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No: the unbounded length of the domain matters. Let $\varphi (t)=\max \{1-|t|,0\}$ and define $f_n(x)=n^{-1}\varphi (x/n)$ on $\mathbb{R}$. These are continuous functions with compact support, and ${\sup }_{\mathbb{R}}|f_n|=1/n\to 0$, so the sequence has uniform convergence to zero. However, the change of variables $t=x/n$ givesThe integral of the limit is zero. The supports widen as the heights decrease; a uniform bound on the heights gives no bound on the total area over an infinite interval.

Let $f_n$ be the continuous triangular bumpand set $f_1=0$. At $x=0$ every term is zero, and each fixed $x>0$ eventually lies outside the support, so $f_n\to 0$ pointwise. The triangle has base $2/n$ and height $n$, henceThis concentrating triangular bump shows that pointwise convergence does not preserve integrals, even for continuous functions on a compact interval.

The integrand is a continuous function, so the fundamental theorem of calculus differentiates the upper limit directly:No explicit antiderivative is needed.

Let $p(x)=x^3-5x-1$. Its values at useful points areSince a polynomial is a continuous function, the intermediate value theorem gives one polynomial root in each of the disjoint intervals $(-3,-2)$, $(-1,0)$ and $(2,3)$. A nonzero cubic polynomial has at most three distinct polynomial roots. Hence it hasThese three distinct polynomial roots already exhaust its degree of a polynomial, so each has multiplicity of a root one.

For $a=b=1$, the stationary points found in parts (a) and (b) have temperatures $0$, $1$, $1/4$ and $1/5$. The unit sphere is compact and $T$ is a continuous function, so the extreme value theorem ensures that a hottest point exists; it must be one of those stationary points. ThusFor a direct bound, put $s=x^2\in [0,1]$. The identity $(y^2-z^2{)}^2\ge 0$ gives $y^2{z}^2\le (y^2+z^2{)}^2/4=(1-s{)}^2/4$, andsince $1-s^2-(1-s{)}^2/4=(1-s)(5s+3)/4\ge 0$. Equality forces $s=1$, giving exactly the two points above.

By the Heine-Borel theorem, a noncompact subset $X\subseteq \mathbb{R}$ is either unbounded or not closed. If $X$ is unbounded, the restriction $f(x)=x$ is a continuous function with unbounded range.

Otherwise $X$ is bounded and not closed. Choose $a\in \overline{X}\setminus X$, and defineIts denominator is nonzero on $X$, so it is continuous. Since $a$ is in the closure, for every $n$ there is $x_n\in X$ with $0<|x_n-a|<1/n$. Hence $f(x_n)>n$, proving that this continuous function is unbounded. These two cases establish unbounded continuous witnesses to noncompactness of a real subspace.

We can arrange that neither the supremum nor the infimum is attained. By part (d), choose a continuous function $f$ on $X$ with unbounded range, and put $u=|f|$. Choose $x_n\in X$ so that $t_n=u(x_n)$ increases strictly to infinity. Construct a continuous function $h:[0,\infty )\to \mathbb{R}$ withkeep it constant up to $t_1$ and interpolate linearly on each $[t_n,t_{n+1}]$. Since $t_n\to \infty$, the joining points have no finite accumulation, so this defines a continuous function everywhere on its domain.

Now setBy function composition, $g$ is continuous, and $-1<g(x)<1$ for every $x$. Along the even and odd chosen points,Thus ${\sup }_{X}g=1$ and ${\inf }_{X}g=-1$, and neither is a value of $g$. This proves the bounded continuous witness with neither extremum attained construction, including for subsets that are bounded but not closed.

Define $g(p)=f(p)-f(-p)$ on the sphere. It is continuous and satisfies $g(-p)=-g(p)$. If $g(p)=0$ at a chosen point, we are done. Otherwise choose a unit vector $q$ perpendicular to $p$ and follow the half great circleThe Pythagorean theorem shows that $\gamma (t)$ lies on the sphere, and its endpoints are $p,-p$. The continuous function $g\circ \gamma$ has opposite signs at the endpoints. The Intermediate value theorem therefore gives a zero $t_{\ast }$, soThis is a scalar equal-value assertion for an antipodal pair; it uses only a path and the Intermediate value theorem.

Expanding the characteristic polynomial givesTherefore $a=-2z$ and $b=z^2-x^2-y^2$. Both are polynomial functions of the three real coordinates, so they are continuous functions, including on the open subset $X$ of invertible matrices.

