayed product ends at $m-1$, but the hypotheses contain $m+1$ alternating intervals indexed by $0,\dots ,m$. On the first $m$ intervals the printed polynomial has sign $(-1{)}^i$, while on the last interval it has sign $(-1{)}^{m-1}$ rather than the required $(-1{)}^m$. For example, when $m=1$ the printed product is the constant $Q=1$, which cannot reduce a negative extremum of $g$ on the final interval. The intended polynomial isWe prove the stated conclusion with this correction.

Each root $w_j$ lies in the gap between the $(j-1)$st and $j$th active intervals. Therefore $Q$ has sign $(-1{)}^i$ throughout $[u_i,v_i]$. In particular, wherever $|g|=M$, the numbers $Q$ and $g$ have the same sign, and $Q$ is nonzero.

LetThese are compact sets. On their union, the continuous function $|Q|$ has a positive minimum. Hence, throughout some open neighbourhood $U$ of $K_{+}\cup K_{-}$, subtracting a sufficiently small positive multiple $\eta Q$ moves $g$ strictly towards zero and gives $|g-\eta Q|<M$. On the compact complement $[0,1]\setminus U$, continuity gives a uniform margin $|g|\le M-ϵ$ for some $ϵ>0$. Taking alsoshows that $|g-etaQ|<M$ there. Thusfor every sufficiently small $\eta >0$.

It remains to prove necessity in the Chebyshev alternation theorem. Let $P$ be a best approximation of degree at most $n$, put $g=f-P$, and let $M=\Vert g{\Vert }_{\infty }$. If $g$ has no sequence of $n+2$ extrema with alternating signs, its sets of positive and negative extrema can be collected, from left to right, into $m+1$ alternating compact groups with $m\le n$. Choose disjoint intervals $[u_i,v_i]$ containing those groups and separated by regions on which $|g|<M$. After replacing $g$ by $-g$ if necessary, they satisfy the displayed hypotheses. The corrected construction produces a polynomial $Q$ of degree $m\le n$ and a small $\eta >0$ such thatThis contradicts the assumed optimality of $P$. Therefore every best polynomial has the required $n+2$ alternating extrema.

LetThe function $h$ is continuous, withThe intermediate value theorem gives $c\in [0,1]$ such that $h(c)=0$, and hence $f(c)=c$. This proves the fixed-point property of a closed interval.

Now suppose $0,1\in f([0,1])$. For any $y\in [0,1]$, choose $a,b\in [0,1]$ with $f(a)=0$ and $f(b)=1$. Applying the intermediate value theorem to $f$ on the interval with endpoints $a,b$ gives a point $x$ with $f(x)=y$. Thus $f$ is a surjective function.

Define the continuous functionAt the mesh points its values telescope:If one term is zero, the result follows immediately. Otherwise the terms cannot all have the same sign, so there are mesh points at which $h$ is positive and negative. The intermediate value theorem between those points gives an $x\in [0,(n-1)a]$ with $h(x)=0$. Thus the telescoping increment lemma yields

Let $f$ be continuous on $[0,1]$. Compactness makes it uniformly continuous. Given $\epsilon >0$, choose $\delta >0$ such thatChoose a partition with mesh smaller than $\delta$. The oscillation $M_j-m_j$ on every subinterval is then at most $\epsilon$, soThe Riemann integrability criterion proves that every continuous function on $[0,1]$ is Riemann integrable.

A function $f:(X,d_X)\to (Y,d_Y)$ is uniformly continuous if for every $\epsilon >0$ there is $\delta >0$ such thatfor all $x,x^{\prime }\in X$.

Suppose $f_n\to f$ uniformly and every $f_n$ is uniformly continuous. Given $\epsilon >0$, choose $N$ such thatfor every $x$. Uniform continuity of $f_N$ supplies $\delta >0$ such that $d_X(x,x^{\prime })<\delta$ impliesThe triangle inequality then gives $d_Y(f(x),f(x^{\prime }))<\epsilon$. Thus the uniform limit theorem for uniformly continuous functions proves that $f$ is uniformly continuous.

Pointwise convergence is insufficient. On $[0,1]$, the uniformly continuous functions $f_n(x)=x^n$ converge pointwise towhich is discontinuous and therefore not uniformly continuous.

