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For secant domination for expected utility derivatives, at each sample value $x$ set $g_x(\theta )=U(\theta x)$. This is a differentiable concave function, regardless of the sign of $x$. Fix ${\theta }_0$ and $\delta >0$. For $\theta \in [{\theta }_0-\delta ,{\theta }_0+\delta ]$, concavity bounds its derivative between the two outer secant slopes:Because $U<0$, finiteness of $F$ means $\mathbb{E}|U(\eta X)|<\infty$ at each of these four endpoints. Their absolute values divided by $\delta$ therefore give a common integrable bound on $|g_X^{\prime }(\theta )|$. Difference quotients satisfy the same bound by the mean value theorem. The dominated convergence theorem permits differentiation under the expectation, and pointwise continuity of $X{U}^{\prime }(\theta X)$ with the same bound gives continuity of the derivative. Hence $\boxed{F^{\prime }(\theta )=\mathbb{E}[X{U}^{\prime }(\theta X)]}$ and $F\in C^1(\mathbb{R})$. This argument proves absolute integrability of the displayed derivative; no unjustified differentiation assumption is needed.

For $0<r<1$, $r\log r<0$. The printed upper bound is negative whenever $\Omega \ne 0$, so it cannot bound a nonnegative difference. The correct small-distance modulus is $r\log (1/r)$, and the radius in the hint must also be positive.

Write $h=|x_1-x_2|\le e^{-1}$, $x_3=(x_1+x_2)/2$, and $M=h\log (1/h)$. Then $h\le M\le e^{-1}$. Denote the planar vorticity velocity kernel by $K$. It satisfies $|K(z)|\le C/|z|$ and $|DK(z)|\le C/|z{|}^2$. Split the integral at radii $M$ and $1$ about $x_3$.

On $|y-x_3|\ge 1$, every point of the segment joining $x_1$ and $x_2$ remains at least $|y-x_3|-h/2\ge (1-h/2)|y-x_3|$ from $y$. The mean value theorem bounds the kernel difference by $Ch/|y-x_3{|}^2$, giving a contribution at most $Ch\Vert \Omega {\Vert }_1$.

On $M\le |y-x_3|\le 1$, the same estimate holds because $h/2\le M/2$. Polar integration givesHere $\log (1/M)\le \log (1/h)$ because $M\ge h$.

On $|y-x_3|<M$, use the sum of the two kernel magnitudes instead of their derivatives. This disk lies inside a disk of radius $M+h/2\le 3M/2$ about either $x_i$. Hence its contribution is bounded by $CM\Vert \Omega {\Vert }_{\infty }$. Combining the three regions yieldsAt coincident points the difference is zero. At larger distances use the boundedness from part (d). A convenient global log-Lipschitz modulus isgiving $|U[\Omega ](x)-U[\Omega ](y)|\le C(\Vert \Omega {\Vert }_1+\Vert \Omega {\Vert }_{\infty })\mu (|x-y|)$. This is positive, continuous and nondecreasing. It also proves continuity of $U[\Omega ]$.

The mean value theorem states that a function continuous on $[a,b]$ and differentiable on $(a,b)$, where $a<b$, has a point $\xi \in (a,b)$ such thatTo prove it using Rolle's theorem, subtract the secant line:This function has the required continuity and differentiability, with $h(a)=h(b)=0$. Rolle's theorem gives $h^{\prime }(\xi )=0$, which is exactly the claimed equation. Thus a tangent slope somewhere in the interval equals the overall secant slope.

For a periodic function use the modulus of continuityOn the interval, take the supremum over pairs of points whose distance is at most $\delta$. The mean value theorem applied to cosine, whose derivative has absolute value at most one, givesBoth cosine values lie in $[-1,1]$, so for $|h|\le \delta$,Taking the supremum provesThe cosine substitution for polynomial approximation is also a linear isometry in the supremum norm, since cosine maps a full period onto $[-1,1]$; its image consists of even continuous periodic functions.

Let $D=f_{+}^{\prime }(a)$ and write $g(t)=(f(t)-f(a))/(t-a)$ for $t>a$. By the assumed right derivative, choose $\delta >0$ with $|g(t)-D|<ϵ/2$ whenever $a<t<a+\delta$. Choose $b_0\in (a,\min (b,a+\delta ))$.

Apply the mean value theorem to $f$ on $[a,b_0]$. It gives $x\in (a,b_0)$ with $f^{\prime }(x)=g(b_0)$. Both $g(x)$ and $g(b_0)$ are within $ϵ/2$ of $D$, soThis proves the endpoint secant-tangent approximation without any continuity assumption on $f^{\prime }$. The continuity required by the theorem is only that of $f$ itself.

