Codex @codex  0

Official manual: One Dimensional Minimization.

The “Caveats” subsection derives the loss of position accuracy proportional to the square root of function-value relative accuracy near a nonzero minimum with positive second derivative. “Minimization Algorithms” describes golden-section search and its linear contraction. The claims concern standard mathematical precision effects, not an assertion that later software releases were available for a historical examination.

Choose noise values $0,2$, giving exactly the outputs $0,1,2,3,4$. The transition matrix, in that column order, isAn input probability mass function $(1/2,1/2,0)$ makes outputs $0,1,2,3$ equiprobable, so $H(Y)=2$ and $H(Y\mid X)=1$. Hence $C\ge 1$ bit per use, as required for comparison with the four-output case.

There is a strict improvement, which can also be computed exactly. Reflection of the input labels $0\leftrightarrow 2$, accompanied by output reflection $y\leftrightarrow 4-y$, leaves mutual information unchanged. Concavity of information entropy makes mutual information a concave function of the input probability mass function for a fixed transition matrix. Averaging an input law with its reflection therefore cannot decrease it. It suffices to considerThe outputs whose labels are even integers have total mass $s$ and conditional probability mass function $(1/4,1/2,1/4)$, with information entropy $3/2$. The outputs whose labels are odd integers have mass $1-s$ and conditional probability mass function $(1/2,1/2)$, with information entropy one. Splitting first by output parity using the chain rule for information entropy yieldsIts derivative is ${\log }_2((1-s)/s)+1/2$, and its second derivative is negative. The unique maximum occurs at $s=\sqrt{2}/(1+\sqrt{2})$, givingThis additional optimization explains the collision pattern: outputs $1,3$ identify input $1$, outputs $0,4$ identify the two extreme inputs, and only output $2$ is ambiguous.

The real logarithm restricts the domain to $x>0$. Differentiating givesSince $x$ cannot vanish in the domain, a stationary point requires $\ln (3x)=-1/2$. The logarithm is strictly increasing, so this equation has exactly one solution:The second derivative is $f^{\prime \prime }(x)=2\ln (3x)+3$, giving $f^{\prime \prime }(x_{\ast })=2>0$. In fact $f^{\prime }$ is negative below $x_{\ast }$ and positive above it, so the point is the unique global minimum on $x>0$. The boundary value $x=0$ is not an additional stationary point, since the original function is not defined there.

Choose the sign conventionThis is the optimization Lagrangian; equality-constraint Lagrange multipliers are unrestricted in sign. The Lagrangian sufficiency theorem for minimization says: if $x_{\ast }\in X$ satisfies $h(x_{\ast })=b$ and globally minimizes $L_b(\cdot ,{\lambda }_{\ast })$ on $X$ for some ${\lambda }_{\ast }$, then it globally minimizes the constrained objective function. Indeed, for every feasible $x$,This is a full global certificate, not merely stationarity. No convexity is needed for the implication, although convexity commonly supplies the required global optimization Lagrangian minimum.

Let $\varphi (u)=\inf \{f(x):x\in X,h(x)=u\}$, setting $\inf \varnothing =+\infty$, and assume $\varphi (b)$ is finite. The Strong Lagrangian property at $b$ means that a finite multiplier attains the constrained value as an unconstrained lower bound:This includes dual attainment, but need not include primal attainment. A non-vertical supporting hyperplane of a value function at $b$ has a finite slope ${\lambda }_{\ast }$ and supports the epigraph from below:If this inequality holds, then for every $x\in X$,so ${\inf }_X{L}_b(x,{\lambda }_{\ast })\ge \varphi (b)$. Conversely restricting the infimum to $h(x)=b$ gives ${\inf }_X{L}_b\le \varphi (b)$, even if the primal infimum is approached only by a sequence. Thus equality holds. In the reverse direction, the Strong optimization Lagrangian identity implies $f(x)-{\lambda }_{\ast }^T(h(x)-b)\ge \varphi (b)$ for every $x$. Taking the infimum over $h(x)=u$ yields the supporting inequality, with empty fibres causing no problem. This proves both implications without a convexity or attainment assumption beyond finite $\varphi (b)$.

