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A cubic spline belongs to $C^2[a,b]$ and is a polynomial of degree at most three on each interval separated by the spline knots. A natural cubic spline is additionally an affine function on the exterior intervals $[a,x_1]$ and $[x_n,b]$. On $[x_1,x_n]$ this is equivalent to the endpoint conditions $s^{\prime \prime }(x_1)=s^{\prime \prime }(x_n)=0$; these conditions remain part of the definition when an outer knot coincides with $a$ or $b$.

Let $s$ be the natural cubic spline interpolant of the prescribed values, and let $v$ be any other interpolating $C^2$ function. Set $u=v-s$, so $u(x_i)=0$. On the interval $[x_i,x_{i+1}]$, integration by parts twice givesbecause the fourth derivative of each cubic piece is zero. The second boundary term is zero because $u$ vanishes at the spline knots. After summing the first terms, the interior contributions cancel by continuity of $s^{\prime \prime }$ and $u^{\prime }$, and the two remaining terms vanish by the natural endpoint conditions. On the exterior intervals $s^{\prime \prime }=0$. Thus ${\int }_a^b{s}^{\prime \prime }{u}^{\prime \prime }=0$, and the second derivative roughness penalty satisfiesIf equality holds, the continuous nonnegative function $(u^{\prime \prime }{)}^2$ has zero integral, so $u^{\prime \prime }=0$ everywhere. Hence $u$ is an affine function, and its zeros at two distinct spline knots force $u=0$. This proves the minimum roughness property of the natural cubic spline interpolant and its equality case: the unique minimum-roughness interpolant is $s$.

For the fixed-design nonparametric regression fit, use the penalized least squares criterionLet $s_z$ be the natural cubic spline interpolant with value vector $z=(z_1,\dots ,z_n{)}^T$. Replacing $v$ by $s_z$ for $z_i=v(x_i)$ leaves the residual sum unchanged and reduces the second derivative roughness penalty, strictly unless $v=s_z$. It therefore remains to minimize the finite-dimensional quadratic formThe supplied spline roughness penalty matrix $K$ is positive semidefinite. For every nonzero vector $w$,Consequently $I+\lambda K$ is a positive-definite matrix, and $Q_{\lambda }$ is strictly convex. Its normal equations for linear least squares have the unique solutionThe preceding strict reduction also proves uniqueness over the whole class $C^2[a,b]$, not just over the natural cubic splines. This is the cubic smoothing spline. If the residual criterion is divided by $n$ instead, the corresponding formula is $(I+n\lambda K{)}^{-1}Y$, reflecting a change in the numerical convention for the smoothing parameter.

The independence and specified error mean and variance are not needed for existence or uniqueness of this minimizer. They do imply, with smoothing matrix $A_{\lambda }=(I+\lambda K{)}^{-1}$ and $g_X=(g(x_i){)}_i$, that $\mathbb{E}\hat{z}=A_{\lambda }{g}_X$ and $\mathrm{Cov}(\hat{z})={\sigma }^2{A}_{\lambda }^2$. The null space of $K$ consists of sampled affine functions: these have zero roughness, and conversely zero roughness forces their natural cubic spline interpolant to be an affine function. Thus these functions are reproduced without bias of an estimator, while other components are smoothed.

Write $S_n(t)={\sum }_{r=-n}^n{a}_r{e}^{irt}$. First we prove decay of coefficients of an almost-everywhere convergent trigonometric series rather than assuming the series can be integrated term by term. For $n\ge 1$, $T_n(t)=S_n(t)-S_{n-1}(t)=a_n{e}^{int}+a_{-n}{e}^{-int}$ tends to zero outside the finite set $E$. Suppose $M_n=|a_n|+|a_{-n}|$ does not tend to zero, and choose a subsequence with $M_n\ge \epsilon >0$. Then $|T_n/M_n|\le 1$ and $T_n/M_n\to 0$ almost everywhere. The dominated convergence theorem givesorthogonality of the two distinct Fourier modes instead makes the integral $(|a_n{|}^2+|a_{-n}{|}^2)/M_n^2\ge 1/2$, a contradiction. Thus $\boxed{a_n,a_{-n}\to 0}$.

