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For nondegenerate independent and identically distributed arrivals with mean below $c$, suppose the positive root is interior to the moment-generating function domain. Then ${\Lambda }_A^{\prime }({\theta }_{\ast })>c$. The Fenchel–Young inequality gives $t{I}_A(c+1/t)\ge {\theta }_{\ast }$ for every $t>0$, and equality occurs at $t=[{\Lambda }_A^{\prime }({\theta }_{\ast })-c{]}^{-1}$. The exponential workload bound from cumulative arrival moments gives the upper logarithmic tail bound for every $\theta <{\theta }_{\ast }$. A horizon of the indicated length times buffer size gives the matching lower bound by Cramér's theorem. A positive interior root is an assumption; if arrivals never exceed capacity, positive workload cannot arise.

A proper lower-semicontinuous convex functional on a real Banach space equals its biconjugate, with the canonical embedding in the bidual understood. One inequality follows from the Fenchel–Young inequality; separation of a point below the closed convex epigraph gives the opposite inequality. This is the closed-convex analogue of recovering a function from all its affine supporting functions.

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.

Write $A_n$ for work arriving in slot $n$ and $c$ for the service rate of a queue. For a work-conserving single-server queue, the Lindley recursion is $Q_{n+1}=(Q_n+A_{n+1}-c{)}^{+}$. For stable independent and identically distributed random variables with mean $\mu <c$, its stationary workload supremum has the formDifferent scaling limits of a queue concern different quantities growing large; they need not share either a probability speed or a likely path to overflow.

For large-buffer queue scaling, fix the input law and $c>\mu$ and let the buffer size of a queue $b\to \infty$. Assume finite exponential moments around a positive root ${\theta }_{\ast }$ of ${\Lambda }_A(\theta )=c\theta$. An overflow accumulated in $n\simeq tb$ slots requires an average input about $c+1/t$, so Cramér's theorem assigns logarithmic cost $bt{I}_A(c+1/t)$. Optimizing the duration gives the Cramér-Lundberg workload exponentFor an interior root with ${\Lambda }_A^{\prime }({\theta }_{\ast })>c$, Fenchel–Young inequality verifies the minimization: the lower bound is ${\theta }_{\ast }$ because $I_A(x)\ge {\theta }_{\ast }x-{\Lambda }_A({\theta }_{\ast })$, and equality occurs at $t=[{\Lambda }_A^{\prime }({\theta }_{\ast })-c{]}^{-1}$. The exponential workload bound from cumulative arrival moments gives the upper exponent for every $\theta <{\theta }_{\ast }$; an exponentially tilted positive-drift path or the duration calculation gives the matching lower exponent. In effective bandwidth notation, ${\Lambda }_A({\theta }_{\ast })/{\theta }_{\ast }=c$. This regime measures long accumulation into a large buffer with a fixed positive stability margin.

For many-source queue scaling, superpose $L$ independent copies of an input process, give the server capacity $Lc$, and give it buffer size $Lb$. For each lookback horizon $n$, the average of the $L$ source cumulative arrivals has a large deviation principle with rate function $I_n$. Overflow at that horizon costs $L{I}_n(b+cn)$. Optimizing the lookback horizon suggestsFor independent slots, $I_n(x)=n{I}_A(x/n)$, giving $V(b)={\inf }_{n\ge 1}n{I}_A(c+b/n)$. The integer horizon is important in a slotted system. For temporally dependent sources, use the cumulative-input cumulant-generating function, rather than multiplying a one-slot formula by $n$. The heuristic needs uniform control of the tail of the stationary supremum to justify optimizing an unbounded set of horizons; stability and suitable exponential-moment estimates provide this in the usual models. The growing quantity here is the number of independent sources, while per-source capacity and buffer remain fixed.

For moderate-deviation queue scaling, consider a bufferless queue with aggregate Poisson input of mean $L\lambda$ and capacity $L\lambda +C{L}^{(1+\beta )/2}$, $0<\beta <1$. The spare capacity is much larger than the ordinary central limit theorem scale $\sqrt{L}$ but much smaller than $L$. The Poisson sample-mean rate function near its minimum has expansionWith $h=C{L}^{(\beta -1)/2}$, the cost $L{I}_P(\lambda +h)$ is asymptotic to $C^2{L}^{\beta }/(2\lambda )$. Hence Poisson moderate deviations give $\log \mathbb{P}(\mathrm{overflow})\sim -C^2{L}^{\beta }/(2\lambda )$. The quadratic cost comes from a shrinking neighborhood of the mean, rather than from a fixed macroscopic overload. The regime is useful for a large aggregate whose service margin lies between its typical fluctuation scale and its mean scale.