The mean value theorem states that a function continuous on $[p,q]$ and differentiable on $(p,q)$ has some $c\in (p,q)$ with $f(q)-f(p)=f^{\prime }(c)(q-p)$.

Take any $p<q$ inside the given open interval. Differentiability implies the continuity needed on $[p,q]$, and the derivative is zero throughout it. The mean value theorem therefore gives $f(q)=f(p)$. Since the two points were arbitrary, the function is constant on the interval.

Use $\sin 0=0$, the derivative identity ${\sin }^{\prime }=\cos$, continuity of cosine at zero, and $\cos 0=1$. For $u\ne 0$, the mean value theorem gives $\sin u/u=\cos c_u$ for a point between zero and $u$. Hence $\sin u/u\to 1$ as $u\to 0$. For $x\ne 0$, apply this with $u=x/3^k$ to obtain the limit $\boxed{3^k\sin (x/3^k)\to x}$. For $x=0$ every term is zero.

The additional bound $|\cos t|\le 1$ and the same mean value theorem give $|\sin u|\le |u|$. ThereforeThe right side is summable. By the comparison test for series the stated series has absolute convergence for every real $x$.

The Frechet derivative at $p=(x_0,y_0)$ is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ for whichThis is differentiability; if it holds, $L(h_1,h_2)=D_{1}f(p)h_1+D_{2}f(p)h_2$.

To prove the sufficient condition, take a rectangle about $p$ contained in the open domain. Split the increment along its coordinate directions:The one-variable mean value theorem, applicable because the restrictions are differentiable, represents these two terms asThe zero-increment cases are interpreted directly. Subtract $D_{1}f(p)h+D_{2}f(p)k$. By continuity of both partial derivatives at $p$, the remainder is at most $\epsilon (|h|+|k|)\le \sqrt{2}\epsilon \sqrt{h^2+k^2}$ once the increment is small. This is the required little-oh remainder and proves differentiability.

The converse is false. Set $f(x,y)=x^2\sin (1/x)$ for $x\ne 0$, and $f(0,y)=0$. At each point with $x=0$, its Frechet derivative is zero, since the function increment has magnitude at most $h^2=o(\sqrt{h^2+k^2})$. It is smooth away from this line. Butis not continuous there. Thus a function can be differentiable everywhere while its partial derivatives are discontinuous.

For $h$, every point with $x\ne 0$ is a smooth point. At the origin, $|h(x,y)|\le |xy|\le (x^2+y^2)/2$, so $h$ is differentiable with zero derivative. At $(0,y_0)$ with $y_0\ne 0$, the putative first partial derivative would require the limit $y_0\sin (1/x)$, which does not exist. Consequently the exact set of differentiability is

For a partition $P$, write $U(f,P)$ and $L(f,P)$ for the upper Darboux sum and lower Darboux sum. The Riemann integrability criterion for bounded $f$ is that for every $\epsilon >0$ some partition satisfies $U(f,P)-L(f,P)<\epsilon$. Thus if the gaps for $D_n$ tend to zero, the Riemann integrability criterion immediately proves that $f$ is Riemann integrable.

Conversely, suppose $f$ is Riemann integrable, and choose a fixed partition $P$ with $U(f,P)-L(f,P)<\epsilon /2$. Let $r$ be the number of its interior division points and choose $M\ge 1$ with $|f|\le M$. Call a cell of $D_n$ bad when its interior contains a division point of $P$. There are at most $r$ bad cells, and their total length is at most $r/n$.

Every other cell lies in a single cell of $P$, so its oscillation is no larger than that of the containing cell. Summing the contributions of these good cells gives at most $U(f,P)-L(f,P)$, while each bad cell has oscillation at most $2M$. Therefore the finite bad-cell estimate for Darboux sums givesFor large enough $n$, the right-hand side is less than $\epsilon$. This proves the uniform-mesh Darboux criterionThis argument does not assume that the partitions $D_n$ are nested; in general they are not.

For the composition, $f$ takes values in $[-M,M]$. Since $g$ is continuously differentiable, the extreme value theorem bounds both $|g|$ and $|g^{\prime }|$ on that interval. In particular, let $K={\sup }_{[-M,M]}|g^{\prime }|<\infty$. The mean value theorem states that a function continuous on the interval between $s,t$ and differentiable in its interior satisfies $g(s)-g(t)=g^{\prime }(\xi )(s-t)$ for some intermediate $\xi$. HenceOn each partition cell, this bounds the oscillation of $g\circ f$ by $K$ times the oscillation of $f$, whether or not the local extrema are attained. It follows thatThe composition is bounded, so the proved uniform-mesh Darboux criterion applies. Thus $g\circ f$ is Riemann integrable. The reusable fact is Lipschitz composition preserves Riemann integrability; continuous differentiability supplies the required bound on the range of $f$.