On $X$, $b=\det A\ne 0$. Its image contains positive values, for example $b(I)=1$, and negative ones, for example $b(\mathrm{diag}(1,-1))=-1$. If $X$ were connected, its continuous real image would be an interval containing both signs and hence zero. That is impossible. Thus $X$ is not connected. More precisely its determinant image is $\mathbb{R}\setminus \{0\}$: take $x=y=0$ for a positive value and $z=y=0$ for a negative one.

Morera's theorem states that a continuous function on a domain is holomorphic if its contour integral around every triangle whose filled-in interior lies in the domain is zero.

The uniform limit $f$ is a continuous function, by the uniform limit theorem. For any such triangle $T$, the estimation lemma givesEach integral of $f_n$ vanishes by the Cauchy integral theorem. Hence the integral of $f$ vanishes too, and Morera's theorem proves $f$ is holomorphic on $D$. This is the basic locally uniform convergence of holomorphic functions argument; the given uniform convergence on all of $D$ is more than needed.

The answers, in the order completeness, sequential compactness, connectedness, and path connectedness, areDefine $F(0)=0$ and $F(x)=\sqrt{x}\sin (1/x)$ for $x>0$. The estimate $|F(x)|\le \sqrt{x}$ proves continuity at zero by the squeeze theorem; elsewhere it follows from the usual continuous functions. Thus $X$ is the graph of a function $F$ on the closed set $[0,\infty )$.

To check closedness directly, if $(x_n,F(x_n))\to (x,y)$ in the Euclidean plane, then $x\ge 0$ and continuity gives $y=F(x)$. Therefore $X$ is a closed subset of the complete metric space ${\mathbb{R}}^2$, hence is a complete metric space. The sequence $(n,F(n))$ has no convergent subsequence, since its first Cartesian coordinates tends to infinity, so sequential compactness fails.

For any two points $(u,F(u)),(v,F(v))$, the mapis a continuous path in $X$ joining them. Thus $X$ is path-connected, and consequently connected. The factor $\sqrt{x}$ is crucial: the oscillations shrink to the single point $(0,0)$.

The answers areWrite $M_n(f,g)={\sup }_{0\le t\le n}|f(t)-g(t)|$. The given metric describes locally uniform convergence: $d(f_j,f)\to 0$ if and only if $M_n(f_j,f)\to 0$ for every fixed $n$. Indeed,so convergence in $d$ forces $M_n\to 0$. Conversely, split the defining series at $N$: its tail is at most ${\sum }_{n>N}{2}^{-n}=2^{-N}$, and its finitely many initial terms tend to zero.

If $(f_j)$ is a Cauchy sequence in $d$, the same single-term estimate makes it a Cauchy sequence in the supremum norm on every $[0,n]$. The completeness of the continuous-map space in the uniform metric gives a continuous function $F_n$ on $[0,n]$ with $f_j\to F_n$ uniformly there. Uniqueness of the limit of a sequence gives $F_{n+1}{|}_{[0,n]}=F_n$, so these functions define a single continuous function $f$ on $[0,\infty )$. The preceding finite-head and tail argument proves $d(f_j,f)\to 0$. Thus the space is complete.

The constant functions $f_j(t)=j$ have $d(f_j,f_k)=1$ whenever $j\ne k$. They have no convergent subsequence, so the space is not sequentially compact.

For fixed $f,g$, set $\gamma (s)=(1-s)f+sg$. On $[0,n]$,as $s^{\prime }\to s$. The equivalence with locally uniform convergence proves that $\gamma$ is a continuous path for $d$. Hence the space is path-connected and connected.

A linear spline is a continuous function whose restriction to every interval $[x_{i-1},x_i]$ is a polynomial of polynomial degree at most one. With $h_i=x_i-x_{i-1}$ and ${\ell }_i=\ell (x_i)$, linear interpolation givesIntegrating the two linear polynomials and adding the contributions givesThis is the composite trapezoidal rule applied to the nodal values, and it is exact for the linear spline itself, including nonuniform spline knots.

A cubic spline on this partition of an interval is a function $S\in C^2[x_0,x_n]$ that restricts to a polynomial of polynomial degree at most three on every interval $[x_{i-1},x_i]$. Thus its second derivative is continuous and piecewise of polynomial degree at most one: $S^{\prime \prime }$ is a linear spline on the same spline knots. Its third derivative is constant on each open interval and can jump across the spline knots; its fourth derivative vanishes inside each open interval.