Define on the real line the twice-integrated continuous functionThe series has both absolute convergence and uniform convergence because its coefficients are bounded. Its symmetric second quotient is therefore, for nonzero $h$,We next prove that this tends to zero whenever $t\notin E$, lifted periodically to the line. Let $b_0=a_0$ and $b_n=a_n{e}^{int}+a_{-n}{e}^{-int}$ for $n\ge 1$, so the partial sums of the $b_n$ are exactly $S_n(t)\to 0$. Put $k=|h|/2$ and $w_n=w(nk)$. Summation by parts givesThe boundary term vanishes because $S_n(t)$ is bounded and $w_n\to 0$ for fixed nonzero $k$. MoreoverIndeed $w^{\prime }(x)=O(x)$ near zero and $|w^{\prime }(x)|\le 2/x^2+2/x^3$ for $x\ge 1$. Each fixed weight difference tends to zero as $k\to 0$. Splitting the last series at an index beyond which $|S_n(t)|<\eta$ therefore bounds its upper limit by $C\eta$. Letting $\eta \to 0$ proves the claimed zero second quotient. This explicitly proves the needed Riemann summation of a convergent trigonometric series, despite the oscillations in its weights.

We also need the elementary fact that a continuous function with zero symmetric second derivative at every interior point of an interval is affine. For a real-valued function, subtract its endpoint chord on any compact subinterval $[a,b]$. If the difference were positive somewhere, adding $\epsilon (t-a)(t-b)$ for sufficiently small $\epsilon >0$ would retain a positive maximum at an interior point. At that maximum each symmetric second difference is nonpositive, while its quotient tends to $2\epsilon >0$, a contradiction. Thus the function lies below its chord. Apply the argument to its negative to get the reverse inequality. Applying this to real and imaginary parts proves the complex-valued case as well. This establishes that vanishing symmetric second derivative forces an affine function. Hence $F$ is an affine function on each open interval between the finitely many exceptional points in any period.

It remains to exclude corners at those points. Multiplying the formula for the second quotient by $h$, and using $h=2k$ and the evenness of $w$, givesfor every $t$, by the infinite-series version of part (i), since $a_n,a_{-n}\to 0$. continuity extends the adjacent affine pieces to each exceptional point, and the fact that symmetric second differences detect an affine corner makes their slopes and intercepts agree. Thus $F$ is globally an affine function on the real line.

The periodic part of its definition gives $F(t+2\pi )-F(t)=2\pi a_{0}t+2{\pi }^2{a}_0$. An affine function has constant increments, so $a_0=0$. The resulting $F$ is both an affine function and a periodic function, hence constant. Finally, absolute uniform convergence justifies integrating its defining series against $e^{-irt}$, and for $r\ne 0$ the resulting Fourier coefficient is $-a_r/r^2$. A constant has all these coefficients zero. Therefore $\boxed{a_r=0\text{ for every }r\in \mathbb{Z}}$, proving uniqueness of a trigonometric series outside a finite set with all exceptional points accounted for.

An affine function preserves mixtures, soThereforeThe positive mixture weight preserves the strict sign; the common component cancels. This is the independence axiom for lottery preferences for the affine utility on probability measures.

Indifference means equality of the representing utility values. Since $U_0(\lambda )>U_0(\mu )>U_0(\nu )$, setApplying the affine function identity gives $U_0(p\lambda +(1-p)\nu )=U_0(\mu )$, so $\mu \sim p\lambda +(1-p)\nu$. This proves mixture solvability of affine preferences. The displayed weight is unique because the two endpoint utilities differ.

First, equality of two $U_0$ values forces equality of the corresponding $V_0$ values: neither strict comparison holds for $U_0$, and the same is then true for $V_0$ in both directions. Thus $V_0=g\circ U_0$ for a well-defined strictly increasing function $g$ on $J=U_0(\mathcal{P})$. Since $\mathcal{P}$ is a convex set and $U_0,V_0$ are affine functions, $J$ is a real interval andIf $J$ contains $u<v$, let $a=[g(v)-g(u)]/(v-u)>0$ and $b=g(u)-au$. The mixture identity gives $g(w)=aw+b$ for $w\in [u,v]$. For $w>v$, express $v$ as a convex combination of $u$ and $w$ and solve the same identity for $g(w)$; for $w<u$, express $u$ as a mixture of $w$ and $v$. Hence the formula holds on all of $J$, not merely between the chosen anchors. We obtain positive affine uniqueness of affine preference representations:If $U_0$ is constant, the same comparisons force $V_0$ to be constant; choose $a=1$ and the appropriate $b$. If the probability-measure set is empty, the assertion is vacuous. These degenerate cases do not require distinct anchors.

