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Substitute the dual linear program from the previous part and optimize jointly over $x,t,s$. The result isThe objective is linear, and the constraints are equalities or inequalities between affine functions, so this is a linear program. For each fixed feasible $x$, minimizing over $t,s$ gives exactly $f_k(x)$ by the threshold formula for the sum of the largest components; hence the reformulation preserves the optimal value and optimal $x$ whenever an optimum exists.

With $h=\Delta x$, $\tau =\Delta t$, $r=\tau /h^2$ and $c=\alpha \tau /h$, the Forward Euler method givesFor a Fourier mode $e^{im\theta }$, the amplification factor is $G(\theta )=1-4r{\sin }^2(\theta /2)+ic\sin \theta$. Set $s={\sin }^2(\theta /2)$. ThenThe expression in braces is an affine function of $s$, so it is nonpositive for all $s\in [0,1]$ precisely when its two endpoint values are nonpositive. For $\tau >0$, these conditions are $c^2\le 2r$ and $r\le 1/2$. Thus Forward Euler stability for centered advection-diffusion requires exactlyFor $\alpha =0$, only the first condition remains. If either bound fails, a band of Fourier modes has $|G|>1$, giving instability under von Neumann stability analysis. At equality the amplification bound still holds, so the endpoints are included. This is discrete L2 norm stability; nonnegative stencil weights would impose the additional spatial restriction $|\alpha |\Delta x\le 2$.

Nash's theorem says that every finite game, including a two-player game with two pure choices each, has at least one Nash equilibrium in mixed strategies.

For player $1$, holding $p_2$ fixed, the expected value of the payoff isIt is an affine function of $p_1$. Therefore player $1$ chooses $p_1=1$ when $1-(U-1)p_2>0$, chooses $p_1=0$ when it is negative, and is indifferent when it vanishes. By symmetry, player $2$ has the analogous best response.

If $U<2$, then $1-(U-1)p>0$ for every $p\in [0,1]$, so becoming a Dark Lord with probability one is a strictly dominant strategy. The unique equilibrium isIf $U>2$, each player is indifferent when the other chooses $p=1/(U-1)$. This gives one interior mixed equilibrium, while the two asymmetric pure equilibria arise because a player facing a certain peasant chooses Lord and a player facing a certain Lord chooses peasant. The three equilibria areConsequently $\boxed{U_0=2}$. At the threshold itself there is a continuum of equilibria with $p_1=1$ or $p_2=1$, consistent with the question's strict inequalities.

The empirical risk isFor $g(z)={\log }_2(1+e^{-z})$,so $g$ is a convex function. Each signed margin $y_i{x}_i^T\beta$ is an affine function of $\beta$, and composition with an affine function preserves convexity. Equivalently,is a positive semidefinite matrix. Therefore $f$ is convex.

Suppose $|Dv|\le M$. The coefficient matrices in part (ii) then have uniform ellipticity constants depending only on $M$. Applying the Harnack inequality for uniformly elliptic divergence-form equations to the nonnegative solutions $M+D_k{v}_R$ and $M-D_k{v}_R$ gives a scale-independent oscillation contractionScaling back,For fixed $r$, iterate this estimate with $R=4^{m}r$ and use the global bound $|D_{k}v|\le M$ to obtain ${\mathrm{osc}}_{B_r}{D}_{k}v\le 2M{\theta }^m\to 0$. Every partial derivative is therefore constant, so $v(x)=a\cdot x+b$ is an affine function. This is a bounded-gradient Bernstein theorem for entire minimal graphs.

Each map $x\mapsto ⟨a_i,x⟩+b_i$ is an affine function. For $0\le t\le 1$, the pointwise maximum of convex functions satisfiesso $f$ is convex.

A vector $g$ is a subgradient of a convex function $f$ at $x$ whenfor every $y$. Choose any active index $j\in I(x):=\{i:f(x)=⟨a_i,x⟩+b_i\}$. Thenand hence $\boxed{a_j\in \partial f(x)}$. More generally, every convex combination of the active vectors is a subgradient, and in fact

Write $z_i(x)=⟨a_i,x⟩+b_i$. The log-sum-exp function is convex and composition with the affine functions $z_i$ preserves convexity, so $f_{\beta }$ is convex. Directly, its Hessian matrix will also be shown positive semidefinite in part d.

