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A convex function satisfies $g(tx+(1-t)y)\le tg(x)+(1-t)g(y)$ for all $x,y$ in its domain and $0\le t\le 1$. If the second derivative exists on an interval, this is equivalent to $g^{\prime \prime }\ge 0$ there. The equivalence follows from the nondecreasing first derivative, or from the supporting-line inequalities obtained by integration. HereThus it is a strictly convex function, and in particular convex.

For the Legendre transform, maximize $px-f(x)$ over $x>0$. Its derivative is $p-1-\log x$, and strict concavity makes the stationary point $x=e^{p-1}$ the unique maximizer. Substitution gives the convex conjugate of x log xThe range of $f^{\prime }$ is all of $\mathbb{R}$, so all real dual arguments have this finite value. Every point of the original graph is supported by its tangent of slope $f^{\prime }(x)$; the conjugate supremum therefore recovers $f$ from these supporting lines. Equivalently, the inverse derivative relation for this strictly convex differentiable Legendre transform gives $f^{\ast \ast }=f$ on $x>0$. If an extended-real function on all of $\mathbb{R}$ is wanted, its closed convex extension has value zero at $x=0$ and infinity for $x<0$; biconjugation includes that endpoint closure.

Fix $(x,y)\in U$ and a sufficiently small rectangle inside $U$. Its rectangular incrementcan be evaluated twice using the fundamental theorem of calculus, givingDivide by $hk$ and let $h,k\to 0$. Continuity of the mixed partial derivatives makes the two limits $D_2{D}_{1}f(x,y)$ and $D_1{D}_{2}f(x,y)$. They are equal everywhere on $U$.

Repeated interchange of adjacent partial derivatives reduces every order-$m$ derivative of a smooth function to $D_1^j{D}_2^{m-j}f$, $0\le j\le m$. Hence there are at most $m+1$ distinct functions. This maximum is attained by $f(x,y)=e^{x+2y}$, whose listed partial derivatives are the distinct functions $2^{m-j}{e}^{x+2y}$. The answer is $m+1$, rather than the $2^m$ possible written orders.

For the supplied $f$, $f(t,t)=1/2$ for $t\ne 0$, whereas $f(0,0)=0$. It is not even continuous at the origin, so $f$ is neither differentiable nor infinitely differentiable there.

For $g$, $|g(x,y)|\le |xy|\le (x^2+y^2)/2$. Thus $g(x,y)=o(\sqrt{x^2+y^2})$, proving Fréchet differentiability at the origin with derivative zero. However,including zero at the origin. Therefore $D_2{D}_{1}g(0,0)=-1$ and $D_1{D}_{2}g(0,0)=1$. $g$ is differentiable but not infinitely differentiable at the origin: it cannot have continuous second partial derivatives near that point.

The denominator vanishes when $z=(2j+1)i\pi /2$. For $R>0$, only $z_0=i\pi /2$ lies inside the rectangle. This is a simple pole, since the denominator derivative there is $2$. Its residue isThe positive orientation and the residue theorem consequently give

For the Dirichlet Green's function, solve $-D{G}_{xx}=\delta (x-\xi )$ with zero endpoint values. Away from $\xi$, $G$ is linear. Continuity at $\xi$ and the derivative jump $G_x(\xi +,\xi )-G_x(\xi -,\xi )=-1/D$ giveFor nonzero Dirichlet boundary conditions, subtract their linear interpolant. The resulting boundary value problem has solutionIn the final heat equation, a steady state has zero time derivative. Here $l=1$, $f(x)=x$, $\alpha =1/D$, $\beta =2/D$. Integrating $u^{\prime \prime }=-x/D$ twice and imposing the endpoint values givesIts second derivative and both boundary conditions verify the answer directly.

Put $p=\sqrt{2mE}/\hslash$ and $\gamma =mk/{\hslash }^2$. Integrating the Schrödinger equation across the delta potential gives continuity of $\psi$ andFor incidence from the left, write $\psi =e^{ipx}+r{e}^{-ipx}$ on $x<0$ and $\psi =t{e}^{ipx}$ on $x>0$. Matching gives $t=1+r$ and $ip(t-1+r)=2\gamma t$, henceThe incident and transmitted probability currents have the same velocity factor, so the reflection coefficient and transmission coefficient areFor $k<0$, an even bound state has $\psi =A{e}^{-\kappa |x|}$, $\kappa =\sqrt{-2mE}/\hslash$. The derivative jump now gives $-2\kappa A=2\gamma A$, so $\kappa =-\gamma >0$ andThere is exactly one positive value of $\kappa$, hence exactly one even bound state. Its normalized amplitude is $A=\sqrt{\kappa }$. An odd decaying solution would have to vanish at zero and is therefore trivial.