For a heavy-traffic limit of a queue, instead hold the input law fixed with mean $\mu$ and positive finite variance ${\sigma }^2$, and let $c_{\epsilon }=\mu +\epsilon d$ with $d>0$ and $\epsilon \downarrow 0$. In $t/{\epsilon }^2$ slots the net input has mean about $-dt/\epsilon$ and random fluctuation of size $\sigma \sqrt{t}/\epsilon$. Thus the functional central limit theorem, followed by the Skorokhod reflection map on the half-line, suggestswhere $R$ is reflected Brownian motion with negative drift $-d$ and variance parameter ${\sigma }^2$. The stationary law of negatively drifted reflected Brownian motion has tail $\mathbb{P}(R_{\infty }>b)=e^{-2db/{\sigma }^2}$. When passing the stationary distribution through the limit is justified, this predictsFor dependent input satisfying a suitable invariance principle, replace ${\sigma }^2$ by its long-run variance of a stationary process. This regime is appropriate near critical load and on long time scales; a functional transient limit alone does not automatically justify the stationary-limit interchange.

A different large-service regime holds the input law fixed. With Poisson distribution of mean $\lambda$ and bufferless capacity $c\to \infty$, let $m=\lfloor c\rfloor +1$. The single-point lower bound $\mathbb{P}(X=m)$ and the optimized Chernoff bound for $X\ge m$, using Stirling's formula, give the Poisson large-service overflow asymptoticHere the service rate grows while the input mean does not. A quadratic moderate-deviation approximation would describe the wrong part of the Poisson tail.

To choose a regime, compare the spare capacity, typical fluctuations, number of sources and buffer size before selecting a formula. Use the large-buffer exponent for a fixed stable input and a long overload episode; use the many-source variational cost when independent flows and resources grow together; use moderate deviations for an intermediate safety margin; use the heavy-traffic diffusion when the stability margin tends to zero; and use the fixed-input Poisson tail when service alone becomes very large. Check temporal dependence and exponential-moment assumptions. For heavy-tailed subexponential distributions, a single exceptionally large arrival can dominate overflow, so a finite exponential workload root and a Brownian approximation to the far tail need not apply. These comparisons explain why the same nominal queueing theory system can have different useful asymptotic descriptions.

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.

Let $C=\mathrm{conv}\{a,b\}$. Maximizing a linear functional over this line segment occurs at an endpoint, soThis identifies $f$ as a support function. Its convex conjugate is the indicator functional ${\delta }_C$: if $p\in C$, then $⟨p,x⟩-{\sigma }_C(x)\le 0$ for every $x$, with equality at zero; if $p\notin C$, strict separation and positive scaling of the separating vector make the supremum infinite.

For a convex function, equality in the Fenchel–Young inequality characterizes its subdifferential. HenceThis is support-function subgradients as exposed faces. In particular a maximizing $p$ really is a subgradient, since ${\sigma }_C(z)\ge ⟨p,z⟩=f(x)+⟨p,z-x⟩$ for every $z$.

Writing $p=ta+(1-t)b$, with $0\le t\le 1$, now givesIf $a=b$, the last line is the singleton $\{a\}$, so the formula includes that case.

Compute the convex conjugate of $f_z$ at a general pair $(v,y)$. With $w=z+u-x$, the independent variables become $(x,w)$ and $u=w+x-z$, soTherefore the Lagrange dual function isUsing strong duality from the preceding solution givesEquivalently the conjugate of an infimal convolution is $F^{\ast }=k^{\ast }+h^{\ast }$, and the continuous convex $F$ equals its biconjugate.

For practical subgradient computation, minimize the known convex dual objectiveEvery optimizer, and only an optimizer, belongs to $\partial F(z)$ by Fenchel–Young inequality. The full characterization isThis is infimal-convolution dual subgradients; the set is nonempty because $F$ is finite convex everywhere. If both conjugates are differentiable at the optimizer, solve $\nabla k^{\ast }(y)+\nabla h^{\ast }(y)=z$. For nonsmooth conjugates, use the displayed aggregate subdifferential or a convex minimization algorithm. Replacing it by $\partial k^{\ast }(y)+\partial h^{\ast }(y)$ requires the usual subdifferential sum rule qualification; it is not automatically justified solely by knowing the two conjugates.

LetThe proximal operator optimality condition saysBy subgradient inversion under convex conjugacy, which uses Fenchel–Young inequality and $f^{\ast \ast }=f$ from the Fenchel-Moreau theorem,Consequently $z-p=\tau v\in \tau \partial f^{\ast }(p)$, meaning $p=B_{\tau f^{\ast }}(z)$. Uniqueness of the backward subgradient step identifies the result:This is the scaled Moreau decomposition. It requires one backward step on $f$, with reciprocal parameter $1/\tau$ and scaled input $z/\tau$, followed by a scalar multiplication and subtraction. No separate proximal computation of the convex conjugate is needed.

