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A locally compact Hausdorff space is regular. Cover $K$ by finitely many open sets whose compact closures lie in $U$; their union $V$ satisfies $K\subset V\subset \overline{V}\subset U$, with $\overline{V}$ compact. The compact Hausdorff space $\overline{V}$ is normal, so the Urysohn lemma gives a continuous function $g:\overline{V}\to [0,1]$ equal to $1$ on $K$ and $0$ on $\overline{V}\setminus V$. Extend $g$ by zero outside $\overline{V}$. The boundary values make the extension continuous, and its support is a compact subset of $U$.

For the sum $F=f+g$:

For the product $G=fg$, the answer is no in both cases:

Every displayed $f$ and $g$ is a continuous function and is nonzero at points arbitrarily close to $a$, so the extra condition in the question is satisfied.

An $m$ by $n$ matrix has matrix rank $m$ exactly when at least one of its $m$ by $m$ minors has nonzero determinant. For each multi-index $I$, the setis open in the vector space of all real $m$ by $n$ matrices, because the determinant is continuous. Their union is $X_{m,n}$, so $X_{m,n}$ is itself open in ${\mathbb{R}}^{mn}$. It therefore inherits the standard smooth manifold structure and has dimension $mn$. This is the full-row-rank matrix manifold.

On the unit ball, part 1 controls the $L^2$ norm. Outside it, $|x{|}^{\alpha }\ge 1$, soHence $\Vert u{\Vert }_2\le C\Vert u{\Vert }_{\Sigma }$, proving that the embedding $\Sigma \hookrightarrow L^2$ is continuous.

For compactness, let $(u_n)$ be bounded in $\Sigma$. On each ball, it is bounded in $H^1$, so the Rellich-Kondrachov compactness theorem gives a subsequence convergent in local $L^2$. The tail estimateis uniform in $n$ and tends to zero as $R\to \infty$. A diagonal argument therefore gives convergence in all of $L^2$. This is the compact embedding of a confining-potential energy space.

Let $K$ contain the support of $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose a cutoff function that equals one near $K$. It makeswell-defined, and differentiating the parameter under the pairing givesThus the Fourier transform of a compactly supported distribution is a smooth function. Since a compactly supported distribution has finite order, some $N$ and $C$ satisfyApplying this estimate to the exponential yields $|\hat{u}(\lambda )|\le C^{\prime }⟨\lambda {⟩}^N$.

Now take $v\in {\mathcal{E}}^{\prime }(X)$, multiply by a cutoff function supported in $X$ and equal to one near $\mathrm{supp}v$, and regard the result as an element of ${\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose $m$ so large thatis Lebesgue integrable. The inverse Fourier transform $f={\mathcal{F}}^{-1}g$ is a bounded continuous function, and the Fourier transform of a derivative givesas a distributional identity. This is the Bessel potential proof of the structure theorem for compactly supported distributions.

The function $f$ itself need not have compact support. Choose another cutoff $\rho \in \mathcal{D}(X)$ equal to one near $\mathrm{supp}v$. Then $v=\rho (1-\Delta {)}^{m}f$. Repeatedly usingexpresses $v$ as a finite sum ${\sum }_{\beta }{\partial }^{\beta }{f}_{\beta }$, where every coefficient $f_{\beta }$ is continuous and compactly supported in $X$.

Applying the distributional Leibniz rule twice givesEliminating the middle term provesChoose $\rho \in \mathcal{D}(\mathbb{R})$ with $\rho (0)=1$. Parts i and ii and the identity just proved, with $\phi =\rho$ and $u=xH$, yieldThus explicit continuous functions of compact support are

Changing variables givesThe average of a continuous function over $[y,y+h]$ tends to $f(y)$ as $h\to 0$. The two terms therefore tend to $f(x)$ and $f(a)$, proving the stated limit.

Fix a continuous function $f$ and $\epsilon >0$. By the Weierstrass approximation theorem, choose a polynomial $p\in {\mathcal{P}}_m$ such thatFor every $n\ge m$, exactness gives $I_n(p)=I(p)$. Positivity of the quadrature weights and exactness on the constant polynomial giveConsequentlySince $\epsilon$ is arbitrary, $I_n(f)\to I(f)$. This is the convergence of positive quadrature rules.

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.