The feasible triangle has verticesThese come from the three pairs of active boundary lines and satisfy the remaining inequalities. Its denominator $3x_1+x_2+2$ is positive at each vertex, with minimum $12/5$, so is positive throughout the triangle because it is an affine function.

For a direct linear programming optimality certificate, add twice the second inequality to the third to get $-x_1-3x_2\le 2$. Thus $x_1+3x_2+2\ge 0$, equivalently $2(x_1-x_2)\le 3x_1+x_2+2$. Division by the positive denominator gives an objective at most $1/2$. Both inequalities used in the bound are equalities at $v_3$, where the first inequality is also satisfied. ThereforeEquality requires the two bounding inequalities to be tight, so this optimizer is unique.

First apply Lebesgue decomposition to the derivative measure relative to ${\mathcal{L}}^n$:Here $\nabla u$ is the almost-everywhere approximate gradient, rather than an assertion that $u$ belongs to $W^{1,1}$. Split the singular part into its jump part of a BV derivative and Cantor part of a BV derivative:An approximate jump point has a unit normal ${\nu }_u$ and distinct finite BV traces on a hypersurface $u^{+},u^{-}$, obtained as mean limits on the corresponding two half-balls. Their set $J_u$ is the jump set of a BV function, countably $(n-1)$-rectifiable. Reversing the normal swaps the BV traces on a hypersurface and leaves the displayed measure unchanged. The approximate discontinuity set $S_u$ differs from $J_u$ only by an ${\mathcal{H}}^{n-1}$-null set.

The remaining $D^{c}u=D^{s}u-D^{j}u$ is singular to Lebesgue measure and gives zero mass to every set with sigma-finite ${\mathcal{H}}^{n-1}$ measure. It is diffuse rather than a second jump contribution. In one dimension the three parts are illustrated by an affine function, a step function and the Cantor function, respectively. Countably many jumps therefore do not imply that the singular derivative has no Cantor part of a BV derivative.

Use the standard setting of a compact metrizable convex set $K$ in a Hausdorff locally convex real topological vector space, with its metrizable topology. Write $A(K)$ for the continuous affine real functions on $K$. The affine upper envelope of a bounded real function isConstants make this infimum finite, and $f\le \overline{f}$. An infimum of affine continuous majorants is concave and upper semicontinuous, hence Borel. For continuous $f$ it is the upper concave envelope appropriate to barycentric measures; it is not the pointwise maximum of $f$ and a selected affine function.

For a fixed probability $\mu$, define $p(g)={\int }_K\overline{g}d\mu$ on real $C(K)$. Affine majorants showThus $p$ is sublinear and $p(a)=\int ad\mu$ for affine $a$. On the span of $f$, the linear functional taking $f$ to $p(f)$ is dominated by $p$: the negative-scalar condition follows from $-p(f)\le p(-f)$. For $f=0$ start from the zero subspace. The Hahn-Banach theorem extends it to a linear functional $L$ on $C(K)$ with $L\le p$ and $L(f)=p(f)$.

If $g\ge 0$, then $p(-g)\le 0$, so $L(g)\ge 0$. Also $p(1)=1$ and $p(-1)=-1$, forcing $L(1)=1$. Positivity gives $|L(g)|\le \Vert g{\Vert }_{\infty }$, and the Riesz-Markov-Kakutani representation theorem produces a Borel probability $\nu$ with $L(g)=\int gd\nu$. ThereforeFor affine $a$, testing both $a$ and $-a$ gives $\int ad\nu =\int ad\mu$: the two measures have the same barycenter. This proves the requested supporting measure lemma for affine upper envelopes.

Choquet's theorem: every $x\in K$ has a Borel probability measure $\lambda$ concentrated on the extreme points of $K$ whose barycenter is $x$; equivalently,Concentration is a measure-one assertion, not a claim that the topological support must be a closed subset of $\mathrm{Ex}K$.