In the shrinking finite square well, its integrated potential is $k=-2Ul$, not $-Ul$: the full width is $2l$. With $Ul$ fixed and $l\to 0$, $0\le (\alpha l{)}^2\le 2mU{l}^2/{\hslash }^2\to 0$. The matching condition then bounds $\beta$ by $(2mUl/{\hslash }^2)(1+o(1))$, so $E$ remains bounded. Consequently $El\to 0$, and the previous matching condition givesUsing $E=-{\hslash }^2{\beta }^2/(2m)$ recovers the same bound-state energy. All other even levels disappear as the width shrinks.

The time-independent Schrödinger equation gives the even interior wavefunction $\psi =A\cos (\alpha x)$, with ${\alpha }^2=2m(U+E)/{\hslash }^2$. Outside, normalizability of a bound state requires $\psi =C{e}^{-\beta (|x|-l)}$, with ${\beta }^2=-2mE/{\hslash }^2$.

Continuity of the wavefunction and its derivative at $x=l$ gives $C=A\cos (\alpha l)$ and $-\beta C=-\alpha A\sin (\alpha l)$. A nonzero solution cannot have $\cos (\alpha l)=0$, so division givesAt the formally allowed endpoint $E=-U$, the even interior solution is constant; derivative matching would force the exterior amplitude to vanish, leaving the zero solution. Nontrivial bound states therefore have $-U<E<0$. The matching formula is the even-state condition for a finite square well.

Both maxima exist by continuity on the compact set $[a,b]$. Their sum is nonnegative and vanishes only for $f=0$. The identities $\Vert cf{\Vert }_1=|c|\Vert f{\Vert }_1$ and the triangle inequality follow by applying the corresponding supremum norm facts to $f$ and $f^{\prime }$. This proves that $\Vert \cdot {\Vert }_1$ is a norm.

For $m=(a+b)/2$, $|\Phi (f)|=|f^{\prime }(m)|\le \max |f^{\prime }|\le \Vert f{\Vert }_1$. Hence the first unit ball has image contained in $[-1,1]$: derivative evaluation is a bounded linear functional for this norm.

For the supremum norm alone, take $f_N(x)=\sin (N(x-m))$. These smooth functions satisfy $\Vert f_N{\Vert }_0\le 1$, but $\Phi (f_N)=N$. Thus the second image is unbounded. The distinction is the unboundedness of derivative evaluation in the supremum norm, not a failure of differentiability of the individual functions.

An $n$-dimensional smooth manifold is a Hausdorff space with a countable topological base, equipped with a smooth atlas of homeomorphisms from open subsets onto open subsets of ${\mathbb{R}}^n$, whose overlap maps are smooth diffeomorphisms. The smooth structure is the maximal smooth atlas compatible with these manifold charts. Here manifolds have no boundary unless specified otherwise.

For a space with all the requested topological properties but no such atlas, take the topological tripod: three closed intervals joined at one endpoint. It is a compact connected subspace of the plane with its induced metric. A countable base of planar rational balls restricts to a countable base, the metric makes it Hausdorff, and compactness gives a finite subcover of every open cover, hence a locally finite refinement and paracompactness.

At an interior point of any arm, arbitrarily small neighborhoods are intervals. A coordinate ball in dimension at least two would remain connected after deleting its center; an interval does not. Dimension zero would make the space discrete. Thus any possible connected manifold structure would have dimension one. But a sufficiently small neighborhood of the junction, minus the junction, has three components, whereas an interval chart has two. This contradiction rules out even a topological manifold structure, and hence any smooth one.

The product smooth structure uses manifold charts $(\varphi ,\psi ):U\times V\to \varphi (U)\times \psi (V)\subset {\mathbb{R}}^{m+n}$. Transition maps act separately in the two coordinate blocks and are smooth with smooth inverses. Products of countable bases give a countable base, and the product remains Hausdorff. Therefore $\boxed{\dim (M\times N)=\dim M+\dim N}$ with this natural smooth structure.