Let $M=f(x)={\max }_i{z}_i(x)$. Factoring $e^{\beta M}$ out of the sum givesAt least one term in the sum is $1$, while every term is at most $1$. Thereforeand henceThus $f_{\beta }$ is a uniform smooth maximum of the affine pieces of $f$.

Insert the linear initial profile into the transformed integral from (b):The constant part integrates to $ax$. The other part is an absolutely integrable odd function, so its integral vanishes. HenceAs a direct check, $u_t=u_{xx}=0$ and $u(x,0)=ax$. Thus a linear temperature profile is stationary on the whole line. More generally, every affine function is preserved by this heat-kernel solution.

The divergence isFor the tetrahedral domain, convenient integration bounds areConsequently its volume integral isPutting $h=2(1-x-y)$, the inner integral is $(2y+6)h+h^2=(16-4x)(1-x-y)$. ThereforeThusThere is an independent centroid check. The vertices are $(0,0,0),(1,0,0),(0,1,0),(0,0,2)$, giving volume $1/3$ and centroid $(1/4,1/4,1/2)$. An affine function averages to its value at the centroid, so $I=(1/3)(6+2/4+2/2)=5/2$. This is the integral of an affine function over a simplex.

Interpret the stated transition as an upper critical miscibility temperature: cooling increases the effective $\chi$ through ${\chi }_c$. Above $T_c$, the homogeneous mixing free energy is convex throughout $0<\varphi <1$, with no two-phase common tangent. Below $T_c$, it acquires a region with $f^{\prime \prime }<0$, bordered by the spinodal. A lower straight line touches the curve at two binodal compositions ${\varphi }_{-}$ and ${\varphi }_{+}$; this common-tangent construction gives the equilibrium polymer-poor and polymer-rich phases.

The precise raw curve need not have two equal minima: subtracting an affine function of $\varphi$ changes reference chemical potentials without changing curvature or coexistence compositions. The free-energy panels above subtract the tangent at ${\varphi }_c$ from the upper-temperature curve and the common tangent from the lower-temperature curve. This makes the common tangent and the convex-versus-nonconvex distinction visible.

If the mean initial composition ${\varphi }_0$ lies inside the low-temperature spinodal region, an arbitrarily small separation into equal-volume regions with compositions ${\varphi }_0\pm \delta \varphi$ changes the mean free energy bySmall fluctuations therefore grow spontaneously: this is spinodal decomposition. Interfacial costs limit very short-wavelength fluctuations, and the polymer-poor and polymer-rich domains subsequently coarsen towards their binodal compositions.

If ${\varphi }_0$ lies just outside a spinodal branch but still inside the binodal, then $f^{\prime \prime }({\varphi }_0)>0$. Small fluctuations increase free energy, so the uniform mixture is metastable. Separation requires a sufficiently large nucleus to overcome the interfacial barrier, followed by nucleation and growth. There is no guarantee that this happens rapidly. If the initial composition is outside the binodal as well, it remains a stable uniform phase after cooling.

For a mean composition in the coexistence interval, conservation of polymer fixes the volume fraction $\alpha$ of the rich phase by the lever rule:The temperatures alone do not determine which mechanism occurs without the initial composition. Numerical free-energy curves additionally require $\chi (T)$; the plotted example uses $\chi (T)={\chi }_c{T}_c/T$ with absolute temperatures $T_c=298.15\mathrm{K}$, $T_1=303.15\mathrm{K}$ and $T_2=293.15\mathrm{K}$.

The Pearson correlation coefficient is the covariance divided by the product of the standard deviations. For real random variables with finite, strictly positive variances,It is the expected value of the product of the two centered variables after dividing each by its standard deviation. The Cauchy-Schwarz inequality gives $-1\le {\rho }_{X,Y}\le 1$. This removes measurement units, while retaining the direction and strength of the linear relationship. Constant random variables have zero standard deviation, so their Pearson correlation coefficient is undefined.

