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The Lagrangian sufficiency theorem for a maximization problem says the following. Suppose the constraints are $g_i(q)\ge 0$ and $h_j(q)=0$, and defineIf a feasible $q^{\ast }$ satisfies complementary slackness ${\lambda }_i{g}_i(q^{\ast })=0$ and globally maximizes $L(\cdot ,\lambda ,\mu )$, then $q^{\ast }$ globally maximizes $f$. Indeed, for every feasible $q$,This proves the theorem. In a concave maximization problem, stationarity and the boundary optimality conditions ensure the required global maximum of the Lagrangian.

For the problem at hand, retain $x,z\ge 0$ as the domain and attach a multiplier $\lambda$ to the equality:For $\lambda \ge 1$, this is concave in $(x,y,z)$ on the domain. Its separate maximizers satisfywhile the $x$ term is $(1-\lambda )x$. Thus $x$ may be positive only when $\lambda =1$; when $\lambda >1$, its maximizing value is $x=0$.

Iftake $\lambda =1$. Then $y=1$, $z=a^2-1$, and feasibility fixesThe Lagrangian sufficiency theorem gives the maximum

Ifthen $\lambda >1$ and $x=0$. The equality constraint becomessoConsequentlyThe assumptions $a\ge 1$ and $b\ge 1/2$ ensure $z\ge 0$. At the stationary point $\sqrt{1+z}=ay$, so the maximum value isThe two formulas agree at $b=a^2-1/2$.

Let $w=N(U)/N(R)$, so $1-w=N(V)/N(R)$. Since class counts add,The function $g(p)=p(1-p)$ is concave function on $[0,1]$. Jensen inequality therefore giveswhich is precisely $Q\le 0$. Thus an axis-aligned split cannot increase the weighted empirical Gini impurity.

For $a<b$, choose the uniform partition of an interval into $n$ pieces and its midpoint tagsTheir arithmetic mean is exactly $(a+b)/2$. Apply the given Jensen inequality with equal weights ${\lambda }_i=1/n$:Multiplying by $b-a$ gives a midpoint Riemann sum on the left. Its mesh of a partition is $(b-a)/n\to 0$, so the Riemann sum convergence theorem givesFor $a=b$ the equality is immediate. Together with part i, this is the Hermite-Hadamard inequality with the directions appropriate to a concave function.

The stated weighted inequality is the Jensen inequality for a concave function. Assume $a<b$; if $a=b$, both sides of the required inequality are zero. Write $x=\lambda a+(1-\lambda )b$ with $\lambda =(b-x)/(b-a)$. Applying the weighted inequality at the two endpoints givesThus the affine function $\ell$ joining the endpoint values lies below $f$. By monotonicity of the Riemann integral,This is the lower bound in the Hermite-Hadamard inequality for a concave function.

The cycloid has limiting endpoints $(0,0)$ and $(2\pi a,0)$, with its highest point $(\pi a,2a)$ at $t=\pi$. The transformation $t\mapsto 2\pi -t$ sends $x$ to $2\pi a-x$ and preserves $y$, so the arch is symmetric about $x=\pi a$. Its slope decreases from $+\infty$ through zero to $-\infty$, giving vertical limiting tangent lines at the cusps and a horizontal tangent line at the top. More explicitly,so the arch is a concave function throughout its interior. The endpoint cusps are limiting points, excluded by the open parameter interval.

Within the requested interval, the function is symmetric about $x=\pi /2$ because $\sin (\pi -x)=\sin x$. Its derivative $\cot x$ is positive before $\pi /2$, zero there, and negative afterwards. The negative second derivative shows that this is a strictly concave function. Consequently its unique maximum is $(\pi /2,0)$. It tends to $-\infty$ at both ends, giving vertical asymptotes $x=0$ and $x=\pi$.

The chord through the endpoints is the straight line $y=(x-1)\log 2$. At $\xi =1/\log 2$, the tangent line isThese two straight lines have the same slope, as required by the mean value theorem. Since $f^{\prime \prime }(x)=-1/x^2<0$, the natural logarithm is a concave function: its graph lies above the chord and below this tangent line.

We first establish the lower truncation identity as well. The pair $(-X,-Y)$ has the same hypotheses as $(X,Y)$, because conditional expectation is linear and negation preserves equality of probability distributions. Apply part (ii) to this pair and multiply by $-1$ to getFor every real $t$,Using this identity, part (ii), the lower truncation identity, and linearity of conditional expectation givesIn particular the argument does not assume that a function obtained by scalar clipping is a convex function or a concave function.