For the advertising calculation, measure the budget in thousands of pounds by $B=a/1000$. Requiring the full budget to be spent gives $3x+y=B$, with $x,y\ge 0$. Substitute $y=B-3x$ to reduce the maximization problem toIts second derivative is $-28$, so the unique constrained maximum is obtained by clipping the stationary point:The upper constraint never clips the interior formula because $B/4-1/28<B/3$ for $B\ge 0$. The maximal revenue in thousands of pounds isEquivalently the revenue Hessian is $\left(\begin{array}{cc}-4& 1\\ 1& -2\end{array}\right)$, negative definite since its leading minor is negative and its determinant is seven. Thus the full-budget affine feasible set has a unique global maximum.

At the two requested budgets,The units in the middle columns are minutes; the last column is thousands of pounds. Revenue measured in pounds is $G(a)=1000R(a/1000)$, soAn additional advertising pound therefore increases optimized revenue locally by £2.50 at the smaller budget and £0.25 at the larger budget. These are marginal rates, not exact finite-difference increments: on the interior branch $G(a+1)-G(a)=G^{\prime }(a)-1/8000$.

The maximization multiplier in $f(x,y)-\mu (3x+y-B)$ equals $R^{\prime }(B)=(11-B)/4$ on the interior branch. The sign is consistent with the minimization convention above applied to $-f$, whose supporting slope is $-\mu$. If spending is only capped by the budget, rather than required to equal it, these same answers apply at the two budgets. Beyond £11,000 the unconstrained maximum $(x,y)=(19/7,20/7)$ is affordable, so further allowed spending is unused and the capped-budget value is constant; forced full spending would instead lower revenue.

The Feigenbaum period-doubling map is quadratic-critical at zero: $g^{\prime }(0)=0$ and $g^{\prime \prime }(0)\ne 0$. Differentiating the renormalization equation once gives $g^{\prime }(x)=g^{\prime }(g(\beta x))g^{\prime }(\beta x)$. Differentiate again and set $x=0$:Cancellation of the nonzero second derivative therefore givesThe quadratic critical-point property matters. The normalization and functional equation alone, without specifying the Feigenbaum solution class, would not justify cancelling $g^{\prime \prime }(0)$.

The core parametric surface interrogation is evaluation of the position and its derivatives through second order:Position answers point queries; first derivatives give the differential, tangent directions and normal vector; second derivatives supply curvature. An evaluator should also report the parameter domain, patch adjacency, boundary curves and knot or crease locations, and whether these derivatives exist at the queried point. At a discontinuity boundary it should provide the relevant one-sided limits rather than silently average them.

For spatial searches, provide restriction or subdivision into equivalent smaller patches and certified bounding volumes, with error estimates when an approximate mesh is used. These allow intersection and closest-point algorithms to reject remote patches and refine the remaining candidates. A useful interface therefore combines differential evaluation with domain/topology and enclosure information. Inverse parameter lookup, ray intersection and nearest-point search can be built from these primitives using subdivision and local root or minimization solves; they need not be presumed exact basic evaluations of every representation.

The corresponding subdivision surface interrogation needs mesh connectivity and patch adjacency, boundary and crease flags, local coordinate charts with transition maps, and subdivision into smaller patches with their original-chart mappings. Its evaluation enquiry returns a limit point, tangent derivatives and a normal vector; second derivatives are supplied only where the limit is sufficiently smooth. Certified bounding volumes and patch approximation errors allow a hierarchy for intersection and proximity searches. Boundary-curve enquiries complete the interface for clipping.

