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Chebyshev's equal-ripple criterion says that a polynomial $q$ of degree at most $m$ is a best uniform approximation to a continuous function $F$ if and only if the error $F-q$ attains alternating extrema of maximum magnitude at $m+2$ ordered points.

Since $T_n$ has leading coefficient $2^{n-1}$, the polynomial $2^{1-n}{T}_n$ is monic and alternates between $\pm 2^{1-n}$ at $t_j=\cos (j\pi /n)$. Therefore

At each $t_j$, both $T_n-f$ and $T_n+f$ have sign $(-1{)}^j$, because $|f(t_j)|<1$. Each has a root in every interval between consecutive extrema, hence all its at most $n$ roots lie in $(-1,1)$. For $t>1$ both polynomials remain positive, giving $|f(t)|<T_n(t)$. For $t<-1$ both have sign $(-1{)}^n$, giving $|f(t)|<|T_n(t)|$. On $[-1,1]$ the hypothesis gives $|f|<1$. Thus

False. Continuous functions can converge pointwise to the non-Riemann-integrable Dirichlet function; for example enumerate the rationals and use continuous narrow bumps whose first $n$ peaks cover the first $n$ rationals while widths shrink suitably pointwise off $\mathbb{Q}$.

A standard Brownian motion $(B_t{)}_{t\ge 0}$ satisfies $B_0=0$ almost surely, has almost surely continuous sample paths, and has independent stationary increments such that

The evaluation map $f\mapsto f(1)$ is continuous on $C[0,1]$ in the uniform norm. The continuous mapping theorem applied to part (c) therefore givesThis recovers the central limit theorem from the functional version.

For the level-$N$ dyadic partition, writeThe dyadic partitions are nested, so the triangle inequality makes $V_N(t)$ nondecreasing in $N$, and $\Vert f_t\Vert ={\lim }_N{V}_N(t)$.

Fix $t_1<t_2$. As the mesh tends to zero, the last dyadic point before $t_1$ approaches $t_1$. Refining from there to $t_2$, the triangle inequality says that the added variation is at least $|f(t_2)-f(t_1)|$ minus the two endpoint errors, which tend to zero by continuity. ThereforeSince $\Vert f\Vert <\infty$, both limits are finite and may be subtracted, giving

No. The point evaluations ${\delta }_x$ satisfy $\Vert {\delta }_x-{\delta }_y\Vert =2$ for $x\ne y$ (choose a continuous function taking values $1,-1$ at the two points). Thus $C[0,1{]}^{\ast }$ is nonseparable, whereas ${\ell }_1$ is separable.

Suppose the continuous function $f$ did not have uniform continuity on $[a,b]$. Negating its definition gives ${\epsilon }_0>0$ such that for every $n\ge 1$ there are $x_n,y_n\in [a,b]$ withThe Bolzano-Weierstrass theorem, which states that every bounded sequence of real numbers has a convergent subsequence, gives $x_{n_k}\to p$. The inequalities $a\le x_{n_k}\le b$ imply $p\in [a,b]$. By the triangle inequality,Since continuous functions preserve convergent sequences, both $f(x_{n_k})$ and $f(y_{n_k})$ tend to $f(p)$. Their difference therefore tends to zero, contradicting its absolute value being at least ${\epsilon }_0$. This proves the Heine-Cantor theorem in the required setting: $f$ has uniform continuity on $[a,b]$.

Choose $h_n=\min \{1/n,(b-c)/2\}>0$. Then $h_n\to 0$ and $[c,c+h_n]\subset (a,b)$. The mean value theorem states that a continuous function on a closed interval, differentiable in its interior, has an interior derivative equal to its endpoint difference quotient. It gives $x_n\in (c,c+h_n)$ withHere $x_n\ne c$, $x_n\to c$, and the right-hand side tends to $f^{\prime }(c)$ by differentiability at $c$. If $L={\lim }_{x\to c}{f}^{\prime }(x)$ exists, the same values $f^{\prime }(x_n)$ must tend to $L$. Uniqueness of the limit of a sequence givesThus a derivative cannot have a removable discontinuity at a point where the original function is differentiable. The proof also excludes an infinite limit of a function there, since the displayed difference quotients have a finite limit of a sequence.