Define the tangent space by point derivations: a tangent vector at $x$ is an $\mathbb{R}$-linear map $D$ on germs of smooth functions at $x$, satisfying $D(ab)=a(x)D(b)+b(x)D(a)$. Addition and scalar multiplication preserve this rule, so these derivations form a vector space. In coordinates $u^1,\dots ,u^n$, the local identityfollows by integrating the derivative of $h$ along the coordinate line segment. Derivations annihilate constants, so it gives $Dh={\sum }_{i}D(u^i){\partial }_{i}h(u(x))$. The coordinate derivations ${\partial }_i{|}_x$ are independent since they evaluate the coordinate functions as ${\delta }_i^j$. ThusFor a smooth map, define the differential of a smooth map intrinsically by $(f_{\ast }D)(h)=D(h\circ f)$. It is again linear and satisfies the derivation rule at $f(x)$, so it maps $T_{x}M$ into $T_{f(x)}N$. Its coordinate matrix is the Jacobian of the coordinate expression of $f$, independently of the charts by the intrinsic definition and chain rule.

Finally, a zero differential forces each target coordinate function to have all derivatives zero on a small connected source coordinate ball mapping into one target chart. Integration on straight segments makes those functions constant there. Hence $f$ is locally constant. Each nonempty fiber is both open and closed, so connectedness gives the zero-differential constancy theorem, $\boxed{f\text{ is constant}}$.

Combining metric compatibility with the torsion-free condition forces the Koszul formula:Nondegeneracy of $g$ proves uniqueness of the Levi-Civita connection. For existence, use the right side to define $2g({\nabla }_{X}Y,Z)$. Expanding brackets shows that this expression is $C^{\infty }$-linear in $Z$, so it defines a smooth one-form; the musical isomorphism gives the required vector field. The same expansion shows ${\nabla }_{aX}Y=a{\nabla }_{X}Y$, additivity, and ${\nabla }_X(aY)=X(a)Y+a{\nabla }_{X}Y$, establishing the connection rules. Subtracting the formulas with $X,Y$ exchanged gives ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$. Adding the formulas pairing ${\nabla }_{X}Y$ with $Z$ and ${\nabla }_{X}Z$ with $Y$ gives metric compatibility. Thus this construction has both required properties.

In coordinate vector fields the brackets vanish. Consequently the Christoffel symbols and coordinate derivative areRepeated indices are summed. The displayed connection type in the source is best understood in its standard form on two vector fields; if the first input is an individual tangent vector at a point, the output lies in the tangent fiber at that point rather than in the space of global sections.

For the parallel metrics with a common Levi-Civita connection, let $\gamma$ be a piecewise smooth path from the point $x$ of equality to any $y$. Connected smooth manifolds admit such paths because coordinate balls are path connected. Parallel transport $P_{\gamma }$ for the common connection is invertible and preserves both metrics. ThereforeEvery pair of tangent vectors at $y$ arises this way, so $\boxed{\tilde{g}=g}$ everywhere.

Dropping the agreement at one point removes the conclusion. For any constant $c>0$, $\tilde{g}=cg$ has the same Christoffel symbols. Nor must the two metrics be proportional: on ${\mathbb{R}}^n$, $n\ge 2$, the constant metrics ${\sum }_i(d{x}^i{)}^2$ and $2(d{x}^1{)}^2+{\sum }_{i\ge 2}(d{x}^i{)}^2$ both have zero connection coefficients. In general write $\tilde{g}(u,v)=g(Au,v)$. Since both metrics are parallel, $0=({\nabla }_X\tilde{g})(Y,Z)=g(({\nabla }_{X}A)Y,Z)$, so $\nabla A=0$. Conversely, a positive $g$-self-adjoint parallel $A$ makes the same torsion-free connection compatible with $\tilde{g}$, proving equality of their Levi-Civita connections. Thus one-point agreement specifies $A=I$ and forces it everywhere; without it, nontrivial parallel choices can remain.

The Bonnet-Myers theorem says that a connected geodesically complete $n$-dimensional Riemannian manifold, with $n\ge 2$ and $\mathrm{Ric}\ge (n-1)\kappa g$ for a constant $\kappa >0$, hasThe dimension and positive lower bound are essential: in dimension one the Ricci condition is vacuous, and a zero lower bound does not imply bounded diameter.