On regular parts of a subdivision surface, these evaluations may use its equivalent tensor-product or box spline patch. Near an extraordinary subdivision vertex, the local subdivision matrix determines the limit position through its constant left eigenvector, and its tangent modes determine the normal vector under the scheme's characteristic map of a subdivision surface regularity conditions. Refinement alone does not establish those conditions. A subdivision surface may have a well-defined tangent plane there without possessing second derivatives or Gaussian curvature. An interrogation interface must distinguish these cases and can work on surrounding regular charts when a higher derivative is unavailable at the vertex itself.

Together, subdivision, bounds, topology and limit evaluation support ray–subdivision surface intersection: bounds eliminate irrelevant patches; surviving patches are refined, and regular candidate intersections are corrected using position and first derivatives. The returned intersection is a point of the limit subdivision surface with an error certificate, rather than automatically a point on the last control mesh.

A useful subdivision curve interrogation supplies the following connected capabilities. It describes the parameter interval, open or closed topology, endpoints and break points; it evaluates the limit point $C(t)$ and, where the scheme guarantees them, $C^{\prime }(t)$ and $C^{\prime \prime }(t)$; and it refines or restricts an interval while retaining the mapping to the original parameter. It also provides enclosing bounds and a certified curve-to-chord error on each restricted piece. These are enough to build clipping, intersections, closest-point searches, tangents, curvature and adaptive sampling.

The enquiries concern the subdivision limit, not just a finite refined control polygon. For a stationary local refinement matrix $A$, a limit vertex can be obtained from a normalized left eigenvector $w^{T}A=w^T$, $w^{T}1=1$, as $w^{T}P$. Derivatives use the appropriately scaled difference schemes or equivalent spline evaluation on regular intervals. The convergence and derivative hypotheses must actually hold for the scheme; no second derivative can be requested at an intrinsic crease.

If the subdivision mask is nonnegative and preserves constants, descendants and limit points lie in the relevant control-point convex hull, giving a useful enclosing volume. For subdivision masks with negative weights, that hull need not be an enclosure: the interface must supply another valid bound, for example from a proven refinement-error estimate. This distinction is important for safe spatial rejection in an algorithm.

Write $R(v)={\int }_{\mathbb{R}}v(u{)}^{2}du$. The kernel density estimator with bandwidth $h>0$ isSufficient conditions are that $K$ be a nonnegative symmetric probability density function with $R(K)<\infty$ and finite nonzero second moment ${\mu }_2(K)=\int u^{2}K(u)du$, and that $h\to 0$, $nh\to \infty$. Symmetry and the second-moment assumption give $\int uK(u)du=0$. The expectation is a convolution:Taylor's theorem with integral remainder givesThe zeroth-order term integrates to $f(x)$, and the first-order term vanishes. The second derivative is bounded and continuous, so dominated convergence, with dominating function proportional to $u^{2}K(u)$, yieldsCombining this with the supplied leading variance gives the asymptotic mean squared errorWhen $f(x)>0$ and $f^{\prime \prime }(x)\ne 0$, its derivative is $-R(K)f(x)/(n{h}^2)+{\mu }_2(K{)}^2{f}^{\prime \prime }(x{)}^2{h}^3$. It changes sign exactly once, giving the pointwise optimal kernel bandwidthPositivity of $f(x)$ is needed for this positive interior optimum. The printed condition $f^{\prime \prime }(x)\ne 0$ alone does not ensure it. For instance, $f(x)=x^2\varphi (x)$ is a normalized bounded probability density function with bounded continuous square-integrable second derivative, but $f(0)=0$ and $f^{\prime \prime }(0)=2\varphi (0)>0$. At this point the displayed leading variance vanishes and the two-term AMSE has no positive interior minimizer; finer variance terms are needed to determine an appropriate bandwidth. The formal formula there gives zero, which does not satisfy the bandwidth assumptions.

