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An affine minorant of $f$ is a function $x\mapsto ⟨a,x⟩+b$ that is everywhere at most $f(x)$. Every proper lower-semicontinuous convex function has such a minorant, by separating a point below its closed epigraph. Adding a positive quadratic then yields a coercive function, which establishes existence in proximal minimization.

For a proper extended-real function with an affine minorant, biconjugate gives its greatest lower-semicontinuous convex minorant. Geometric separation of its epigraph turns containing closed half-spaces into affine lower bounds. Vertical half-spaces are recovered by adding multiples of a domain-separating inequality to one fixed affine minorant.

Convex analysis studies convex sets and convex functions, including epigraphs, convex conjugates and subdifferentials.

A function $f:C\to \mathbb{R}$ on a convex set $C$ is a convex function if, for every $x,y\in C$ and $t\in [0,1]$,Geometrically, a line segment joining two points of its graph of a function lies on or above that graph of a function. The convex set condition ensures that the intermediate input stays in the domain of a function. An affine function satisfies the defining inequality with equality.

For the quadratic function $f(x)=x^2$, the inputs $1$ and $3$ have equally weighted arithmetic mean $2$. Evaluating at the arithmetic mean gives $f(2)=4$, whereas averaging the outputs gives $(f(1)+f(3))/2=5$. Averaging first saves one unit in this example. More generally, direct expansion givesThis identity proves convexity and also shows exactly when its defining inequality is strict.

For a differentiable convex function, the tangent line in one dimension, or the corresponding supporting affine function in several dimensions, lies below the entire graph of a function. This is the subgradient inequality. The equivalent geometric statement is that the epigraph, consisting of points on or above the graph of a function, is a convex set. These characterizations are developed in Boyd and Vandenberghe's convex-function slides, sections 3.1–3.4.

Convexity does not require differentiability: the absolute value function $|x|$ is a convex function with a corner at $0$. When a function is twice continuously differentiable on an open interval, it is a convex function exactly when its second derivative is nonnegative throughout that interval. In several dimensions, the corresponding criterion on an open convex set is that its Hessian matrix is positive semidefinite everywhere. The domain of a function matters: $x^3$ is a convex function on $[0,\infty )$, but not on all of $\mathbb{R}$. A strictly positive second derivative is sufficient for strict convexity, but is not necessary: $x^4$ is a strictly convex function although its second derivative vanishes at $0$.

The usefulness of convexity is that local information can certify a global conclusion. Every local minimum of a convex function on a convex set is a global minimum: if a lower value existed elsewhere, the defining inequality would produce lower values arbitrarily close to the proposed local minimum. For a differentiable convex function on an open convex set, a zero gradient therefore certifies a global minimum. Strict convexity gives uniqueness when a global minimum exists; it does not supply existence by itself. These facts explain the role of convex functions in convex optimization.

Minimizing ${\max }_i(a_i^{T}x+b_i)$ over unrestricted $x$ is a linear program after introducing an epigraph variable $t$ and inequalities $a_i^{T}x+b_i\le t$. Its optimization Lagrangian has finite infimum over $x,t$ exactly when the nonnegative multipliers sum to one and their weighted slopes sum to zero. The displayed Lagrangian dual problem maximizes the resulting constant lower bound. It identifies the role of a zero-slope convex combination of the affine functions.

The epigraph is $\{(u,t):E(u)\le t\}$. It is a convex set exactly when $E$ is a convex function, and is a sequentially closed set in the product topology exactly when $E$ is sequentially lower semicontinuous.

The Fenchel–Young inequality says that every input and every slope obeywhere $f$ is any proper extended-real function and $f^{\ast }$ is its convex conjugate. No convexity assumption on $f$ is needed for this inequality: the defining supremum is at least the particular value $p\cdot x-f(x)$. If either term on the left is $+\infty$, the inequality still holds, but cannot be an equality because the dot product on the right is finite.

For a proper convex function, equality is equivalent to the subgradient conditionTo see this, with $f(x)$ finite, equality says that $x$ attains the supremum defining $f^{\ast }(p)$, so $p\cdot z-f(z)\le p\cdot x-f(x)$ for every $z$. Rearranging gives the subgradient inequality $f(z)\ge f(x)+p\cdot (z-x)$. Geometrically, the affine minorant ${\ell }_p(z)=p\cdot z-f^{\ast }(p)$ touches $f$ at $x$.