A function $f:U\to \mathbb{C}$ is holomorphic when it is complex differentiable at every point of the open connected set $U$. Morera's theorem states that a continuous function on a domain is holomorphic if its integral around every triangle whose interior lies in the domain is zero.

The integrand $e^{tz}/(1+t^2)$ is continuous jointly in $(t,z)$ on $[0,1]\times \mathbb{C}$, so the displayed integral defines a continuous function. For any triangle $T$, Fubini's theorem and the Cauchy integral theorem giveMorera's theorem therefore proves that $f$ is entire.

The function $1/z$ is holomorphic on $\mathbb{C}\setminus \{0\}$ but has no antiderivative there, because an antiderivative would integrate to zero around every closed curve whereasThis is the standard period obstruction to a holomorphic antiderivative.

Let $E_n$ and $E$ be the projection-valued measures of $A_n$ and $A$. Using the projection $P_n:\mathcal{H}\to {\mathcal{H}}_n$, defineThe scalar spectral measures converge weakly whenfor every bounded continuous function $f$ and every $v,w\in \mathcal{H}$. By the spectral theorem for normal operators on a separable Hilbert space, this is equivalent to

The assumed moment identities say precisely that this convergence holds for every monomial $f(x)=x^m$. It follows by linearity for every polynomial. For $v=w$, the $m=2$ identity givesso Markov inequality makes the positive measures $({\mu }_v^{(n)})$ tight. Higher even moments similarly control the tails of any fixed polynomial.

Given a bounded continuous $f$ and $ϵ>0$, choose $R$ so that the measure tails are uniformly small. The Weierstrass approximation theorem supplies a polynomial $p$ withMoment convergence handles $p$; tightness and a sufficiently high even moment handle the two tails. Hence $\int fd{\mu }_v^{(n)}\to \int fd{\mu }_v$. The polarization identity then gives the same conclusion for ${\mu }_{v,w}^{(n)}$. This proves weak convergence of scalar spectral measures.

The assertion fails if only $m=1$ is assumed. Let $\mathcal{H}={\ell }^2(\mathbb{N})$, let ${\mathcal{H}}_n=\mathrm{span}\{e_1,\dots ,e_n\}$, take $A=0$, and letThe reversal matrices are self-adjoint unitaries. For fixed $v,w\in {\ell }^2$,because the finite head of one vector is paired with the vanishing tail of the other. Thus the $m=1$ condition holds. However, $A_n^2=I_{{\mathcal{H}}_n}$, soin general. Taking $f(x)=x^2$ shows that the spectral measures do not converge weakly.

The Hahn-Banach theorem says that if $Y$ is a linear subspace of a real normed space $X$ and $g:Y\to \mathbb{R}$ is bounded and linear, then there is a bounded linear extension $G:X\to \mathbb{R}$ with $\Vert G\Vert =\Vert g\Vert$.

For $x\in X$, the map $\hat{x}(f)=f(x)$ is linear on $X^{\prime }$ andso $\hat{x}\in X^{\prime \prime }$ and $\Vert \hat{x}\Vert \le \Vert x\Vert$. The assignment $Jx=\hat{x}$ is visibly linear. If $x\ne 0$, define $g$ on $\mathrm{span}\{x\}$ byIt has norm one, and Hahn--Banach extends it to an $f\in X^{\prime }$ of norm one with $f(x)=\Vert x\Vert$. ThereforeThus the canonical embedding into the bidual is an isometry and

On $C([0,1])$, point evaluation $\ell (f)=f(0)$ has norm one for the $L^{\infty }$ norm, because a continuous function's essential supremum equals its supremum. Hahn--Banach extends it to $\Lambda \in (L^{\infty }{)}^{\prime }$. If $\Lambda$ were represented by some $g\in L^1$, thenfor every continuous $f$. Choose continuous $0\le f_n\le 1$ with $f_n(0)=1$ and support in $[0,1/n]$. Absolute continuity of the integral makes the right side tend to zero, while the left side is always one, a contradiction. This singular functional on L infinity proves

The real Stone-Weierstrass theorem says that if $X$ is compact Hausdorff and $A\subset C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb{R})$.

Let $\overline{A}$ be the uniform closure. Polynomial approximation to $\sqrt{t}$ on a bounded interval shows that $|f|\in \overline{A}$ whenever $f\in \overline{A}$. Hence $\overline{A}$ is closed underGiven $h\in C(X)$, separation and the constants provide, for each $x,y$, a function in $A$ agreeing with $h$ at $x$ and $y$. Compactness first combines finitely many such functions by minima to obtain one that agrees at $x$ and lies below $h+\epsilon$ everywhere; a second finite cover and maxima produces $f\in \overline{A}$ with $|f-h|<\epsilon$. Thus $h\in \overline{A}$.

