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For the unlabelled Euler-Lagrange assertion, use its integrated form $f^{\prime }(u^{\prime })=c$ almost everywhere. This follows from the first variation $\int f^{\prime }(u^{\prime }){\varphi }^{\prime }=0$ for all compactly supported tests. If $u\in C^1$, the left side is continuous and so equals $c$ everywhere. Its derivative values lie in the finite fiber $(f^{\prime }{)}^{-1}(c)$, while the continuous image of the interval under $u^{\prime }$ is connected. Thus $u^{\prime }$ is constant. Endpoint values fix it to $m=(\beta -\alpha )/(b-a)$, giving exactly the affine $\bar{u}$. Conversely that affine function has constant $f^{\prime }({\bar{u}}^{\prime })$, so is a solution. This proves uniqueness among $C^1$ solutions; the scalar integrand is the one-dimensional case of the notation in the source.

If $f$ is strictly convex, Jensen's inequality applied to $u^{\prime }$ givesEquality in the strict inequality can occur only when $u^{\prime }$ is almost everywhere constant. Absolute continuity and the endpoints then give $u=\bar{u}$. Hence the affine function is the unique global minimizer.

A global minimizer has $I(\bar{u})\le I(v)$ for every admissible $v$. Global minimality implies strong local minimality, which implies weak local minimality. The reverse implications are not valid in general.

Where $f_v$ exists, the strengthened Weierstrass condition says that for some $\delta >0$, the excessis nonnegative for every $x\in [a,b]$, every trial slope $q\in \mathbb{R}$, and all $y,p$ with $|y-\bar{u}(x)|<\delta$, $|p-{\bar{u}}^{\prime }(x)|<\delta$. This is a neighborhood condition in height and reference slope, not just an inequality on the reference curve.

The classical Weierstrass sufficient condition, with the usual smooth integrand regularity in addition to continuity, states that an admissible $C^1$ curve satisfying the Euler-Lagrange equation, a coercively positive second variation, and this strengthened condition is a strong local minimizer. A strict excess away from the reference slope gives the corresponding strict conclusion. The differentiability assumptions are part of that classical sufficient theorem; continuity alone does not define its first and second variations.

The inequality $\cos t\ge 1-t^2/2$ givesThe elementary endpoint estimate $|u(x){|}^2\le x{\int }_0^1{u}^{\prime 2}$ implies $\int u^2\le \frac{1}{2}\int u^{\prime 2}$. ConsequentlyThus zero is the unique global minimizer, and in particular a strict strong and strict weak local minimizer.

For a fixed nonzero $u_0\in \mathcal{A}$,The quartic coefficient is positive. Therefore the energy is unbounded below and zero is not a global minimizer, for every value of $\lambda$.

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.