Write $C_a(t)=(t\wedge a)\vee (-a)$. The bounded random variables $C_a(X)$ and $C_a(Y)$ have the same probability distribution, and the displayed identity identifies the latter as the conditional expectation of the former. They are square-integrable, so part (i) gives $C_a(X)=C_a(Y)$ almost surely. Take $a=1,2,\dots$ and discard the countable union of exceptional null sets. At every remaining outcome, let $a\to \infty$ to obtainThus conditional expectation preserving a distribution forces equality even when only integrability is assumed.

The function ${\varphi }_a(t)=t\wedge a$ is a concave function, and $X\wedge a$ is integrable since $|X\wedge a|\le |X|+a$. The concave form of the conditional Jensen inequality givesThe random variables $X$ and $Y$ have the same probability distribution, so $\mathbb{E}[X\wedge a]=\mathbb{E}[Y\wedge a]$. Using the law of total expectation, the nonnegative difference between the right and left sides of the inequality has zero expectation. It must vanish almost surely. ThereforeThe equality comes from the distributional assumption as well as the conditional Jensen inequality.

On $I=[1,7/5]$, $f^{\prime }<0$ and $f^{\prime \prime }<0$, since $x^2<2$. Hence $|f^{\prime }|$ increases and has minimum $|f^{\prime }(1)|=14$. Alsowith equality at $x=1$. Therefore

There is a unique root $r$ in $I$: $f(1)=2>0$, $f(7/5)=-5.678464<0$, and the function is strictly decreasing. Moreover $f(6/5)=-1.382016<0$, so $r<6/5=x_0$.

It remains to justify that the iteration stays where the error estimate applies. Because $f$ is a concave function, its graph lies below each tangent line. At any $x\in [r,6/5]$, the tangent has nonpositive value $f(x)$ at $x$, negative slope, and nonnegative value at $r$. Its zero $N(x)=x-f(x)/f^{\prime }(x)$ therefore satisfiesBy mathematical induction, the Newton root-finding iteration stays in $[r,6/5]$ and decreases to some limit $L\ge r$. Continuity in $x_{k+1}-x_k=-f(x_k)/f^{\prime }(x_k)$ implies $f(L)=0$, so uniqueness gives $L=r$.

Now the supplied error bound applies at every step:This proves at least quadratic convergence, with an invariant interval and a nonvanishing derivative explicitly established.

Here a dense $100000$-by-$100000$ matrix has $10^{10}$ entries, about $80$ GB even with eight-byte storage per entry. Use limited-memory BFGS on the negative of the objective, or equivalently its positive-definite approximate-metric form of gradient ascent. It stores only a fixed small number $\ell$ of vector pairs, costing $O(\ell d)$ memory rather than $O(d^2)$.

Let $G(x)={\sum }_{i=1}^{100}{e}^{e^{-\Vert x-a_i{\Vert }^2}}$ and put $c_i(x)=e^{e^{-\Vert x-a_i{\Vert }^2}}{e}^{-\Vert x-a_i{\Vert }^2}$. Direct use of the chain rule givesIf $H_n$ denotes the positive-definite inverse-Hessian matrix approximation for $-G$, the iteration iswith a line search increasing $G$ and the usual safeguarded limited-memory updates. No exact Hessian matrix is required. With fixed memory and the usual bounded-metric and line search hypotheses, expect local linear convergence near a nondegenerate strict maximum; fixed-memory limited-memory BFGS does not generally inherit the superlinear theorem for full-memory BFGS.

Unlike the preceding objective, $G$ is not generally a concave function. Thus this is a method for finding a local maximum; comparison of different converged candidates or a global search is needed to certify a global maximum for arbitrary centres. The symbolic problem supplies no assumption that would justify an unconditional global-convergence claim.^[ref]

The denominator is positive throughout the domain. Holding $y$ fixed, differentiate with respect to $x$:At the interior point $(1,1/2)$ this second partial derivative is $-2/9<0$, so the restriction to a horizontal line is locally strictly concave, ruling out convexity. At the interior point $(1,2)$ it is $4/9>0$, ruling out concavity. Restrictions of a convex function or a concave function to every line segment must retain that property. Consequently $k$ is neither convex nor concave on its stated domain.

The function taking the lesser value at every common-domain point. It forms the lower envelope of two graphs. For finitely many affine functions it is a concave function: each affine branch is at least the corresponding convex combination of the two endpoint minimum values, so taking the minimum preserves that lower bound. A maximizing mixed strategy in a two-player zero-sum game maximizes such a lower envelope of opposing pure-choice expected payoffs.

Consider the minimization problemThe objective function is a strictly convex function with its unconstrained optimization minimum at $x=2$. If $b<2$, it decreases throughout the feasible set until the bound $x=b$ is reached. If $b\ge 2$, the feasible set already contains $2$. HenceWith $g(x)=x-b$, the optimization Lagrangian and KKT conditions areSubstituting $x_{\ast }(b)$ gives the unique inequality Lagrange multiplierFor $b<2$, the active constraint has ${\lambda }_{\ast }>0$. For $b>2$, there is strict slack $b-x_{\ast }=b-2$ and ${\lambda }_{\ast }=0$. At $b=2$, the active constraint has zero Lagrange multiplier: the boundary happens to pass through the unconstrained optimum. Feasibility and complementary slackness must both be checked; solving only $2(x-2)+\lambda =0$ would allow infeasible candidates.