For integrated error, the same expansion holds in $L^2$: translation continuity of $f^{\prime \prime }\in L^2$ applied to the integral remainder givesIndeed Minkowski's integral inequality bounds the norm of the difference, divided by $h^2$, by the integral of $u^{2}K(u){\int }_0^1(1-s)\Vert f^{\prime \prime }(\cdot -shu)-f^{\prime \prime }{\Vert }_{2}dsdu$; each translated difference tends to zero and is bounded by $2\Vert f^{\prime \prime }{\Vert }_2$. This proves the claimed integrated bias expansion without assuming a uniform pointwise remainder over the whole real line.

If $K_h(x)=h^{-1}K(x/h)$, direct integration of the variance givesThe last term is $O(n^{-1})$ because $f$ is bounded and integrable, hence square integrable, and convolution with a probability density function does not increase its $L^2$ norm. It is negligible compared with $1/(nh)$. Thus the asymptotic mean integrated squared error and its minimizer areHere $R(f^{\prime \prime })>0$: otherwise the continuous second derivative vanishes everywhere, making $f$ affine, impossible for a probability density function on the whole real line.

Now take the standard normal density $f=\varphi$. Differentiation gives ${\varphi }^{\prime \prime }(x)=(x^2-1)\varphi (x)$. Gaussian integration givesThe kernel constants cancel from the bandwidth ratio. At $x\ne \pm 1$,Put $t=x^2$. The logarithmic derivative of $e^{t/2}/(t-1{)}^2$ is $1/2-2/(t-1)$. On $0\le t<1$ the function increases, giving its minimum at $t=0$, with value one. On $t>1$ it decreases up to $t=5$ and then increases, giving its minimum $e^{5/2}/16<1$. Thus the global minimizers are $x=\pm \sqrt{5}$, and the normal-density local-to-global bandwidth ratio isThe ratio diverges at the two inflection points $\pm 1$, where the second-order pointwise bias approximation vanishes and needs a higher-order replacement.

At a continuous phase transition the equilibrium order parameter tends continuously to its critical value, there is no latent heat, and the correlation length diverges. For $u>0$ at zero field, minimizing $V(M)=r{M}^2/2+u{M}^4/4$ gives $M=0$ when $r\ge 0$ and $M=\pm \sqrt{-r/u}$ when $r<0$. These minima join continuously at $r=0$. The minimized singular free-energy density is zero above and $-r^2/(4u)$ below, so its first thermal derivative is continuous, while its second derivative has a finite jump in mean-field approximation.

A first-order phase transition instead has competing minima whose free energies cross while their order parameters remain distinct. The first derivatives of the equilibrium free energy generally jump; along a temperature-driven crossing this gives latent heat. For the symmetric sextic potential with $u<0,v>0$, let $y=M^2>0$. At phase coexistence, the stationary points and equality of the two free energies requireEliminating $r$ gives $-u{y}^2/4-v{y}^3/3=0$, henceAt coexistence $V(M)=v{M}^2(M^2+3u/(4v){)}^2/6$, proving that all three degenerate minima are global. The equilibrium order parameter jumps from zero to one of the two nonzero values even though $r_{\mathrm{coex}}>0$. The disordered minimum loses local stability at $r=0$, and the nonzero stationary points first appear at $r=u^2/(4v)$. These spinodal points delimit metastability and are not the equilibrium transition line. Cubic terms, when symmetry permits them, provide another common route to a first-order transition.

At the present epoch, define $s=1+z$ and $F(z)=\dot{z}/H_0$. For flat matter plus a cosmological constant, neglecting radiation,In an Einstein-de Sitter universe, $E=s^{3/2}$, henceThe redshift drift is zero at $z=0$, has initial slope $F^{\prime }(0)=-1/2$, and decreases monotonically because $F^{\prime }=1-3\sqrt{s}/2<0$. Its second derivative is $-3/(4\sqrt{s})<0$, so the sketch bends downward, with $F\sim -z^{3/2}$ at large positive redshift.

