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Work in the real zero-boundary Sobolev space $X=H_0^1(0,1)$, whose members have square-integrable weak derivatives and zero endpoint trace. Define the bounded bilinear form and linear functionalThe weak formulation is to find $u\in X$ with $a(u,v)=\ell (v)$ for every $v\in X$. Integration by parts recovers the differential equation for a smooth solution; conversely, this identity is the definition of its weak solution.

Use the full $H^1$ norm. The Cauchy-Schwarz inequality gives $|a(u,v)|\le \Vert u{\Vert }_{H^1}\Vert v{\Vert }_{H^1}$ and $|\ell (v)|\le \Vert v{\Vert }_{H^1}/\sqrt{3}$. Also $a(v,v)=\Vert v{\Vert }_{H^1}^2$, so $a$ is a coercive bilinear form with constant one. The Lax-Milgram theorem states that a bounded, coercive bilinear form on a real Hilbert space determines a unique weak solution for every bounded linear functional. It therefore applies here.

The equivalent minimization problem isIf $u$ is the weak solution, symmetry and $a(u,v-u)=\ell (v-u)$ giveThis is positive unless $v=u$, proving a unique global minimizer. Conversely, differentiating $\mathcal{E}(u+tv)$ at $t=0$ gives the weak identity. Thus the unique energy minimizer is exactly the unique weak solution. As a check, the classical representative isIt satisfies both endpoint conditions and $-u^{\prime \prime }+u=x$.

For the stated uniaxial nematic order normalization, the nematic director is an eigenvector of $Q$ with eigenvalue $\lambda$, and the two perpendicular eigenvalues are $-\lambda /2$. ThereforeUsing precisely the Landau-de Gennes free energy convention in the preceding solution givesIf the quartic invariant were instead normalized as $b\mathrm{Tr}{Q}^4/4$, its coefficient would be $\bar{b}=9b/32$; the physical predictions are unchanged after redefining $b$.

For a nonzero global minimizer of the free energy, compare opposite values of the scalar nematic order parameter:The lower one has $\bar{c}\lambda <0$, soThe claim concerns the stable ordered phase, not every metastable stationary point. For $\bar{c}=0$ the two signs are degenerate. For $\bar{c}\ne 0$, equality of the ordered and isotropic free energies, together with stationarity, gives ${\lambda }_t=-\bar{c}/(2\bar{b})$ and ${\bar{a}}_t={\bar{c}}^2/(4\bar{b})>0$. The finite jump ${\lambda }_t$ exhibits the first-order phase transition driven by the cubic invariant.

First expand the variational integrand:Its needed partial derivatives are ${\mathcal{L}}_f=3f^{\prime }+4f$ and ${\mathcal{L}}_{f^{\prime }}=2f^{\prime }+3f-x$. The Euler-Lagrange equation is thereforeThe general solution is $f=A\cosh (\sqrt{2}x)+B\sinh (\sqrt{2}x)-1/4$. Subtracting the two endpoint equations gives $B=0$; adding them gives $A=1/[4\cosh \sqrt{2}]$. HenceThis is the unique stationary solution. It is also the unique global minimizer: after discarding the fixed boundary terms, integration by parts rewrites $I$ as ${\int }_{-1}^1[f^{\prime 2}+2f^2+f-2x^2]dx$. For an endpoint-zero variation $\eta$, the linear part vanishes at the displayed solution andwith equality only for $\eta =0$.

Impose $y=x$ before varying. Writing $\delta ={\delta }_1+{\delta }_2$, the restricted integrand isIts middle term is the total derivative of $\alpha ({\gamma }_1+{\gamma }_2)(x+\beta x^2/2)$, so it can be dropped for fixed endpoints by part (b). The remaining Euler-Lagrange equation is $\delta \ddot{x}=\alpha \beta$. Integrating twice and imposing $x(0)=0$, $x(T)=S$ givesThis is the unique global minimizer among paths on the specified line. To prove it, add any admissible variation $\eta$ with $\eta (0)=\eta (T)=0$. The change in fuel, after dropping the fixed boundary term, isThe first integral vanishes by integration by parts and the equation for $x$. Equality forces $\eta =0$. No monotonicity or restriction to the segment between endpoints was imposed in the question; the minimizing coordinate can initially decrease for sufficiently large $T$.