For the covariance model $Y=2X+Z$, with $X,Z$ independently uniform on $\{-1,0,1\}$, the calculation givesThe positive correlation is less than one because the independent $Z$ term introduces extra variation. For $Y=-2X+Z$ the correlation is $-2/\sqrt{5}$. Both examples appear in the covariance diagram.

For nonzero real $a,c$, the correlation of $aX+b$ and $cY+d$ is $\mathrm{sgn}(ac){\rho }_{X,Y}$. Positive rescaling and added offsets therefore leave the correlation unchanged; reversing one scale reverses its sign. Equality $|{\rho }_{X,Y}|=1$ means that one random variable is an affine function of the other almost surely, with a nonzero slope of the corresponding sign. This is the equality condition in the Cauchy-Schwarz inequality, and includes perfect positive correlation.

Zero correlation says that the centered variables are orthogonal in their square-integrable inner product. It permits nonlinear statistical dependence, as the example $Y=X^2$ in uncorrelated random variables demonstrates. More generally, a correlation is a statement about a joint probability distribution; identifying effects of interventions is the separate task of causal inference. This distinction keeps an observed association from being mistaken for a demonstrated mechanism.

An early source for standardized association is Francis Galton's 1888 paper “Co-relations and their Measurement, chiefly from Anthropometric Data”. He compared stature and forearm measurements and changed each measurement to units of its own variability so the two regression directions could be compared. His scale used probable error, so it should not be identified word for word with the modern standard deviation formula above. ^[ref].

A nonconvex primal problem can have a positive duality gap even when both optima are attained. Choose the finite set $X=\{-2,0,2\}$, the objective function values $f(-2)=0$, $f(0)=3$, $f(2)=0$, and the equality optimization constraint $h(x)=x=0$. The only feasible point is $0$, so $p_{\ast }=3$.

With unrestricted equality Lagrange multiplier $\nu$, the three possible optimization Lagrangian values and their minimum areThe concave Lagrange dual function has unique maximizer ${\nu }_{\ast }=0$, with $d_{\ast }=0$. Thus the duality gap is $3$. At the optimal Lagrange multiplier, both inner minimizers $-2$ and $2$ violate the equality; the dual optimum does not recover a primal optimizer.

Why can no slope close the gap? Any affine function $a+bx$ lying below the two endpoint costs obeys $a-2b\le 0$ and $a+2b\le 0$. Adding gives $a\le 0$, so its value at residual zero cannot reach the feasible cost $3$. A convex combination of the endpoint cost-residual pairs with equal weights has residual zero and average cost zero. This averaged pair is unattainable: the actual decision $x=0$ costs $3$. This geometry explains precisely what the dual bounds fail to distinguish. Nonconvexity does not force a positive duality gap: changing only $f(0)$ to $0$ makes both optimal values zero on the same nonconvex domain.

The earliest approach within distance $d$ is a convex optimization problem with a norm constraint and the same segment and time bounds as contact. An elementary alternative enumerates whether each segment parameter is free or at an endpoint. In each nonsingular active case the free closest-point parameters are affine functions of time, leaving a quadratic distance inequality on a valid time interval. Rank-deficient cases need their null-space constraints retained. Unlike contact, the two segment parameters and time are no longer determined by a vector equality.

Choose the illustrative local modelHere $t$ is the Celsius temperature, measured relative to $0{}^{\circ }\mathrm{C}$, so $\alpha t$ is dimensionless. For this chosen affine function, the derivative and inverse relation areCalculated values are $R(0{}^{\circ }\mathrm{C})=100\Omega$, $R(20{}^{\circ }\mathrm{C})=107.70\Omega$, $R(25{}^{\circ }\mathrm{C})=109.625\Omega$, and $R(50{}^{\circ }\mathrm{C})=119.25\Omega$. These are model outputs, not measurements or an assertion that any real Pt100 follows this line exactly. The number $0.00385$ here is the chosen slope parameter; it is not a universal local derivative for all platinum resistance thermometers. Curvature, manufacturing differences, and the fitted calibration interval matter. A wider-range calibration uses a validated response curve rather than extrapolating this line indefinitely.