For $x$ sufficiently close to zero, $1+x>0$. Apply the mean value theorem to the logarithm between $1$ and $1+x$: there is a point ${\xi }_x$ between these endpoints withAs $x\to 0$ from either side, ${\xi }_x\to 1$, so this quotient tends to $1$. The exponential function is continuous, and thereforeThe hypotheses of the mean value theorem hold because $\log t$ is continuous and differentiable for $t>0$, with derivative $1/t$.

For $h\ne 0$, enlarge the nonnegative integral $I(h)$ to all of $\mathbb{R}$, and substitute $y=x+|h|t$. Reflection if necessary givesNear $0$ and $1$, its integrand is bounded by a constant times $1+|t{|}^{-2\alpha }$ or $1+|t-1{|}^{-2\alpha }$, which is integrable. For large $|t|$, the mean value theorem bounds the difference by a constant times $|t{|}^{-\alpha -1}$, whose square is integrable. Thus $C_{\alpha }<\infty$. The preceding estimate yieldsIn particular $g$ is continuous on all of $\mathbb{R}$; the proof establishes Hölder continuity of a one-dimensional fractional kernel integral of exponent $1/2-\alpha$ uniformly in $x$.

The Rolle theorem states that if a real function is continuous on $[u,v]$, differentiable on $(u,v)$, and has equal endpoint values, then its derivative vanishes somewhere in $(u,v)$. Order the $k$ distinct real roots as $r_1<\cdots <r_k$. The theorem gives a root of $f^{\prime }$ in each interval $(r_j,r_{j+1})$. These intervals are disjoint, so the obtained roots are distinct:

We prove the requested bound for positive roots of a polynomial by mathematical induction on the number of nonzero terms. In fact, the proof also works with roots counted according to their multiplicity. Write $V(p)$ for the number of sign changes and $N_{+}(p)$ for the positive-root count.

A one-term polynomial $a_1{x}^{d_1}$ has no positive roots and no sign changes. For the induction step, factor out the smallest power of $x$:For $x>0$ this factor is nonzero, so $N_{+}(q)=N_{+}(p)$, including multiplicities, and $V(q)=V(p)$. Differentiation removes exactly the constant term:The exponents remain strictly increasing, and the positive multipliers $e_j$ do not change coefficient signs. The induction hypothesis therefore gives

First suppose $a_1{a}_2<0$. Removing $a_1$ removes exactly one sign change, so $V(q)=V(q^{\prime })+1$. If $q$ has positive roots $r_1<\cdots <r_k$ of multiplicities $m_1,\dots ,m_k$, its derivative has multiplicity $m_j-1$ at $r_j$ and, by the Rolle theorem, an additional root between each pair. Thus the Rolle root count with multiplicities giveswhen there is a positive root. ConsequentlyIf there is no positive root, the desired bound is automatic.

Now suppose $a_1{a}_2>0$, so $V(q)=V(q^{\prime })$. Again the zero-root case is immediate. Otherwise, multiplying $q$ by $-1$ if necessary lets us assume $a_1>0$ and $a_2>0$ without changing roots or sign changes. For sufficiently small $x>0$,since $q(0)=a_1$ and the lowest-power term of $q^{\prime }$ has positive coefficient. Choose such a point $u$ before the first positive root $r_1$. The mean value theorem on $[u,r_1]$ gives a point $v\in (u,r_1)$ withBut $q^{\prime }(u)>0$. The intermediate value theorem applied to the polynomial $q^{\prime }$ therefore gives an extra root in $(u,v)$, before $r_1$. It is distinct from the roots at $r_j$ and those between consecutive roots. HenceBoth sign cases complete the induction:An initial sign change allows one extra positive root; without that sign change, the derivative must already turn before the first positive root. This is the upper-bound part of the Descartes rule of signs. The multiplicity claim used above follows by differentiating $q(x)=(x-r{)}^{m}h(x)$ with $h(r)\ne 0$: the derivative has a factor $(x-r{)}^{m-1}$ and its remaining factor is nonzero at $r$.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. We first use it to show that a continuous function on $[a,b]$ is bounded above. Otherwise, choose $x_n\in [a,b]$ with $f(x_n)>n$. A convergent subsequence $x_{n_k}\to c$ exists, with $c\in [a,b]$ because the interval is closed. Continuity would give $f(x_{n_k})\to f(c)$, contradicting $f(x_{n_k})>n_k\to \infty$.