If the proper convex function is also closed, meaning its epigraph is a closed set, or equivalently it is lower semicontinuous, thenThe forward implication holds without closedness; the reverse one requires the additional hypothesis in general. Applying the equality criterion to $f^{\ast }$ relates $x\in \partial f^{\ast }(p)$ to $f^{\ast \ast }(x)+f^{\ast }(p)=p\cdot x$, and the Fenchel-Moreau theorem supplies $f^{\ast \ast }=f$ when $f$ is closed. See Bertsekas, Convex Optimization Theory: A Summary, §5.4.1, Proposition 5.4.3 for precisely this distinction.

A concrete failure of the reverse implication occurs for the indicator functional of a constraint set $f$ of the open interval $(-1,1)$. Its convex conjugate is $f^{\ast }(p)=|p|$: the endpoint gives the supremum, even though it is excluded from the interval. At $(x,p)=(1,1)$, $x\in \partial f^{\ast }(p)=\{1\}$, but $f(1)=+\infty$, so equality fails and $\partial f(1)$ is empty. Closing the interval changes the endpoint value and restores equality there.

For the convex quadratic function $f(x)=x^2/2$, the Fenchel–Young inequality is $x^2/2+p^2/2\ge px$, and equality holds exactly when $x=p$. For the absolute value $f(x)=|x|$, finite equality requires $p=\mathrm{sgn}(x)$ at $x\ne 0$, where $\mathrm{sgn}$ is the sign function; at $x=0$ every $p\in [-1,1]$ gives equality. These are exactly the subgradients of the absolute value, including its entire interval of slopes at the corner.

A closed convex set in a finite-dimensional Euclidean space is the intersection of all containing closed half-spaces. For a point outside a nonempty set, its Euclidean projection onto a convex set gives a separating normal. Applied to an epigraph, this turns geometric separation into affine lower bounds and is a key step in the Fenchel-Moreau theorem.

At a point where $\varphi (b)$ is finite, this is an affine plane supporting the epigraph from below: $\varphi (u)\ge \varphi (b)+{\lambda }^T(u-b)$ for every $u$. A nonzero coefficient of the vertical coordinate permits normalization to the displayed graph form. For a constrained value function, the slope is exactly an attained multiplier in the Strong Lagrangian property. When the value function is differentiable at an interior point of its finite domain, a supporting slope equals its gradient: apply the supporting inequality along both positive and negative multiples of each coordinate direction and take limits.

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.

Use the optimization Lagrangian with the sensitivity sign conventionFor equality constraints the Lagrange multipliers have unrestricted signs. The Lagrangian sufficiency theorem says that if $h(x^{\ast })=b$ and, for some ${\lambda }^{\ast }$, $x^{\ast }$ globally minimizes $L_b(\cdot ,{\lambda }^{\ast })$ over $X$, then $x^{\ast }$ globally minimizes $f$ subject to the constraint. Indeed every feasible $x$ satisfiesThis proves the theorem without a convexity or differentiability assumption. Merely finding a stationary point of the optimization Lagrangian would not establish its required global minimum.

Let $\varphi (u)=\inf \{f(x):x\in X,h(x)=u\}$, with value $+\infty$ when the feasible set is empty, and suppose $\varphi (b)$ is finite. The problem has the Strong Lagrangian property if a finite multiplier $\lambda$ satisfiesThis includes attainment of the dual bound by a multiplier. It does not by itself assert attainment of the primal infimum. For every multiplier, restricting the infimum to feasible $x$ gives ${\inf }_X{L}_b(x,\lambda )\le \varphi (b)$, the relevant weak duality. A non-vertical supporting hyperplane of a value function is a planewith $\varphi (u)\ge \varphi (b)+{\lambda }^T(u-b)$ for every $u$. Its vertical coefficient is nonzero and has been normalized to one; the epigraph lies above it. The two directions of the equivalence are proved in parts (a) and (b).

For the sampling allocation put $A={\sum }_i\sqrt{a_i{v}_i}$. Finite variance requires every $x_i>0$; interpret the objective at a zero coordinate as $+\infty$. An optimum uses all the budget, because scaling every positive $x_i$ up would otherwise reduce the objective while remaining feasible. The Cauchy-Schwarz inequality gives the global boundEquality requires $v_i/x_i$ to be proportional to $a_i{x}_i$. The budget then determines the unique allocation:This is the optimal cost-constrained stratified sampling allocation.