Set $g_j=G_{1/L}(x^j)$, $r_j=\Vert x^j-x^{\ast }{\Vert }^2$ and ${\delta }_j=\varphi (x^j)-\varphi (x^{\ast })$. Here $g_j$ is the proximal gradient mapping, $r_j$ is a squared Euclidean distance, and ${\delta }_j\ge 0$ because $x^{\ast }$ is a global minimizer. The update gives $x^{j+1}=x^j-g_j/L$.

First use the supplied inequality (1) with $z=x^j$:Thus the objective function values, and hence the errors ${\delta }_j$, are nonincreasing. Next use (1) with $z=x^{\ast }$:Expansion of the squared Euclidean norm givesso the last bound is exactly ${\delta }_{j+1}\le L(r_j-r_{j+1})/2$.

The telescoping sum over $j=0,\dots ,k-1$ therefore givesSince each summand is at least ${\delta }_k$, we obtain the proximal gradient objective error boundThe bound holds for the last iterate. The monotonicity of the objective function values is what converts an averaged estimate into this last-iterate estimate.

There is a summation-index misprint. On the native PDF, paper page 3, equation (4) has $X^k$ inside a sum indexed by $j$. The intended telescoping estimate has $X^j$ inside that sum; this is the estimate proved below and used in part (d). The index mismatch is present in the original PDF, rather than being introduced by the transcription.

Take the expectation of the supplied conditional expectation bound (3), using the tower property:Rearranging and forming a telescoping sum givesBecause $x^{\ast }$ is a global minimizer, $f(X^k)\ge f(x^{\ast })$ for every realisation. Hence the corrected equation (4) isAll expectations at a finite iteration are finite here: a deterministic starting point and finitely many choices of the sampled index produce only finitely many possible iterates. The bound controls accumulated squared gradients, with an additive contribution from the stochastic gradient noise floor.

The subgradient inequality for the differentiable convex function $f$ givesThus $y$ is a global minimizer of $\varphi$ with value zero. Subtracting an affine function preserves convexity.

We must also justify using part (a) for $\varphi$, since its partial derivative can differ from that of $f$. First show that condition (6) for a singleton block implies the full coordinate Lipschitz continuity estimateIf ${\partial }_{i}f(z)\ne 0$, choose $\alpha =-t/{\partial }_{i}f(z)$ in (6). If it is zero but ${\partial }_{i}f(z+t{e}_i)\ne 0$, apply the same argument starting at $z+t{e}_i$ and reverse the displacement. If both partial derivatives are zero, the estimate is immediate. The availability of all real $\alpha$ in (6) justifies these choices.

Now ${\partial }_i\varphi (z)={\partial }_{i}f(z)-{\partial }_{i}f(y)$, so differences of this partial derivative along coordinate lines are unchanged. Hence $\varphi$ has the same coordinate Lipschitz continuity constant $L_i$ and satisfies the singleton version of (6). Apply part (a) to $\varphi$ with step size $1/L_i$:Combining with its already established nonnegativity provesThe coordinate estimate survives subtraction of the supporting affine function. This is the step that allows the arbitrary comparison point $y$ to be exploited in part (c).

The supporting inequality in part 2(a) bounds $f$ below by an affine function plus a positive squared Euclidean norm, making it a coercive function. Its continuity gives a global minimizer $x^{\ast }$, which is unique by strict convexity. Let $L_{\max }={\max }_i{L}_i$, $\ell =n/m$ and ${\Delta }_0=f(x^0)-f(x^{\ast })$. Here $m$ is the block size, so its role differs from the number of component functions in Question 2. Use randomized block coordinate descent with uniformly sampled blocks of size $m$, independently at each update. Sampling uniformly from a fixed partition into $\ell$ equal-sized blocks gives the same calculation, since each coordinate has inclusion probability $1/\ell =m/n$.