To prove it, let ${\mathcal{M}}_x$ be the measures satisfying all the displayed affine equalities. It is nonempty because it contains ${\delta }_x$, and it is weakly closed in the compact space $P(K)$, hence compact. Let $h\in C(K)$ be strictly convex, as permitted. Choose $\lambda \in {\mathcal{M}}_x$ maximizing $\int hd\lambda$. Apply the supporting measure lemma with $\mu =\lambda$ and $f=h$. It gives $\nu$ with the same affine integrals, so $\nu \in {\mathcal{M}}_x$, andThus $\overline{h}-h\ge 0$ has integral zero. If $z$ is not extreme, write $z=ty+(1-t)w$ with $0<t<1$ and distinct $y,w\in K$. Strict convexity and every affine majorant giveTherefore the nonnegative gap is strictly positive at every nonextreme point. It is Borel, and $\mathrm{Ex}K$ is Borel by the allowed G-delta set assertion. Its zero integral forces $\lambda (K\setminus \mathrm{Ex}K)=0$, proving Choquet's theorem by strict convexity. No uniqueness is asserted for the representing measure.

For the real $L^{\infty }$ example, give its closed unit ball $K$ the weak-star topology $\sigma (L^{\infty },L^1)$, not the norm topology. The Banach-Alaoglu theorem makes it compact, and separability of $L^1[0,1]$ makes this ball metrizable. Its extreme points are exactly the classes $h$ satisfying $|h|=1$ almost everywhere. Indeed, if $|h|\le 1-\epsilon$ on a positive-measure set, adding and subtracting $\epsilon$ times its indicator decomposes $h$ nontrivially inside the ball. Conversely, if $|h|=1$ almost everywhere and $h=(u+v)/2$ with $|u|,|v|\le 1$, pointwise equality at the endpoints of $[-1,1]$ forces $u=v=h$ almost everywhere. This is the extreme-point criterion for the L-infinity unit ball.

For the hinted case $f=1_A-1_B$, put $C=[0,1]\setminus (A\cup B)$ and $h_{\pm }=1_A-1_B\pm 1_C$. Then $h_{\pm }$ are extreme andhas barycenter $f$. If $C$ is null, the two point masses coincide.

For a general real $f$, choose a measurable representative in $[-1,1]$ and set, for $0\le t\le 1$,Every $H_t$ is extreme. The map into the weak-star compact ball is Borel: for each $u\in L^1$, the function $t\mapsto \int u(s)H_t(s)ds$ is measurable by joint measurability and integration; a countable dense family of such tests generates the ball's topology and Borel sigma-algebra. The measure is independent of changes to $f$ on a null set. For each $s$,The integrand paired with $u$ is dominated by $|u|$, so Fubini's theorem yieldsThese continuous linear tests define the weak-star barycenter, hence that barycenter is $f$. Together with $\nu (\mathrm{Ex}K)=1$, this is the required threshold Choquet representation in L-infinity.

If $L^{\infty }$ is instead taken over complex scalars, its extreme points satisfy the same unit-modulus condition. Write $f(s)=r(s)\zeta (s)$, with $r=|f|$ and $|\zeta |=1$, choosing $\zeta =1$ where $f=0$. Replace the threshold family by $\zeta (s)(21_{\{t\le (1+r(s))/2\}}-1)$. Its members have unit modulus and its average is $f$, so the same pushforward and Fubini argument gives a representing measure in the real locally convex interpretation of the complex ball.

The chain rule for Fréchet derivatives states that if $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, thenThe composition on the right is a composition of linear maps, equivalently a product of Jacobian matrices in coordinates.

Apply the chain rule to the affine function $x\mapsto (x,c-x)$. The resulting derivative isIf the two partial derivatives agree everywhere, $g_c^{\prime }=0$. The mean value theorem makes $g_c$ constant on $\mathbb{R}$. Define $h(c)=f(c,0)$, which is itself differentiable by the chain rule. Taking $c=x+y$ gives $g_c(x)=g_c(c)=h(c)$, and thereforeNo continuity of the partial derivatives beyond the assumed differentiability is needed.

With $K\ge 2$ distinct knots ${\kappa }_1<\cdots <{\kappa }_K$, a natural cubic spline is a $C^2$ function on the real line, cubic on each interval between consecutive knots and affine outside the two outer knots. The latter condition gives $m^{\prime \prime }({\kappa }_1)=m^{\prime \prime }({\kappa }_K)=0$.