For an inequality ${\sum }_i{a}_i{x}_i\le b$, our sign convention uses $\lambda \le 0$, so $L_b=f-\lambda ({\sum }_i{a}_i{x}_i-b)$ is a lower bound on $f$ at feasible points. Its coordinate stationary equations are $-v_i/x_i^2-\lambda a_i=0$, and the displayed allocation givesThe function $v_i/x_i+(-{\lambda }^{\ast })a_i{x}_i$ has a strict global minimum at $x_i^{\ast }$: its second derivative is $2v_i/x_i^3>0$ and its value diverges at both endpoints of $(0,\infty )$. Thus the multiplier also supplies the Lagrangian sufficiency theorem certificate, with the tight inequality giving complementary slackness.

Finally, for $b+\delta b>0$,Hence the first-order change in minimal variance is $\boxed{{\lambda }^{\ast }\delta b}$. Increased resources decrease the variance. With the alternative convention $L=f+\mu ({\sum }_i{a}_i{x}_i-b)$, the multiplier is $\mu =-{\lambda }^{\ast }>0$ and the sensitivity is $-\mu \delta b$.

We first prove the algebraic extension step needed for Hahn-Banach separation theorem. Suppose $p$ is a finite sublinear functional on a real vector space and $g$ is a linear functional on a vector subspace $W$, with $g\le p$ there. For $v\notin W$, choose a real number $c$ betweenThis interval is nonempty: for any $w,z\in W$,Taking $w=0$ or $z=0$ also bounds the endpoints by finite numbers. Define $\tilde{g}(w+tv)=g(w)+tc$. For $t>0$, the upper bound for $c$, applied to $w/t$, proves $\tilde{g}(w+tv)\le p(w+tv)$. For $t<0$, writing $t=-s$ and using the lower bound at $w/s$ proves the same inequality. The case $t=0$ is the original domination. Thus a dominated linear functional extends over one further direction. Order all dominated extensions by extension of their domains. A chain has its union as an upper bound, so Zorn's lemma supplies a maximal extension. The one-direction argument shows its domain is the whole vector space. This proves the required algebraic Hahn-Banach theorem, rather than assuming the separation result.

Choose $a\in A$, and put $C=A-a$. This is a radially open convex set containing zero. It is absorbing: along each line through zero, sufficiently small multiples of any vector lie in $C$. Its Minkowski functionalis finite and nonnegative. Positive homogeneity follows by rescaling $t$. If $x\in sC$ and $y\in tC$, convexity gives $x+y\in (s+t)C$, proving subadditivity by taking infima. Moreover,Indeed, $p(x)<1$ gives $x\in sC$ for some $s<1$, hence $x\in C$ by convexity and $0\in C$. Conversely, radial openness at $x\in C$ permits $(1+ϵ)x\in C$ for some $ϵ>0$, giving $p(x)<1$.

Since $A\cap U=\varnothing$, $a\notin U$ and $u-a\notin C$ for every $u\in U$. Define a linear functional on $U+\mathbb{R}a$ byIt is well defined because $a\notin U$. For $t<0$, write $s=-t$; then $p(u+ta)=sp(u/s-a)\ge s=g(u+ta)$. For $t\ge 0$, domination follows from $g(u+ta)\le 0\le p(u+ta)$. The extension step therefore gives $L:V\to \mathbb{R}$ with $L\le p$, $L{|}_U=0$ and $L(a)=-1$. If $x\in A$, thenso $L(x)<0$. ConsequentlyBecause $L(a)\ne 0$, this kernel of a linear map has codimension one and is a hyperplane. This establishes the geometric Hahn-Banach theorem for radially open sets without assuming a norm or a topology on $V$.

For the convex function, use its strict epigraph in $V\times \mathbb{R}$. It is nonempty and convex, and does not contain $(0,0)$. It is also radially open: on every affine line, a finite convex function is continuous. To justify this last fact, convexity orders secant slopes; on a smaller interval the slopes are bounded above and below by slopes to two fixed exterior endpoints, giving a local Lipschitz bound. Thus $t\mapsto f(x+ty)-\beta -ts$ is continuous, and a strict negative value persists near $t=0$.

Apply the separation theorem with the zero subspace. A nonzero separating linear functional has form $L(x,\beta )=g(x)+c\beta$. Its values on the convex epigraph cannot have both signs, since a line segment would then meet its kernel. Choose the sign so that $L>0$ on the epigraph. At $x=0$ and $\beta >0$ this gives $c>0$. Letting $\beta$ decrease to $f(x)$ yields $g(x)+cf(x)\ge 0$. ThereforeThe resulting linear functional need not be nonzero: for example, the zero functional is the only linear minorant of $f(x)=x^2$ on $\mathbb{R}$.