Let $\beta$ be a common upper bound on the block Lipschitz constants. Part (c) permits $\beta =m{L}_{\max }$. With step size $1/\beta$, part (a) givesSumming squared partial derivatives according to their inclusion probabilities,For a strongly convex function, part 2(a) also yields the strong convexity gradient inequalityIndeed, substitute $y=x^{\ast }$ into the supporting inequality and complete the square in $x^{\ast }-x$: the expression $\nabla f(x{)}^{T}d+\gamma \Vert d{\Vert }^2/2$ is at least $-\Vert \nabla f(x){\Vert }^2/(2\gamma )$ for every $d$.

Combining the last three estimates gives a contraction in conditional expectation:By the tower property and repeated application,An optimization epoch counts $n$ coordinate partial derivative evaluations: it comprises $\ell$ block updates or one full gradient update. With sampling with replacement, an optimization epoch is this work measure; it need not visit every coordinate. Consequently the block method needsto bring the expected objective function gap below $\epsilon$, for $0<\epsilon <{\Delta }_0$, up to integer rounding.

For the steepest descent method, use a global Lipschitz gradient constant $L_f$. Part (c) with the full block permits $L_f={\sum }_i{L}_i\le n{L}_{\max }$. The descent lemma with step size $1/L_f$, together with the strong convexity gradient inequality, givesAn exact line search along the negative gradient does at least as well in objective function decrease per update. Ignoring its additional work, the resulting epoch complexity isThe common coordinate-based guarantees favour blocks by a factor $n/m=\ell$ in epochs. In fact, with the conservative constants $\beta =m{L}_{\max }$ and $L_f=n{L}_{\max }$, both methods have the same guaranteed contraction per update, while a block update uses only $m$ coordinate partial derivatives.

This compares the bounds justified by the given constants. For a particular objective function, sharper constants give the comparison $\beta /\gamma$ against $L_f/\gamma$; an actual speedup of $\ell$ is not universal. Computing a block coordinate gradient also need not cost exactly $m/n$ of a full gradient evaluation. The epoch complexity comparison assumes that coordinate partial derivative counts are a useful measure of work, and does not itself establish a wall-clock comparison.

For a convex function $h$ with a well-defined proximal operator, and a differentiable function $f$, the proximal gradient mapping is the displacement of one proximal gradient method update divided by its positive step size $\alpha$. Writing $p=x-\alpha G_{\alpha }(x)$, the subgradient optimality condition gives $G_{\alpha }(x)-\nabla f(x)\in \partial h(p)$. For convex functions $f,h$, its zeros are precisely the global minimizers of $f+h$, by the subgradient optimality condition and subdifferential sum rule. When $h=0$, it is the ordinary gradient.

Target: convex-optimization.bigb, under Proximal gradient method.

For $\varphi =f+h$, with $f$ a convex function having an $L$-Lipschitz gradient and $h$ a proper convex function with a well-defined proximal operator, the proximal gradient method with step size $1/L$ satisfies the displayed bound whenever a global minimizer $x^{\ast }$ exists. The one-step subgradient inequality and descent lemma giveThe objective function values are nonincreasing. A telescoping sum over $j=0,\dots ,k-1$, followed by comparison of the last error with the arithmetic mean of the errors, proves the result. The bound controls the last objective function error without requiring a strongly convex function or a unique minimizer.

Target: convex-optimization.bigb, under Proximal gradient method.

For a differentiable strongly convex function with modulus $\gamma >0$ and global minimizer $x^{\ast }$, the supporting inequality givesCompleting the square shows that the last two terms are at least $-\Vert \nabla f(x){\Vert }^2/(2\gamma )$, which proves the result. Combining this with the descent lemma converts squared gradient decrease into a contraction of the objective function gap.

Target: mathematical-optimization.bigb, under Line search.