An independent certificate follows by completing the square:The Lagrange dual function is a concave function of $\lambda$. Its maximum over $\lambda \ge 0$ occurs at ${\lambda }_{\ast }(b)$ and equals $V(b)$. Thus every case has an explicit lower bound attained by a feasible point; the KKT conditions provide a certificate that can be checked without searching all real numbers.

In independent bond percolation on a graph, let $c_n$ be the number of $n$-step self-avoiding walks from a fixed graph vertex and ${\kappa }_n$ the number that are open. Then $\mathbb{E}{\kappa }_n=c_n{p}^n$. The Jensen inequality for the concave function $t^{1/n}$ gives $\mathbb{E}({\kappa }_n^{1/n})\le p{c}_n^{1/n}$. No independence between different self-avoiding walk indicators is needed. When $c_n^{1/n}\to \mu$, the limit superior is at most $p\mu$.

Let a differentiable concave function on all of ${\mathbb{R}}^d$ be strictly increasing in each coordinate separately. Every coordinate derivative is strictly positive. Indeed it is nonnegative by monotonicity and nonincreasing along its coordinate by concavity; a zero value would force zero derivative on the remaining positive ray and hence a constant segment. The global-domain assumption matters at endpoints of restricted domains.

The tangent line to a differentiable function at $a$ is its affine first-order approximation $f(a)+f^{\prime }(a)(x-a)$. For a differentiable concave function, it is an upper bound throughout a convex interval: the secant-slope inequality and passage to the derivative give $f(x)\le f(a)+f^{\prime }(a)(x-a)$. This upper bound constructs valid envelopes in adaptive rejection sampling.

As a quadrature rule, the trapezoidal rule replaces an integrand by the linear interpolation joining its two endpoint values:For nonnegative endpoint values, this is the area of a trapezoid; in general it is a signed area. The quadrature nodes are $a,b$, each with quadrature weight $(b-a)/2$. Unlike the midpoint quadrature rule, it uses endpoint samples, which can be reused between adjacent panels.

Every affine function $f(x)=\alpha x+\beta$ agrees with its linear interpolation, so the trapezoidal rule integrates it exactly. On $[0,2]$, $f(x)=2x+1$ gives $T(f)=(2/2)(1+5)=6=I(f)$. On $[0,1]$, the polynomial $f(x)=x^2$ instead gives $T(f)=1/2$, exceeding $I(f)=1/3$ by $1/6$. Its degree of precision of a quadrature rule is exactly one.

If $f\in C^2[a,b]$, its signed approximation error satisfiesA convex function lies below its endpoint chord, so the trapezoidal rule overestimates its integral; a concave function gives the reverse inequality. The error formula assumes that the second derivative exists and is a continuous function throughout the closed interval.^[ref] A short derivation is the identityobtained by two integrations by parts. The nonnegative Peano kernel has integral $(b-a{)}^3/12$, and the weighted mean value theorem for integrals gives the formula.

The trapezoidal rule for an ordinary differential equation averages vector fields at the old and new states and is A-stable.

This time-stepping use applies the same trapezoidal rule to the integral of the derivative in an ordinary differential equation. That connection explains its place under implicit Runge-Kutta methods; its quadrature behavior and the nonlinear issue in trapezoidal rule fails B-stability concern different uses of the same endpoint averaging.

Choose two Ising spins with Hamiltonian $H=-J{s}_1{s}_2$, where $J$ independently takes $J_0$ or $2J_0$ with equal prior probabilities and $J_0>0$. This original finite example has no geometrical frustration and no phase transition. Its four spin assignments give $Z_J=4\cosh (\beta J)$. ThereforeThe concave function $\log$ puts $F_{\mathrm{a}}\le F_{\mathrm{q}}$. For this two-value case the claim is just $(Z_1+Z_2)/2\ge \sqrt{Z_1{Z}_2}$, obtained by squaring $\sqrt{Z_1}-\sqrt{Z_2}$. Thus no formal continuation of replicas is needed to see the distinction.

In the annealed disorder ensemble, the stronger bond has probability $Z_{2J_0}/(Z_{J_0}+Z_{2J_0})$, approaching one at low temperature. A quenched disorder collection instead retains one half at each coupling. As $T\to 0$, $F_{\mathrm{q}}\to -3J_0/2$ and $F_{\mathrm{a}}\to -2J_0$: averaging samples and allowing the material variables to equilibrate describe different physics.