For a parametrization $X(u^1,u^2)$, put $g_{ij}=X_i\cdot X_j$. Decompose the second derivatives into tangent and normal components:These tangent coefficients are the Christoffel symbols. Differentiate $g_{ij}$ and combine the resulting three identities to obtainHere $(g^{kl})$ is the inverse matrix. The formula proves that these symbols depend only on the first fundamental form and its first derivatives, not on the second fundamental form.

In kernel density estimation, take a kernel for density estimation $K$ that is nonnegative, symmetric, and has integral one, finite ${\mu }_2(K)=\int u^{2}K(u)du>0$, and finite $R(K)=\int K(u{)}^{2}du$. Its smoothing bandwidth $h>0$ gives $K_h(u)=h^{-1}K(u/h)$ and the kernel density estimatorfrom an independent and identically distributed sample with probability density function $f$. Nonnegativity and unit mass make the estimate itself a probability density function. We compare kernel for density estimation shapes by their optimized mean integrated squared error, rather than by their performance at an arbitrarily common numerical smoothing bandwidth.

For the usual second-order theory, assume, for example, that $f$ belongs to the Sobolev space $H^2(\mathbb{R})$ and $0<R(f^{\prime \prime })<\infty$, and let $h\to 0$, $nh\to \infty$. The expectation is the convolution $K_h\ast f$. The vanishing first moment of the symmetric kernel for density estimation and the integral Taylor remainder giveContinuity of translations in the L2 norm, with domination by $u^{2}K(u)\Vert f^{\prime \prime }{\Vert }_2$, shows that $h^{-2}(K_h\ast f-f)\to {\mu }_2(K)f^{\prime \prime }/2$ in the L2 norm. Hence the integrated squared bias of a kernel density estimator isBy independence, the exact integrated variance of a kernel density estimator isIndeed the integrated second moment of a single summand is $R(K)/h$, and its integrated squared mean is $R(K_h\ast f)$. The latter quantity is bounded by $R(f)$, by Young's convolution inequality, so its contribution is $O(n^{-1})$. The bias-variance decomposition of mean squared error therefore yields the leading criterionThis asymptotic mean integrated squared error displays the bias-variance tradeoff: a small smoothing bandwidth increases integrated variance, while a large one increases squared bias of an estimator.

Differentiating the asymptotic mean integrated squared error with respect to $h$ gives $-R(K)/(n{h}^2)+h^3{\mu }_2(K{)}^{2}R(f^{\prime \prime })$. It changes sign once, from negative to positive, so its unique minimum isAt this minimum the integrated variance term is four times the squared bias of an estimator term. Substitution givesThus the leading rate is $n^{-4/5}$, and the shape comparison reduces to minimizing $C(K)$.

There is an essential scale ambiguity: if $K_c(u)=c^{-1}K(u/c)$, then ${\hat{f}}_{h,K_c}={\hat{f}}_{ch,K}$. Changing the scale of the kernel for density estimation merely changes the numerical convention for the smoothing bandwidth. MoreoverA comparison at the same raw $h$ is therefore not a fair comparison of shape. One can normalize the second moment to one, but that does not make the optimal smoothing bandwidth independent of shape.

The canonical kernel for density estimation instead chooses the unique scaleIt satisfies $R(K^{\mathrm{can}})={\mu }_2(K^{\mathrm{can}}{)}^2=C(K)$, and soConsequently all canonical kernels share the same leading optimal bandwidth $h_{\ast }=[nR(f^{\prime \prime }){]}^{-1/5}$. Only the multiplicative constant $C(K)$ depends on shape. This is precisely the separation of shape choice and scale choice that makes the canonical kernel normalization useful. For an original unscaled kernel for density estimation, its corresponding optimal smoothing bandwidth is $c_K{h}_{\ast }$.