There are $4(K-1)$ coefficients for the interior cubic pieces. Matching values and the first two derivatives at the $K-2$ interior knots imposes $3(K-2)$ independent constraints, and the two natural endpoint conditions impose two more. The affine continuations are then determined by endpoint values and slopes. The dimension of a vector space is thereforeOne explicit basis uses $1,x$ and $d_j(x)-d_{K-1}(x)$ for $j=1,\dots ,K-2$, whereThe cubic and quadratic terms cancel in each difference above ${\kappa }_K$, making it affine there; below ${\kappa }_1$ the truncated powers vanish. The changes in third derivative at individual knots show independence of these basis functions. This count uses the linear independence of the spline constraints and counts the listed outer knots as boundary knots. If a convention instead lists $K$ interior knots and two additional boundaries, the dimension is $K+2$. With only one listed knot and linearity on both sides, the $C^2$ condition forces a globally affine function, so the exceptional dimension is two.

For $\lambda >0$, minimize the smoothing spline objective over the Sobolev space of functions with square-integrable second derivative on the observed interval:The roughness penalty measures curvature; its null space consists of affine functions. To see why the minimizer is a natural cubic spline, let $g$ be the natural spline interpolating the values of any competitor $m$ at its distinct observation locations, and put $h=m-g$. Integration by parts twice on each knot interval gives $\int g^{\prime \prime }{h}^{\prime \prime }=0$: $g^{\prime \prime \prime \prime }=0$ on each interval, $h$ vanishes at the knots, $g^{\prime \prime }$ is continuous, and $g^{\prime \prime }$ vanishes at the outer endpoints. Hencewhile both functions have the same residual sum of squares. This proves the reduction to a finite-dimensional spline space.

For distinct $X_i$ and $n\ge 2$, use its $n$-dimensional basis, write $B_{ij}=b_j(X_i)$, and define the positive semidefinite matrixThe penalized least squares problem is $\Vert Y-B\beta {\Vert }^2+\lambda {\beta }^T\Omega \beta$, soUse linear continuation outside the observation interval. The fitted-value matrix is $S_{\lambda }=B(B^{T}B+\lambda \Omega {)}^{-1}{B}^T$; its trace gives the effective degrees of freedom. As $\lambda$ decreases to zero, the natural spline approaches interpolation, with trace $n$; as it grows without bound, the fit approaches ordinary least squares on $1,x$, with trace two. For $\lambda =0$ the criterion alone does not determine values between observations; the natural interpolant is the limiting convention. Repeated observation locations are combined by using distinct knots and their multiplicities as weights, rather than pretending the spline design has $n$ independent interpolation values.

The command fits a Gaussian generalized additive mixed modelThe population growth curve $s$ is a centered penalized regression spline with a cubic regression basis. The incubator term remains a random slope without a random intercept, because ~0+hours removes the intercept. A population linear component is allowed within the smooth's unpenalized part; curvature is penalized rather than being required.

A gamm fit returns a list with gam and lme components. Plot the population smooth and pointwise uncertainty bands withInspect whether the fitted curve departs materially from an affine function, particularly in regions supported by observations. The effective degrees of freedom near one suggest that the penalization has selected an approximately linear centered smooth. Bands and partial residuals help distinguish convincing curvature from noisy deviations, although pointwise bands are not a simultaneous test of linearity. A formal comparison would need to account for both smoothing selection and the incubator dependence. The smooth plot tests the visual plausibility of a common straight growth curve while retaining incubator-specific slope variation.

Use the natural cubic spline convention appropriate to the interior design knots: the spline is linear on $[a,x_1]$ and $[x_n,b]$. We also need $n\ge 2$. These points will be important for uniqueness.

A cubic spline is a function in $C^2[a,b]$ whose restriction to each of $[a,x_1]$, $[x_1,x_2]$, through $[x_n,b]$ is a polynomial of degree at most three. A spline knot is a junction between consecutive pieces. For a natural cubic spline, the two exterior pieces are affine functions; equivalently, one uses the natural endpoint conditions $g^{\prime \prime }(x_1)=g^{\prime \prime }(x_n)=0$ on the spline over $[x_1,x_n]$ and then extends linearly with matching derivatives. Merely prescribing $g^{\prime \prime }(a)=g^{\prime \prime }(b)=0$ on otherwise arbitrary exterior cubic pieces is a different condition and does not provide the unique interpolant invoked in this question.

The natural cubic spline interpolant $N\mathbf{v}$ to values $\mathbf{v}=(v_1,\dots ,v_n{)}^T$ is the natural cubic spline satisfying $(N\mathbf{v})(x_i)=v_i$. Its existence and uniqueness for $n\ge 2$ are the interpolant fact allowed in the PDF. The interpolation map $N$ is a linear map: a linear combination of interpolants satisfies the same spline, boundary and value conditions, so uniqueness identifies it with the interpolant to the corresponding linear combination of value vectors.