Now suppose the norm and continuity hypotheses hold. Domination at $x$ and $-x$ givesIf $x_n\in \ker l$ and $x_n\to x$ in norm, apply this inequality to $x-x_n$. Since $l(x-x_n)=l(x)$ and both bounding values tend to $f(0)=0$, we obtain $l(x)=0$. The kernel of a linear map is therefore closed.

Finally, a linear functional with closed kernel on a normed vector space is continuous. If $l=0$, this is immediate. Otherwise choose $v$ with $l(v)=1$ and put $M=\ker l$. Closedness gives $d=\mathrm{dist}(v,M)>0$. For $l(x)\ne 0$, $x/l(x)-v\in M$, so $d\le \Vert x\Vert /|l(x)|$. The same resulting bound holds when $l(x)=0$:This proves continuity and explains exactly how closedness of the kernel is used.

Introduce a free scalar $t$. The epigraph reformulation is the linear programFor each fixed $x$, the smallest feasible $t$ equals the maximum of the affine expressions, so the reformulation preserves the optimum.

Associate nonnegative multipliers $y_i$ with these inequalities. Its optimization Lagrangian isBecause $x$ and $t$ are unrestricted, the infimum is finite precisely when the coefficients of both vanish. Therefore the Lagrangian dual problem isA feasible dual vector is a convex combination of the slopes whose mean slope is zero. It gives a constant lower bound on the maximum of the affine functions. The primal is always feasible, by taking $x=0$ and sufficiently large $t$. If it has a finite optimum, linear programming duality gives attainment and equality with this dual; otherwise the dual is infeasible and the primal is unbounded below.

The Legendre-Fenchel transform and biconjugate areThe Fenchel-Moreau theorem says that every proper convex function which is lower semicontinuous equals its biconjugate. More generally, if a proper extended-real $f$ has an affine minorant, thenwhere the right side is the greatest lower semicontinuous convex minorant. This is biconjugation as closed convexification. The affine-minorant hypothesis ensures that the closed convexification is proper; an unqualified statement including arbitrary improper functions would need separate conventions.

Here are the essential proof steps. The Fenchel–Young inequality gives $f^{\ast \ast }\le f$. Every term in the supremum defining $f^{\ast \ast }$ is an affine minorant of $f$, and conversely any affine minorant $⟨p,x⟩+a\le f(x)$ has $a\le -f^{\ast }(p)$. Thus $f^{\ast \ast }$ is exactly the supremum of all affine minorants, hence is convex and lower semicontinuous.

Put $g=\mathrm{clconv}f$. Its epigraph is the closed convex hull of the epigraph of $f$. Applying the half-space representation of a closed convex set in ${\mathbb{R}}^{n+1}$ recovers that epigraph from its containing half-spaces. A containing half-space writtenhas $q\le 0$, since epigraphs extend upwards. If $q<0$, it is precisely the epigraph inequality of an affine minorant, $t\ge (⟨v,x⟩-a)/(-q)$.

Vertical half-spaces with $q=0$ must also be accounted for. Choose one affine minorant ${\ell }_0$ of $g$, which exists because $g$ is proper and closed: strictly separate $(x_0,g(x_0)-1)$ from its epigraph for a finite domain point $x_0$; the separating coefficient of $t$ cannot be zero, since that would not distinguish points with the same $x_0$. If a vertical containing inequality is $⟨v,x⟩\le a$, thenis still an affine minorant on the domain of $g$. At a point violating the vertical inequality, these minorants tend to infinity as $L\to \infty$. Thus vertical domain restrictions are also recovered by the supremum of affine minorants.

Consequently $g$ equals that supremum. Every affine minorant of $g$ is below $f$, while the epigraph of any affine minorant of $f$ contains its closed convex epigraph hull. Therefore $f$ and $g$ have the same affine minorants, completing $g=f^{\ast \ast }$. The theorem in the previous solution supplies the geometric separation step behind biconjugation.

For an extended-real function define its Fenchel conjugate by $f^{\ast }(p)={\sup }_x(⟨p,x⟩-f(x))$ and its biconjugate by $f^{\ast \ast }(x)={\sup }_p(⟨p,x⟩-f^{\ast }(p))$. The Fenchel-Moreau theorem states, in the standard proper-envelope setting,Here the right side is the largest lower semicontinuous convex function below $f$, equivalently the function whose epigraph is the closed convex hull of $\mathrm{epi}f$. It is enough to assume $f$ is proper and has an affine minorant, ensuring this envelope is proper. In particular, for a proper convex function that is lower semicontinuous, $\boxed{f^{\ast \ast }=f}$.