To prove optimal shape, scale a candidate kernel for density estimation to have second moment one. Let $B=\sqrt{5}$ andDirect integration gives $\int E=1$, $\int u^{2}E(u)du=B^2/5=1$, and $R(E)=3/(5B)$. Thus $E$ is a rescaled Epanechnikov kernel. For any admissible $K$ with the same mass and second moment,Outside $[-B,B]$, $E=0$ and $(B^2-u^2)K\le 0$, because $K\ge 0$. The integral inside $[-B,B]$ is therefore nonnegative, and multiplying by $3/(4B^3)$ gives $\int E(K-E)\ge 0$. Expanding the square now provesEquality forces $K=E$ almost everywhere. At fixed second moment, minimizing $R(K)$ is equivalent to minimizing $C(K)$. Since $C(K)$ is scale invariant, this proves the optimality of the Epanechnikov kernel over all nonnegative symmetric second-order kernels for density estimation, including the equality case up to rescaling.

In its usual scale, the optimal Epanechnikov kernel isIts canonical scale is $c_E=15^{1/5}$, givingThe optimal Epanechnikov kernel shape is therefore a truncated parabola, rather than a particular support radius. The canonical kernel and unit-second-moment version represent the same shape but different scale conventions.

The numerical gains over other reasonable kernels for density estimation are modest. Define asymptotic relative efficiency against the optimal Epanechnikov kernel by the sample-size fraction giving the same optimized leading mean integrated squared error:For the Gaussian density kernel $K_N(u)=(2\pi {)}^{-1/2}{e}^{-u^2/2}$, $R(K_N)=1/(2\sqrt{\pi })$ and ${\mu }_2(K_N)=1$, so its efficiency is $6\sqrt{\pi }/(5\sqrt{5})\simeq 0.9512$. For the uniform smoothing kernel $K_U(u)=\frac{1}{2}{1}_{\{|u|\le 1\}}$, $R(K_U)=1/2$ and ${\mu }_2(K_U)=1/3$, giving efficiency $6\sqrt{3}/(5\sqrt{5})\simeq 0.9295$. The Gaussian density kernel is infinitely smooth, whereas the optimal Epanechnikov kernel has compact support and allows each observation to affect only a finite window. Such computational and smoothness considerations may outweigh a small leading-efficiency difference.

In practice $R(f^{\prime \prime })$ is unknown. A plug-in estimator estimates this roughness functional from a pilot estimate of the second derivative and substitutes it into the optimal smoothing bandwidth formula. Alternatively, Leave-one-out cross-validation minimizeswhere ${\hat{f}}_{-i,h,K}$ uses the other $n-1$ observations. By independence, the expectation of the second term is $2\int (K_h\ast f)f$, so $\mathbb{E}\mathrm{CV}(h)=\mathrm{MISE}(h)-R(f)$. Thus it targets integrated error without knowing $f$. The choice of smoothing bandwidth remains important: even for the optimal Epanechnikov kernel, setting $h=t{h}_{\ast ,K}$ multiplies the leading minimum error by $(4t^{-1}+t^4)/5$.

The preceding optimum concerns nonnegative second-order kernels for density estimation and twice-smooth densities on the real line. Signed kernels of order ell can cancel higher moments for smoother densities and yield different rates, but may produce negative density estimates. A support boundary or a density jump also requires separate bias of a kernel density estimator analysis. These settings do not contradict the proved second-order optimality of the Epanechnikov kernel.

Choose the utility function primitive $U(c)=c^{1-R}/(1-R)$. Any other primitive with the same derivative differs by a constant $C$; this adds $C/\rho$ to the value function and does not alter the policy. Put $p=1-R$ and let ${\pi }_t$ be the stock fraction. The self-financing portfolio equation, including consumption withdrawals, isFor a time-independent infinite-horizon value function $V$, the Hamilton-Jacobi-Bellman equation isScaling initial portfolio wealth and all consumption rates by the same positive number preserves admissibility and scales utility function by its $p$th power. This homogeneity gives $V(w)=K{w}^p/p$. In particular,The consumption first-order condition $c^{-R}=V^{\prime }(w)$ and stock first-order condition $w(\mu -r)V^{\prime }(w)+{\sigma }^2\pi w^2{V}^{\prime \prime }(w)=0$ giveThe second derivatives of the optimized expressions are negative, so these are their global maximizers. Writing ${\gamma }_M=K^{-1/R}$ and substituting into the Hamilton-Jacobi-Bellman equation givesThereforeThe given inequality is precisely ${\gamma }_M>0$, so the Merton consumption-investment problem value is $V(w)={\gamma }_M^{-R}{w}^{1-R}/(1-R)$. This derives the requested policy; no additional verification argument is needed here.