The conjugate of an infimal convolution is the sum of the conjugates:Here $g^{\ast }(p)=\frac{1}{2}\Vert p{\Vert }_2^2$, and $h^{\ast }(p)={\delta }_{[-1,1{]}^n}(p)$, the indicator functional of the unit infinity-norm ball. ThusThe infimal convolution is finite convex and continuous, so the Fenchel-Moreau theorem applies without a closure defect. By equality in the Fenchel–Young inequality and the subdifferential sum rule,This is precisely the variational characterization of projecting $x$ onto the cube. ConsequentlyFor the primal split, $y_i=\mathrm{clip}(x_i,-1,1)$ and $z_i=\mathrm{sign}(x_i)(|x_i|-1{)}_{+}$. These directly minimize the two scalar terms. The Huber loss is continuously differentiable, including at $x_i=\pm 1$, but its second derivative changes there. The one-dimensional sketch shows a quadratic center joined tangentially to linear tails:

Applied to a discrete gradient, the Huber gradient regularizer penalizes small slopes quadratically and large slopes linearly. Compared with pure squared-gradient smoothing it preserves large edges better; compared with pure total variation denoising it encourages small smooth variations and reduces the strong preference for piecewise-constant plateaus. It can therefore be useful for denoising signals or images containing both smooth regions and sharp transitions. It still penalizes edges and can bias their amplitude, and it does not guarantee complete elimination of staircasing in total variation denoising. The unit threshold must be scaled appropriately for data units and grid spacing.

For a real Banach space $X$ and a proper convex function $E:X\to \mathbb{R}\cup \{+\infty \}$, the convex conjugate is the extended-valued function on the continuous dual spaceHere proper means that $E$ never takes $-\infty$ and is finite somewhere. The pairing is the duality pairing, or the inner product after identification in a Hilbert space. The conjugate is convex and lower semicontinuous because it is a supremum of continuous affine functions. Its definition gives the Fenchel–Young inequality $E(x)+E^{\ast }(p)\ge ⟨x,p⟩$; the Fenchel-Moreau theorem recovers $E$ as $E^{\ast \ast }$ under the stated closed-convex hypotheses.

Sincefor $x>0$, the second-derivative criterion shows that $f(x)=x^a/a$ is strictly convex. Put $b=a/(a-1)$, so $1/a+1/b=1$. For dual variable $p>0$, the stationary equation in the Legendre transform isIt is the unique global maximizer of $px-f(x)$, and henceApplying the same calculation with the conjugate exponent $b$ givesUnder the Legendre-Fenchel transform on all real dual variables, the same supremum additionally gives $f^{\ast }(p)=0$ for $p\le 0$.

Now take $a=r$, $b=s$, and apply the Fenchel–Young inequality $f(x)+f^{\ast }(y)\ge xy$. This yields Young's inequality for productsfor positive $x,y$ and conjugate exponents $r,s$.

Let $\Phi ={\phi }_{\epsilon }$ and let ${\Phi }^{\ast }$ denote its convex conjugate. Its Fenchel–Young gapis nonnegative by the Fenchel–Young inequality, and the hypothesis says $\int Hd{\pi }_{\epsilon }\le \epsilon$.

First justify the integrability needed for the certificate. Finite second moments and the Cauchy-Schwarz inequality imply, for every transport plan $\pi$,Since $\Phi \in L^1(\mu )$ and $H\in L^1({\pi }_{\epsilon })$, the identity ${\Phi }^{\ast }(y)=H(x,y)-\Phi (x)+x\cdot y$ proves ${\Phi }^{\ast }\in L^1(\nu )$ by the marginal distribution property. In particular, no subtraction of infinite integrals is being used.

SetThese are integrable Kantorovich potentials. The Fenchel–Young inequality gives $u(x)+v(y)\le \frac{1}{2}|x-y{|}^2$. More precisely, for every transport plan $\pi$,Taking the infimum in the first inequality therefore provesThe factor $1/2$ in the quadratic cost is essential for this exact gap identity.

The definition of the concave conjugate gives the pointwise Fenchel–Young inequalityApply it to $(X_T,Y_T)$, take expectations, and use part b:If $U^{\prime }(X_T)=Y_T$, the first inequality is equality; if $XY$ is a true martingale, the second is equality.

For a proper lower-semicontinuous convex functional on a real Hilbert space, $p\in \partial J(u)$ means $⟨v,p⟩-J(v)\le ⟨u,p⟩-J(u)$ for all $v$. Taking the supremum gives equality in the Fenchel–Young inequality: $J(u)+J^{\ast }(p)=⟨u,p⟩$. Apply the same argument to $J^{\ast }$ and use the Fenchel-Moreau theorem $J^{\ast \ast }=J$ to obtain the reciprocal subgradient condition.

For a nonempty closed convex set $C$, the finite-valued support function has subgradients exactly at the maximizing points of $C$. This follows from Fenchel–Young inequality and the conjugacy between the support function and the indicator functional of a constraint set. For compact $C$, the exposed maximizing set is nonempty.