By Hopf-Rinow, any two distinct points have a unit-speed minimizing geodesic $\gamma :[0,L]\to M$. Choose parallel orthonormal fields $E_1,\dots ,E_{n-1}$ perpendicular to its tangent $T$. For the endpoint-vanishing fields $V_i(t)=\sin (\pi t/L)E_i(t)$, the second variation of geodesic energy gives nonnegative index formsSumming and using the Ricci lower bound produces the sine index-form bound for positive Ricci curvature:If $L>\pi /\sqrt{\kappa }$, the last expression is negative, a contradiction. This proves the diameter bound. Hopf-Rinow makes closed bounded sets compact, so the entire manifold is compact.

Give the universal cover the pullback metric. Local isometry preserves its Ricci bound, and lifting complete base geodesics proves completeness of the cover. The same diameter and compactness argument applies there. A fiber of the covering is closed and discrete, hence finite in this compact cover; its cardinality is that of the fundamental group. This proves the final assertion.

For a counterexample that also breaks the diameter conclusion, use the incomplete positively curved strip with infinite diameterThe map $\Phi (u,v)=(\cos u\cos v,\cos u\sin v,\sin u)$ is a local isometry to the round unit sphere: its coordinate derivatives are orthogonal, with squared lengths one and ${\cos }^{2}u$. Thus $K=1$ and $\mathrm{Ric}=g$, satisfying the required lower bound with $n=2$, $\kappa =1$.

The meridian $(u,v)=(t,0)$ is a unit-speed geodesic and reaches the excluded boundary at $t=\pi /4$, so this metric is incomplete. Every curve joining $(0,0)$ to $(0,L)$ has length at least $\int \cos u|v^{\prime }|dt\ge |L|/\sqrt{2}$, because $\cos u\ge 1/\sqrt{2}$ on the strip. Therefore $\boxed{\mathrm{diam}M=\infty }$ despite the positive Ricci bound. Completeness cannot be omitted.

The exact normalizing sum can be expensive on a large loss network. The Erlang fixed point approximation replaces joint resource acceptance by a product of marginal acceptances. Here consider unit requirements $A_{jr}\in \{0,1\}$, positive integer capacities and finitely many fixed routes. Let $B_j$ be resource blocking and $q_j=1-B_j$. A route-$r$ call contributes to the offered traffic at resource $j$ after surviving the other resources on its route. Thus the reduced-load approximation iswhere the Erlang B formula isThe estimated route acceptance is ${\prod }_{j\in r}{q}_j$. This is an independence approximation, not an alternative exact factorization of the stationary law in part (i).

Uniqueness follows from a convex potential for the Erlang fixed point. For a single resource, the carried load of an Erlang loss resource is $m_C(v)=v[1-E(C,v)]$. It increases strictly from zero to $C$ as $v$ increases: differentiating the expected value of its upper-truncated Poisson distribution with respect to $\log v$ gives the strictly positive occupancy variance. Blocking also increases strictly, as is evident on dividing the Erlang denominator by its final term.

Use $p_j=-\log q_j\ge 0$. Let $v_C(p)$ be the unique offered load with $1-E(C,v_C(p))=e^{-p}$, and set $g_C(p)=v_C(p)e^{-p}=m_C(v_C(p))$, with $g_C(0)=0$. This function is strictly increasing and tends to $C$. DefineEach exponential term is a convex function, and each integral is a strictly convex function because its derivative is strictly increasing. Therefore $F$ is a strictly convex function. It is also a coercive function on the nonnegative orthant: an unbounded coordinate makes its integral grow asymptotically linearly with positive slope $C_j$. A unique minimizer exists.

If resource $j$ carries some positive offered route, its inward derivative at $p_j=0$ is negative, so its minimizing coordinate is positive. At such a coordinate the first-order equation isDividing by $q_j$ gives precisely $v_j={\sum }_{r:j\in r}{\rho }_r{\prod }_{k\in r\setminus \{j\}}{q}_k$. A resource with no positive offered route uniquely has $p_j=0$, hence $B_j=0$. Thus the minimizer and the fixed point coincide, proving existence and uniqueness of the Erlang fixed point for fixed unit-resource routing. This does not by itself guarantee convergence of every simultaneous substitution algorithm; the uniqueness claim concerns the solution of the equations.