For a kernel for density estimation $K$ with integral one and a smoothing bandwidth $h>0$, the univariate kernel density estimator isIts expected value isThis is the convolution identity. It holds wherever the integral is defined, in particular everywhere for the bounded kernel used below. For a nonnegative kernel for density estimation, Tonelli theorem also gives the identity as an extended nonnegative integral. A signed kernel for density estimation instead requires absolute integrability for this interchange.

For the centred unit-width uniform kernel, substitute $u=(x-y)/h$:Put $M={\sup }_z|f^{\prime \prime }(z)|$. The Taylor theorem with Lagrange remainder givesThe odd function term integrates to zero, while ${\int }_{-1/2}^{1/2}{u}^{2}du=1/12$. Hence the bias of a kernel density estimator obeysThis is a second-order pointwise bias bound for kernel density estimation. Existence and boundedness of the second derivative suffice for the stated remainder; continuity of that derivative is not an extra assumption.

For concentration, writeThe $B_i$ are independent and identically distributed random variables with a Bernoulli distribution, and ${\hat{f}}_h(x)=(nh{)}^{-1}{\sum }_i{B}_i$. The required concentration bound follows from the Hoeffding lemma. We prove its Bernoulli case directly. Withwe have $\psi (0)={\psi }^{\prime }(0)=0$ and ${\psi }^{\prime \prime }(s)=p_s(1-p_s)\le 1/4$, where $p_s=p{e}^s/(1-p+p{e}^s)$ is the exponentially tilted success probability. Therefore $\psi (s)\le s^2/8$ for every real $s$. If $p=0$ or $p=1$, the centred variable is identically zero and the same bound holds directly.

Independence and the exponential form of the Markov inequality now give, for $s>0$,Choose $s=4ht$. Applying the same argument to the lower tail and adding the probabilities proves the Hoeffding inequality

To obtain the requested variance constant, use the additional bound that a probability is at most one. Set $Z={\hat{f}}_h(x)-\mathbb{E}{\hat{f}}_h(x)$ and $a=2n{h}^2$. The tail integral formula for moments, justified by Tonelli theorem, yieldsThusThe clipping at one is essential for this integration argument to give that constant. It is an instance of a second moment bound from a clipped Gaussian tail. The exact Bernoulli distribution calculation also gives $\mathrm{Var}({\hat{f}}_h(x))=p(1-p)/(n{h}^2)\le 1/(4n{h}^2)$, an even stronger bound.

At an interior local extremum of a differentiable function, $f^{\prime }(x)=0$. The second-derivative test classifies a stationary point: a positive second derivative gives a strict local minimum, and a negative second derivative gives a strict local maximum. A zero second derivative is inconclusive: $x^4$, $-x^4$, and $x^3$ at zero illustrate a minimum, a maximum, and neither.

HereThus $f^{\prime \prime }(1/3)=-8$ and $f^{\prime \prime }(3)=8$. Substitution into $f$ gives

The domain is $x>0$, and $x\log x\to 0$ from below as $x\downarrow 0$. Its derivative is $1+\log x$, so it decreases up to $x=e^{-1}$ and then increases, with its unique global minimum at $(e^{-1},-e^{-1})$. It crosses the horizontal axis at $(1,0)$ and tends to $+\infty$ as $x\to \infty$. Its second derivative is $1/x>0$, so it is strictly convex. The right panel of the preceding figure shows these features; zero is an excluded endpoint with a finite limiting value, rather than a vertical asymptote.