Let $r=f-p_n$ be the continuous function describing the error. The preceding orthogonality equations sayIf $r$ vanishes identically, any $n+1$ distinct points work. If $r$ already has at least $n+1$ distinct zeros, the assertion also follows. Suppose instead that it has at most $n$ distinct zeros. Among its interior zeros, list those at which its sign changes as $t_1<\cdots <t_m$, where $m\le n$, and define the polynomialwith $q=1$ if $m=0$. Between consecutive zeros of $r$, its sign is constant by the intermediate value theorem. At each $t_j$, both $r$ and $q$ change sign; at every other zero of $r$, neither sign changes. Hence $rq$ has one fixed sign throughout $[-1,1]$ wherever it is nonzero. Replacing $q$ by $-q$ if necessary makes $rq\ge 0$ everywhere.

Since $r$ is not identically zero and $q$ has only finitely many zeros, $rq$ is strictly positive on some open interval. Positivity of $w$ givescontradicting orthogonality because $q\in {\mathcal{P}}_n$. ThusThis is the interpolation property of weighted least-squares polynomial approximation. It supplies interpolation nodes depending on $f$ and $w$, rather than nodes prescribed in advance.

Write ${\mathcal{P}}_n$ for the vector space of polynomials of polynomial degree at most $n$, and put $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$. As the question calls the weighted integral an inner product, its finiteness is implicit; for example, a positive integrable weight $w$ suffices for all the continuous functions involved here.

For every $q\in {\mathcal{P}}_n$, differentiating the squared weighted norm $\Vert f-p_n-tq{\Vert }_w^2$ at its minimum $t=0$ gives the orthogonality equationsIn particular, the requested linear system isIts Gram matrix $G_{kj}={\int }_{-1}^{1}w(x)x^{k+j}dx$ is a symmetric positive-definite matrix: for any nonzero vector $a$,Indeed a nonzero polynomial is nonzero on some open interval, and the weight is positive. Therefore $G$ is an invertible matrix, proving uniqueness. Conversely, if these equations hold, the Pythagorean theorem in an inner-product space gives, for every $q\in {\mathcal{P}}_n$,so the solution really is the unique best approximation.

When $a=0$, the explicit amplification factor becomesA single Fourier mode has constant modulus in space and is not in the l2 sequence space. Instead choose a whole interval of frequencies on which there is definite growth. LetThe inverse Fourier transform here gives the real sequenceBy the Parseval identity, its squared norm is $\Vert U^0{\Vert }_{{\ell }_2}^2=\Delta x|E|/(2\pi )=\Delta x/2$, finite and strictly positive. These data are also compatible with sampling a bounded continuous function: useThe value at zero is the limit of the displayed quotient, and $u_0(j\Delta x)=U_j^0$.

On $E$ we have ${\sin }^2\theta \ge 1/2$. Therefore the transformed solution and the Parseval identity giveFor each fixed pair $\Delta x,\Delta t>0$, takeTaking square roots proves $\Vert U^m{\Vert }_{{\ell }_2}\ge (1+c{)}^m\Vert U^0{\Vert }_{{\ell }_2}$ for every $m\ge 1$. In particular the norm already increases at the first step, so no positive mesh pair gives practical stability in the contraction sense of (b).

The native PDF places the existence of $c>0$ before the universal quantifier over mesh pairs. If this means a single positive $c$ independent of $\Delta x,\Delta t$, that assertion is false for nonzero data. Indeed, for every datum the same Fourier symbol givesFor any proposed fixed $c>0$, choose $0<\nu <\sqrt{(1+c{)}^2-1}$. The upper bound then contradicts the requested lower bound at $m=1$. The correct quantifiers are $\forall \Delta x,\Delta t>0\exists c=c(\Delta x,\Delta t)>0$, with nonzero square-summable data as constructed above.

Finally, failure of this stronger practical stability of a finite difference scheme should not be confused with every possible refinement notion of numerical stability. If $\Delta t/(\Delta x{)}^2\le K$, then $m\Delta t\le T$ givesThus a finite-time mesh-uniform bound can hold under this severe refinement even though contraction fails on every positive mesh. This is the distinction recorded in centered-advection refinement-dependent stability.

Use the nonnegative increasing truncations $f_n=x^{-2}{1}_{[1/n,1]}$. They increase almost everywhere to the specified function. On each compact interval the continuous function has equal Riemann and Lebesgue integrals, so the fundamental theorem of calculus givesMonotone convergence yields $\int f=+\infty$. Since a nonnegative function is Lebesgue integrable exactly when its integral is finite, this function is not Lebesgue integrable.

False in both cases. Without the zero-at-zero condition, $g=0$ and $f\equiv 1$ already give an integrable input and a non-integrable composition on $\mathbb{R}$.