The space $C_b(\mathbb{R})$ is Banach because a uniform Cauchy sequence converges uniformly to a bounded continuous function. Compactness is essential: the algebra of bounded continuous functions having finite limits at both $+\infty$ and $-\infty$ contains constants and separates points, for example using $\tanh x$, but its uniform closure has the same limiting property. It cannot uniformly approximate $\sin x$, so it is not dense in $C_b(\mathbb{R})$.

Fix $z_0\notin \Gamma$ and put $\delta =\mathrm{dist}(z_0,\Gamma )>0$. If $|z-z_0|<\delta /2$, thenuniformly for $w\in \Gamma$. Multiplication by the bounded continuous function $f$ and integration therefore prove continuity of $F$ at $z_0$.

Take $\Gamma$ to be the positively oriented unit circle and $f\equiv 1$. By the Cauchy integral formula, $F(z)=1$ inside the circle and $F(z)=0$ outside it, so no continuous extension across $\Gamma$ exists.

For the final assertion, cover $K$ by finitely many sufficiently small closed axis-parallel squares whose slightly enlarged squares lie in $\Omega$, choosing the grid so that no boundary meets $K$. Apply the square Cauchy formula to $f$ on every selected square and add the results. Integrals over shared edges cancel with opposite orientations. The remaining finitely many oriented boundary polygons can be split into contours ${\Gamma }_j\subset \Omega \setminus K$, and their sum isfor every $z\in K$.

Pointwise, $\tanh (nx)$ tends to $-1,0,1$ according as $x<0,x=0,x>0$. This limit is discontinuous, whereas a uniform limit of continuous functions is continuous. Convergence is therefore not uniform.

For a harmonic function $u$, letThe divergence theorem givesSince $M(r)\to u(x)$ as $r\downarrow 0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the mean value property for harmonic functions. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. Continuity makes $u$ constant on those spheres, and connectedness propagates that value through the domain. Thus the weak maximum principle for elliptic operators gives

For the derivative estimate, choose $r>0$ smaller than half the distance from ${\Omega }^{\prime }$ to $\partial \Omega$, and let ${\rho }_r$ be a smooth radial mollifier supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u\ast {\rho }_r=u$ on ${\Omega }^{\prime }$. Hence, for every multi-index $\alpha$,so Holder inequality gives

Suppose $u_1$ and $u_2$ are weak solutions with the same trace, and put $w=u_1-u_2\in H_0^1(\Omega )$. The weak formulation permits $w$ itself as a test function, givingThus $w$ is almost everywhere constant, and its zero trace makes that constant zero. This proves uniqueness.

The weak identity also says that $\Delta u=0$ in the sense of distributions. The Weyl lemma therefore gives $u\in C^{\infty }(\Omega )$ and $\Delta u=0$ pointwise. The assumed continuity on $\overline{\Omega }$ retains the prescribed boundary values, so the weak solution is the unique classical solution in $C^{\infty }(\Omega )\cap C^0(\overline{\Omega })$.

Fix a closed ball $\overline{B_R(a)}\subset \Omega$, and let $v$ solve the Dirichlet problem for the Laplace equation in this ball with boundary data $u$. Put $w=u-v$. The function $w$ is continuous, vanishes on the boundary, and inherits the restricted spherical mean identity because $v$ has the full mean value property for harmonic functions.

Suppose $M={\max }_{\overline{B_R(a)}}w>0$. Its maximum set $E$ is a nonempty compact subset of the open ball. Choose $x\in E$ maximizing $|x-a|$. For every sufficiently small radius in the sequence attached to $x$,Equality of the average with the maximum and continuity imply that the whole sphere belongs to $E$. Its point in the direction from $a$ through $x$ lies farther from $a$ than $x$ does; if $x=a$, any point on the sphere does. Both cases contradict the choice of $x$. Hence $w\le 0$, and applying the same argument to $-w$ gives $w=0$.

For $x,y\in X$, the triangle inequality and the fact that $\varphi$ is an isometry giveThus the displacement function is $2$-Lipschitz, hence continuous.