First, $⟨p,x⟩-f^{\ast }(p)\le f(x)$ by the definition of the convex conjugate. The biconjugate is a supremum of continuous affine functions, so it is convex, lower semicontinuous and no greater than $f$. Second, the best intercept for an affine minorant with slope $p$ is $-f^{\ast }(p)$: $⟨p,x⟩+a\le f(x)$ for all $x$ precisely when $a\le -f^{\ast }(p)$. Thus $f^{\ast \ast }$ is the supremum of all affine minorants.

To prove that no part of the closed convex envelope is missed, set $E=\mathrm{cl}\mathrm{conv}(\mathrm{epi}f)$. The half-space representation of a closed convex set from part (a), applied in ${\mathbb{R}}^{n+1}$, separates any $(x_0,t_0)\notin E$ from $E$ by an inequality $a\cdot x+bt\le d$. Because $E$ is upward closed, $b\le 0$. If $b<0$, division by $-b$ gives an affine minorant $\ell$ with $\ell (x_0)>t_0$.

A vertical separator has $b=0$. Let ${\ell }_0$ be an existing affine minorant. Combine $a\cdot x\le d$ with ${\ell }_0(x)\le t$ to obtainSince $a\cdot x_0-d>0$, a sufficiently small positive $\epsilon$ makes this affine function exceed $t_0$. Thus vertical half-spaces can be approximated by nonvertical epigraph supports. Every point below the envelope is excluded by an affine minorant, so the supremum of these minorants is exactly the envelope. This is the decisive use of part (a).

Properness and the minorant convention matter for unrestricted extended-real functions. For example, $f(x)=-x^2$ on $\mathbb{R}$ has no affine minorant; $f^{\ast }\equiv +\infty$ and $f^{\ast \ast }\equiv -\infty$. With the corresponding improper-envelope convention its closed convex envelope is also $-\infty$. The theorem should not silently describe such an envelope as proper. The identically $+\infty$ function is another degenerate case, handled separately by extended-real conventions.

Fix $x$ and minimize $J_x(y)=f(y)+\Vert y-x{\Vert }^2/(2\tau )$. A proper lower semicontinuous convex function has an affine minorant $f(y)\ge a\cdot y+b$, as follows by separating a point below its closed epigraph. HenceThe quadratic dominates the linear term, proving coercivity. Properness supplies at least one finite trial value, lower semicontinuity passes to limits, and finite-dimensional compactness makes a bounded minimizing sequence converge along a subsequence to a minimizer. Thus the proximal operator exists at every $x$.

The subdifferential sum rule applies because the quadratic is finite and continuous everywhere. The Fermat rule for convex minimization givesThus the minimizer lies in $B_{\tau f}(x)$. The uniqueness proved in part (a), or strict convexity of the quadratic sum, now yieldsThis proof displays the separate roles of properness, lower semicontinuity, convexity and finite dimension. In particular, compactness here is not inferred merely from strict convexity.

Use the extended real numbers $\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty ,+\infty \}$. A proper convex function never takes $-\infty$ and is finite somewhere. Its effective domain isFor a proper extended-real function, convexity meansThe endpoint cases are identities; taking $0<t<1$ avoids the ambiguous product $0\cdot (+\infty )$. Equivalently, its epigraph is a convex set. In particular, the effective domain is a convex set. Properness is an additional condition: it rules out the identically $+\infty$ function and all $-\infty$ values, rather than requiring finiteness everywhere.

For $x\in \mathrm{dom}f$, the subdifferential is the set of supporting slopesEach element is a subgradient. Define $\partial f(x)=\varnothing$ outside $\mathrm{dom}f$. The defining subgradient inequality must hold globally, not merely in a neighborhood of $x$. For a differentiable finite convex function, $\partial f(x)=\{\nabla f(x)\}$.

The epigraph isAssume $E$ is $\tau$-sequentially lower semicontinuous. If $(u_n,t_n)$ lies in the epigraph and converges to $(u,t)$ in the product topology, thenThe limit remains in the epigraph, so it is a sequentially closed set.

Conversely, suppose the epigraph is a sequentially closed set and sequential lower semicontinuity fails along $u_n\stackrel{\tau }{\to }u$. Choose a finite real number $a$ with $\mathrm{liminf}E(u_n)<a<E(u)$. There is a subsequence along which $E(u_{n_j})\le a$. Thus $(u_{n_j},a)$ lies in the epigraph and converges to $(u,a)$, which does not lie there. This contradiction provesThis is a sequential statement; identifying it with ordinary closedness requires an appropriate assumption on the topology.