For a separate synthetic thermometer calibration, stipulate $R_a=100.20\Omega$ at $t_a=0{}^{\circ }\mathrm{C}$ and $R_b=119.60\Omega$ at $t_b=50{}^{\circ }\mathrm{C}$. A two-point affine function has slope $0.388\Omega /{}^{\circ }\mathrm{C}$ and givesThus $R=109.90\Omega$ would indicate $25{}^{\circ }\mathrm{C}$. If an additional synthetic check at reference $25{}^{\circ }\mathrm{C}$ instead supplied $110.10\Omega$, the line would indicate $25.515\dots {}^{\circ }\mathrm{C}$. The $0.20\Omega$ residual exposes disagreement that two fitted points cannot reveal. This separate example changes both offset and slope; it is not a replacement specification for the nominal Pt100 local linear model.

A quadrature rule approximates an integral by a finite weighted sum of sampled function values:The quadrature nodes $x_i$ specify where to sample, and the quadrature weights $w_i$ specify how much each sample contributes. The density $\rho (x)$ specifies the target integral, whereas the discrete quadrature weights belong to the approximation. Ordinary numerical integration has $\rho =1$. Gaussian quadrature and the midpoint quadrature rule choose interior quadrature nodes; the trapezoidal rule includes endpoints.^[ref]

For ordinary numerical integration, exactness on the constant function $1$ requires ${\sum }_i{w}_i=b-a$; exactness on $x$ additionally requires ${\sum }_i{w}_i{x}_i=(b^2-a^2)/2$. These two checks make the quadrature rule exact for every affine function. For example, both the midpoint quadrature rule and trapezoidal rule give $6$ for ${\int }_0^2(2x+1)dx$. Testing progressively higher polynomial degrees defines the degree of precision of a quadrature rule. Exactness describes a class of integrands, rather than promising exact results for every function.

Composite midpoint quadrature rules and composite trapezoidal rules apply a simple quadrature rule on every panel of a partition of an interval. This buys accuracy by shrinking the panels while keeping the elementary formula fixed. For continuous functions sampled exactly, their quadrature errors tend to zero as the largest panel width tends to zero. The familiar second-order bounds require stronger assumptions about derivatives. The signed approximation error used here is $Q(f)-I_{\rho }(f)$; some sources define quadrature error with the opposite sign, so the convention must accompany any formula.

A roughness penalty penalizes variation of derivatives, such as $\int (m^{\prime \prime }{)}^2$. The second-derivative penalty vanishes precisely on affine functions and is represented in a chosen basis by a roughness penalty matrix.

The nonnegative functional $J(g)={\int }_a^b(g^{\prime \prime }(x){)}^{2}dx$. Its null space on $C^2[a,b]$ consists of affine functions. The natural cubic spline interpolant uniquely minimizes this penalty among functions matching at least two prescribed knot values. It is a squared second derivative penalty, not the exact geometric curvature energy of a graph.

The Haar wavelet provides an explicit one-vanishing moment comparison:It cancels a constant polynomial, but not a nonconstant affine function. To obtain two vanishing moments by an equally transparent construction, take three adjacent unit intervals with heights proportional to $1,-2,1$:The L2 norm normalization follows from $1^2+(-2{)}^2+1^2=6$. Direct integration givesThus $\eta$ has exactly two vanishing moments. In fact its coefficient is a finite difference of adjacent local integrals:The weights are those of the second-order central difference. The first moment balances the centres of the three intervals; it is not enough merely to balance their signed areas.

This $\eta$ is a normalized analyzing function, with no orthonormal basis assertion. Indeed, its unit translate has nonzero L2 inner product:Hence these translates already fail orthogonality. The Daubechies wavelet construction imposes additional refinement and orthogonality requirements; two vanishing moments and unit L2 norm by themselves do not supply those requirements.

A semidefinite program optimizes a linear function subject to equalities between affine functions and positive semidefinite matrix inequalities. It generalizes linear programming: a diagonal matrix is a positive semidefinite matrix exactly when its diagonal entries are nonnegative. Matrix trace expresses the objective as $⟨C,X⟩=\mathrm{tr}(CX)$ for real symmetric matrices.