Even imposing $f(0)=0$ does not control how fast $f(t)$ approaches zero. Take $g(x)=(1+x^2{)}^{-1}$ and the continuous function $f(t)=\sqrt{|t|}$. The input has integral $\pi$, buthas a divergent integral, because its tails are comparable to $1/|x|$. Thus continuity at zero is weaker than the linear growth estimate that would preserve integrability.

A connected space has no expression as the union of two disjoint nonempty open subsets. A continuous surjection to the two-point discrete space would pull back its singleton sets to such a separation. Conversely, a separation $X=U\bigsqcup V$ defines a continuous surjection by taking the values $0$ on $U$ and $1$ on $V$. Thus these two conditions are equivalent, including for the empty space.

If a subset $S\subseteq \mathbb{R}$ contains $a<c$ but omits some $b$ between them, its intersections with $(-\infty ,b)$ and $(b,\infty )$ separate it. Hence every nonempty connected subset is an interval. Conversely, let an interval $I$ have a separation and choose $a<b$ in its opposite sides. Restrict the associated two-valued continuous function to $[a,b]\subseteq I$ and regard it as a real-valued function. The intermediate value theorem forces the value $1/2$, a contradiction. Singletons and the empty set are connected directly.

For the union assertion, let $u:X\to \{0,1\}$ be continuous. Each restriction $u{|}_{A_i}$ is constant because $A_i$ is connected. Since every pair of the $A_i$ intersects, these constants agree, so $u$ is constant on $X$. There is no continuous surjection, and $X$ is connected. The empty indexing case gives $X=\varnothing$ and is harmless.

To apply this to the plane, for each $t\in \mathbb{R}$ take the union of the horizontal line at height $t$ with the vertical axis. Its two lines are connected and intersect; all such unions share the vertical axis and cover ${\mathbb{R}}^2$. Thus the plane is connected. For the punctured plane, take, for every $r>0$, the circle of radius $r$ together with the positive horizontal ray. The circle is the continuous image of $[0,2\pi ]$, and intersects that connected ray. These unions share the ray and cover ${\mathbb{R}}^2\setminus \{0\}$, so the punctured plane is connected too. Translation gives the same conclusion after removing any one point.

Finally, suppose $g:{\mathbb{R}}^2\to \mathbb{R}$ were a continuous injection. Its image $I$ is a connected interval with at least three points. Choose an interior point $c$ of $I$ and its unique preimage $p$. The restriction of $g$ to ${\mathbb{R}}^2\setminus \{p\}$ is a continuous surjection onto $I\setminus \{c\}$. Its domain is connected, whereas its image is separated by the two half-lines on either side of $c$. This contradicts preservation of connectedness under continuous images. Therefore no continuous injection of the plane into the line exists.

A metric space is sequentially compact if every sequence has a convergent subsequence whose limit belongs to the space. A limit point $\ell$ of $S$ is a point for which every open ball about $\ell$ contains a point of $S$ other than $\ell$.

Let $(x_n)$ lie in a compact metric space. If one value occurs infinitely often, take a constant subsequence. Otherwise its set of values $S$ is infinite. If $S$ had no limit point, every point of the space would have a neighborhood containing at most one member of $S$. A finite subcover of these neighborhoods would force $S$ finite, a contradiction. Thus $S$ has a limit point; the stated subsequence result supplies a convergent subsequence. This proves sequential compactness.

Assume now that $X$ is nonempty, as is necessary for the requested minimum and fixed point. The function $r(x)=d(f(x),x)$ is continuous, as a composition of $x\mapsto (f(x),x)$ with the metric. The extreme value theorem on a compact space implies that $r$ attains its infimum, say at $p$. If $f(p)\ne p$, strict decrease of distances givescontradicting minimality. Thus $f(p)=p$. If $p,q$ were distinct fixed points, then $d(p,q)=d(f(p),f(q))<d(p,q)$, another contradiction. Hence there is exactly one fixed point. This is the fixed point theorem for a strictly distance-decreasing map on a compact metric space.

Compactness is essential. On $\mathbb{R}$, takeHere $f(x)-x=\log (1+e^{-x})>0$, so there is no fixed point. Nevertheless $0<f^{\prime }(x)=e^x/(1+e^x)<1$ everywhere, and the mean value theorem gives $|f(x)-f(y)|<|x-y|$ for distinct $x,y$. There is no uniform Lipschitz constant less than $1$, since $f^{\prime }(x)\to 1$ as $x\to \infty$.