Let $g=N\mathbf{v}$, let $\tilde{g}\in C^2[a,b]$ have the same values, and put $r=\tilde{g}-g$. Then $r(x_i)=0$. We prove the orthogonalityApply integration by parts twice on each polynomial interval. On an interior interval, $g^{\prime \prime \prime \prime }=0$, andThe $g^{\prime \prime \prime }r$ terms vanish because $r$ vanishes at the knots. The $g^{\prime \prime }{r}^{\prime }$ terms cancel at internal knots because $g^{\prime \prime }$ and $r^{\prime }$ are continuous, and vanish at $x_1,x_n$ because $g^{\prime \prime }$ is zero there. On the exterior intervals, $g^{\prime \prime }=0$ identically. Summing therefore gives the claimed orthogonality without assuming that $g^{\prime \prime \prime }$ is continuous across knots.

Expanding the square now gives the minimum roughness property of the natural cubic spline interpolantEquality forces $r^{\prime \prime }=0$ everywhere, since $r^{\prime \prime }$ is continuous. Thus $r$ is an affine function. It vanishes at two distinct design points, so $r=0$, proving equality if and only if $\tilde{g}=g$.

For the penalized fit, take the allowed spline roughness penalty matrix $\Gamma$ to be a symmetric matrix with ${\mathbf{z}}^T\Gamma \mathbf{z}\ge 0$; thus it is a positive semidefinite matrix. If its supplied representative were not symmetric, its symmetric part defines the same quadratic form and is all that is needed. For any $\tilde{g}$, let $\mathbf{v}=(\tilde{g}(x_i){)}_{i=1}^n$. Replacing it by $N\mathbf{v}$ preserves every fitted value and can only decrease the second derivative roughness penalty. Therefore it is enough to minimizeover ${\mathbb{R}}^n$. The matrix $A_{\lambda }=I_n+\lambda \Gamma$ is a positive-definite matrix, sinceCompleting the quadratic or differentiating gives the unique solutionThis is the cubic smoothing spline. More explicitly, the normal equations are $(I_n+\lambda \Gamma ){\hat{\mathbf{v}}}_{\lambda }=\mathbf{Y}$, andFor an arbitrary function with fitted vector $\mathbf{v}$, the earlier orthogonality also givesBoth terms are nonnegative. Equality forces $\mathbf{v}={\hat{\mathbf{v}}}_{\lambda }$ and, by the proved interpolation equality case, $\tilde{g}=N\mathbf{v}={\hat{g}}_{\lambda }$. This proves both existence in $C^2[a,b]$ and uniqueness over the entire stated function class, not just over splines.

The argument is deterministic for each observed $\mathbf{Y}$. The fixed-design nonparametric regression and homoscedasticity assumptions motivate the squared-error loss, but Gaussian errors are not required.

At least two distinct design points are required for uniqueness. Question 4 does not explicitly restate $n\ge 2$, and this hypothesis is necessary. With one design point, every function $\tilde{g}(x)=Y_1+c(x-x_1)$ has zero residual and zero second derivative roughness penalty, for any real $c$. Thus the printed unrestricted uniqueness conclusion would be false for $n=1$. The corrected theorem assumes at least two distinct knots and the stated linear-tail convention. At $\lambda =0$, uniqueness over all of $C^2[a,b]$ also fails; the stipulated $\lambda >0$ eliminates that degeneracy.

Use the value function of variational regularization in its selection-independent formBecause $J\ge 0$, each objective is non-decreasing with $\alpha$, hence so is its infimum. For $0\le t\le 1$, the affine function of the parameter at each fixed $u$ givesThus $\Psi$ is concave.

For continuity, on any interval $0<a\le \alpha <\beta \le b$, comparison using $u_{\alpha }$ givesIt follows that $\Psi$ is a locally Lipschitz function, hence continuous, on $(0,\infty )$. Therefore the value function of variational regularization is non-decreasing, concave and continuous. No uniqueness of minimizers is needed for these conclusions.

For a functional $E$ on a real Banach space, its convex conjugate is defined on the dual space byHere $⟨p,u⟩$ is the dual pairing. For complex spaces, use its real part, equivalently regard the space as real for convex analysis. The convex conjugate is a convex function, being the supremum of affine functions of $p$, even when $E$ itself is not convex.