The continuous function $|h|$ attains a finite maximum $M$ on $[a,b]$ by the extreme value theorem. Additivity and the triangle inequality for the Riemann integral imply, for any $x,y\in [a,b]$,For reversed endpoints, reverse the orientation of the integral before taking its absolute value. Given $\epsilon >0$, choose $\delta =\epsilon /(M+1)$. Then $|x-y|<\delta$ implies $|H(x)-H(y)|<\epsilon$. Thus $H$ is continuous on $[a,b]$, including one-sided continuity at the endpoints; it is even Lipschitz continuous. If $a=b$, the function is defined on one point and the conclusion is immediate.

Write $G={\int }_a^{b}g(t)dt\ge 0$ and $I={\int }_a^{b}f(t)g(t)dt$. By the extreme value theorem, $f$ attains a minimum $m$ and a maximum $M$ on $[a,b]$. Because $g\ge 0$, the monotonicity of the Riemann integral givesIf $G=0$, this forces $I=0$, and every $c\in [a,b]$ satisfies the required identity. This includes $a=b$.

If $G>0$, divide to obtain $m\le I/G\le M$. The intermediate value theorem, applied between points where the continuous function $f$ attains $m$ and $M$, supplies $c\in [a,b]$ with $f(c)=I/G$. Hence the weighted mean value theorem for integrals gives

The original double integral covers the triangle $0\le y\le x\le 1$. By Fubini's theorem, after reversing the order, $y$ runs from $0$ to $1$ and $x$ from $y$ to $1$. Since a continuous function is integrable on this bounded region,For the square pyramid, a horizontal section at height $z$ is a square of side $2(1-z)$: each vertex moves linearly from the base towards $(1,1,1)$. Hence its cross-sectional area is $4(1-z{)}^2$, and its volume is $4{\int }_0^1(1-z{)}^{2}dz$. To use the identity just proved, choose $f(y)=1-y$:

If $f$ is a continuous function, the intermediate value theorem ensures that every value between $f(a)$ and $f(b)$ is attained. Monotonicity excludes values outside that range, so $f([a,b])=[f(a),f(b)]$.

Conversely, suppose this equality of sets holds. At an interior point, if $f^{-}(x_0)<f(x_0)$, any $y$ strictly between these two numbers lies in $[f(a),f(b)]$. Yet inputs below $x_0$ have values at most $f^{-}(x_0)$ and inputs at or above $x_0$ have values at least $f(x_0)$, so no input attains $y$. This contradicts surjectivity. The same argument excludes $f(x_0)<f^{+}(x_0)$, so part (b) gives continuity at every interior point.

At $a$, a strict gap $f(a)<f^{+}(a)$ likewise leaves every value between them unattained; hence $f^{+}(a)=f(a)$, giving right continuity. At $b$, the corresponding argument gives $f^{-}(b)=f(b)$ and left continuity. These are the appropriate endpoint notions on $[a,b]$. Hence

With the strict monotonicity assumed at the start of the question, parts (a) and (c) give a strictly increasing inverse functionIts image of a function is exactly the interval between its endpoint values. Apply part (c) to $g$, with domain $[f(a),f(b)]$, to conclude that $g$ is a continuous function, including at both endpoints.

If the word “increasing” in this subpart were interpreted as merely nondecreasing, an inverse function need not exist: a constant function is a counterexample. The preceding strict hypothesis supplies the required injectivity.

No. Let $E=\mathbb{R}$, $x_0=0$ andThe supremum norm satisfies $\Vert f_n-f{\Vert }_{\infty }=1/n\to 0$, so there is uniform convergence to the continuous function $f$. But every $f_n$ is discontinuous at zero: $f_n(0)=1/n$ whereas $f_n(x)=0$ for every $x\ne 0$. No tail consists of functions continuous at $x_0$.