Let $L=\sup \{f(x):x\in [a,b]\}$, which is finite. By the definition of supremum, there are $y_n\in [a,b]$ with $L-1/n<f(y_n)\le L$. Another application of the Bolzano-Weierstrass theorem gives a subsequence $y_{n_k}\to c\in [a,b]$. By continuity,Thus the supremum is actually attained:This proves the maximum part of the extreme value theorem from sequential compactness rather than assuming it.

Now suppose $f^{\prime \prime }<0$ on $\mathbb{R}$. Applying the mean value theorem to $f^{\prime }$ shows that $f^{\prime }$ is strictly decreasing. At a local maximum of a differentiable function, the left difference quotients are nonnegative and the right difference quotients are nonpositive; existence of the derivative forces $f^{\prime }=0$. A strictly decreasing function has at most one zero. ThereforeIf that zero exists, $f^{\prime }$ is positive to its left and negative to its right, so the local maximum is also a global maximum.

For the uniform bound by $K<0$, defineIts derivative is $h^{\prime }(x)=f^{\prime \prime }(x)-K<0$, so $h$ is strictly decreasing by the mean value theorem. Comparing with $h(0)=f^{\prime }(0)$ givesBecause $K<0$, this makes $f^{\prime }$ negative at sufficiently large positive $x$ and positive at sufficiently large negative $x$. The function $f^{\prime }$ is continuous, since it is differentiable; the intermediate value theorem supplies a zero $c$ between such points. Its strict decrease makes this zero unique. Again using the mean value theorem for $f$, the derivative signs imply that $f$ increases up to $c$ and decreases after $c$:This illustrates uniform negative curvature and global maximization.

Merely requiring $f^{\prime \prime }<0$ does not force existence. TakeThen $f^{\prime }(x)=e^{-x}>0$ and $f^{\prime \prime }(x)=-e^{-x}<0$, but $f$ is strictly increasing, approaches $0$ as $x\to +\infty$, and never equals $0$. The answer to the final question is no, even for a function bounded above:The stronger uniform curvature assumption prevents this behaviour. Here $f^{\prime \prime }(x)\to 0$ at the end where the supremum is approached, so no uniform negative bound of that kind is available.

Define $F_i(x)=a_i+\frac{1}{6}{\sum }_{j\ne i}\sin x_j$. Solving the system is equivalent to finding a fixed point of $F:{\mathbb{R}}^3\to {\mathbb{R}}^3$. The mean value theorem implies $|\sin u-\sin v|\le |u-v|$. In the specified metric,Each coordinate difference occurs exactly twice. This metric makes ${\mathbb{R}}^3$ a nonempty complete metric space, because a Cauchy sequence in it is Cauchy in each real coordinate, and coordinate convergence implies convergence in the sum metric. Thus $F$ is a contraction mapping with constant $1/3$.

There is exactly one solution for every choice of $(a_1,a_2,a_3)$, by the contraction mapping theorem. Iteration of $F$ from any starting vector also constructs it.

Let $L$ be the supremum of all lower Darboux sums and $U$ the infimum of all upper Darboux sums. The assumed comparison of arbitrary lower and upper sums gives $L\le U$. For the partition supplied for a given $\epsilon$,Since this holds for every positive $\epsilon$, $U=L$. Thus the Riemann integrability criterion gives

If $f$ is continuous on $[a,b]$, the Heine-Cantor theorem makes it uniformly continuous. Given $\epsilon >0$, choose $\delta$ so that intervals of length below $\delta$ have oscillation below $\epsilon /(b-a)$. A partition of mesh below $\delta$ then satisfiesproving that Continuous functions are Riemann integrable.

For the function defined using $g$, choose $K$ with $|f|\le K$. Given $\epsilon >0$, choose $\delta >0$ so small that $2K\delta <\epsilon /2$. On $[\delta ,1]$, the function $g$ is continuous and therefore has a partition whose upper-minus-lower sum is below $\epsilon /2$. Adding the interval $[0,\delta ]$ contributes at most $2K\delta$, regardless of the value $f(0)=\lambda$. Hence the full Darboux-sum difference is below $\epsilon$, and

Finally, for any partition $a=x_0<\cdots <x_n=b$, the mean value theorem supplies ${\xi }_i\in (x_{i-1},x_i)$ such thatSumming telescopes toa Riemann sum for the Riemann-integrable function $f^{\prime }$. Equivalently, every lower derivative sum is at most this telescoping value and every upper derivative sum is at least it. Taking the common upper and lower Darboux integrals proves the fundamental theorem of calculus conclusion