By the Heine-Cantor theorem, a continuous function on the compact interval $[a,b]$ is uniformly continuous. Given $\epsilon >0$, choose $\delta >0$ such that $|f(s)-f(t)|<\epsilon$ whenever $s,t\in [a,b]$ and $|s-t|<\delta$.

Choose a finite partition of an interval $a=t_0<t_1<\cdots <t_m=b$ with every width below $\delta$. Define the step functionFor each $t<b$, its distance from the chosen left endpoint is below $\delta$, so $|f(t)-\varphi (t)|<\epsilon$; at $b$ the error is zero. HenceA step function may have separately assigned values at its finitely many partition points; these values do not affect its Riemann integral. If $a=b$, simply take $\varphi =f$ on that singleton.

Let $F=x+y+z^2$ and $g=x^2+y^2+z^2$. The unit sphere is compact, so the continuous function $F$ has a global maximum and global minimum. Since $\nabla g\ne 0$ on the unit sphere, all such points satisfy the Lagrange multiplier equationsThus $x=y$ and either $z=0$ or $\lambda =1$.

If $z=0$, then $x=y=\pm 1/\sqrt{2}$, giving $F=\pm \sqrt{2}$. If $\lambda =1$, then $x=y=1/2$ and $z=\pm 1/\sqrt{2}$, giving $F=3/2$. These exhaust the candidates, so

If there were no interior $x_2$ with $f^s(x_2)\le 0$, the symmetric derivative would be positive everywhere in $(a,b)$. Part ii would make $f$ strictly increasing there. Choose $a<u<v<b$. Continuity at the endpoints and monotonicity on the interior implycontradicting $f(a)=f(b)$. Hence such an $x_2$ exists.

If there were no interior $x_1$ with $f^s(x_1)\ge 0$, then $(-f{)}^s=-f^s>0$ everywhere. Apply the same argument to the continuous function $-f$, whose endpoint values are equal. This is again impossible. ThusThe two points need not be distinct.

For $n=1$, choose $x=0$, $y=1$. For $n\ge 2$, define the continuous functionIts values at the grid points have the telescoping sumIf one summand is zero, we already have a zero of $g$. Otherwise the summands cannot all have the same sign, so two of these values have opposite signs. The intermediate value theorem between their arguments gives $t\in [0,1-1/n]$ with $g(t)=0$. Set $x=t$, $y=t+1/n$. ThenThis is the horizontal chord of reciprocal-integer length property; the telescoping sum explains why reciprocal integers appear.

Rolle's theorem states that a continuous function $g:[u,v]\to \mathbb{R}$ with $u<v$, differentiable on $(u,v)$, and satisfying $g(u)=g(v)$ has $g^{\prime }(c)=0$ for some $c\in (u,v)$.

Fix $x>0$ and set $r=(\sqrt{1+x}-1)/x$. The functionis continuous and differentiable as required, and $g(0)=g(x)=0$. By Rolle's theorem, some $c\in (0,x)$ satisfiesSince $c>0$, this gives $r<1/2$. Multiplying by $x>0$ and adding one proves the strict square root bound

Rearrange the polynomial equation as a fixed point equation:First establish existence and uniqueness of the target root of a polynomial. Since $f$ is a continuous function, $f(0)=-1<0$ and $f(1/7)=1/7^5>0$, the intermediate value theorem gives a root of a polynomial $\xi \in (0,1/7)$. Moreover $f^{\prime }(x)=5x^4+7>0$ for all real $x$, so $f$ is strictly increasing and this is its only real root of a polynomial.

For $x\in [0,1/7]$,so $g$ maps the closed interval into itself. AlsoThe mean value theorem therefore gives $|g(x)-g(y)|\le q|x-y|$ on the closed interval: $g$ is a contraction mapping. For any starting value $x_0\in [0,1/7]$, all iterates remain in that closed interval, and since $g(\xi )=\xi$,Thus the fixed-point iteration converges to the unique real root of a polynomial from every $x_0\in [0,1/7]$. This explicit error estimate proves the needed convergence directly.