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The direct method in the calculus of variations gives the following existence theorem. Suppose bounded sequences in the Banach space $\mathcal{U}$ have $\tau$-convergent subsequences, and $E$ is a proper extended-real function with coercivity and is $\tau$-sequentially lower semicontinuous. Then $E$ has a finite-valued global minimizer.

To prove this, properness makes $m=\inf E<+\infty$. First rule out $m=-\infty$: a sequence with $E(u_n)\to -\infty$ eventually lies in a fixed sublevel set, which is bounded by coercivity. A $\tau$-convergent subsequence would then have a limit $u$ with $E(u)\le -\infty$, contradicting the codomain. Hence $m$ is finite.

Choose a minimizing sequence with $E(u_n)\to m$. It eventually belongs to the bounded sublevel set $\{E\le m+1\}$. Extract $u_{n_j}\stackrel{\tau }{\to }\bar{u}$. By sequential lower semicontinuity,Thus $E(\bar{u})=m$. In particular, a reflexive Banach space with the weak topology supplies the required subsequence property by weak sequential compactness of bounded sequences in a reflexive Banach space. A strictly convex function has at most one minimizer; this is an additional property, not part of the existence theorem.

First interpret the stated well-definedness as including uniqueness of the global minimizer for each positive parameter. Under that assumption, the parameter continuity of variational regularization follows from the direct method in the calculus of variations argument below.

Let $M=D(0)=\Vert f{\Vert }^2/2$. For large $n$, ${\alpha }_n\ge \beta :=\alpha /2$. Comparison with zero, using $J(0)=0$, givesCoercivity of ${\Phi }_{\beta ,f}$ therefore bounds the sequence. Take a $\tau$-convergent subsequence with limit $\bar{u}$. For any $v\in \mathrm{dom}J$, optimality givesThe last two terms vanish. Sequential lower semicontinuity at the fixed parameter $\alpha$ implies ${\Phi }_{\alpha ,f}(\bar{u})\le {\Phi }_{\alpha ,f}(v)$. Thus every cluster point is a global minimizer at $\alpha$. Uniqueness makes it $u_{\alpha }$, and the subsequence property provesIndeed, a subsequence staying outside a neighborhood of $u_{\alpha }$ would have a further convergent subsequence, whose limit must be $u_{\alpha }$, a contradiction.

The listed existence hypotheses alone do not imply uniqueness or convergence of an arbitrary selection. For example, on ${\mathbb{R}}^2$, set $K(x,y)=x$, $f=1$, andThis is nonnegative, convex, continuous and has $J(0)=0$; every ${\Phi }_{\alpha ,f}$ has coercivity. Its minimizers are $(1/(1+2\alpha ),y)$ for all $|y|\le 1$. Along ${\alpha }_n=\alpha +1/n$, selecting $y_n=(-1{)}^n$ prevents convergence. Without uniqueness, the valid conclusion is that every cluster point minimizes the limiting objective.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence.

Condition (i) is insufficient because the allowed limit may be zero. For example, $a_n=1/n$ is nonzero and converges to zero, whereas $1/a_n=n$ does not converge in $\mathbb{R}$.

A compact space is one for which every open cover has a finite subcover. A metric space is sequentially compact when every sequence in it has a convergent subsequence whose limit belongs to the space.

Let $(x_n)$ be a Cauchy sequence in a compact metric space $X$. By sequential compactness of a compact metric space, it has a convergent subsequence $x_{n_j}\to x\in X$. Given $\epsilon >0$, choose $N$ such that $d(x_m,x_n)<\epsilon /2$ whenever $m,n\ge N$, and then choose $j$ such that $n_j\ge N$ and $d(x_{n_j},x)<\epsilon /2$. For every $n\ge N$, the triangle inequality givesThus the whole sequence converges to $x$. Every Cauchy sequence converges in $X$, so $X$ is a complete metric space.

The sum is nonempty because $A$ and $B$ are nonempty. It is a bounded set: if $|a|\le M_A$ and $|b|\le M_B$, then $|a+b|\le M_A+M_B$.

To prove that it is a closed set, let $x_j=a_j+b_j\in A+B$ be a convergent sequence with $x_j\to x$. By the Heine-Borel theorem, $A$ is a compact set, so $(a_j)$ has a convergent subsequence $a_{j_r}\to a\in A$. ThenSince $B$ is closed, $x-a\in B$, and hence $x=a+(x-a)\in A+B$. Therefore $A+B$ is nonempty, closed and bounded, so $A+B\in \mathcal{K}$.

Each $A^{ϵ}$ is continuous, adapted, and increasing. From ucp convergence one can choose a subsequence that converges uniformly almost surely on every compact interval. Its limit $A$ is therefore also continuous and increasing, hence a finite-variation process. The stochastic integral $M_t={\int }_0^t\mathrm{sgn}(B_s)d{B}_s$ is a continuous local martingale, and part e gives the semimartingale decompositionConsequently $|B|$ is a semimartingale. In fact, comparison with the Tanaka formula identifies $A_t$ as the local time of a semimartingale $L_t^0(B)$.

Choosewhere $r_p$ is a normal radius at $p$. Take piecewise smooth curves $c_j$ from $p$ to $q$ such that $L_g(c_j)\to d_g(p,q)$. Each $c_j$ first leaves the normal ball $B_g(p,\delta )$ at a point $p_j$. The Gauss lemma gives $d_g(p,p_j)=\delta$. The geodesic sphereis compact, because the tangent-space sphere is compact and ${\exp }_p$ is defined on it. After taking a convergent subsequence, let $p_j\to p_0$.

The part of $c_j$ after $p_j$ has length at least $d_g(p_j,q)$, so continuity of the Riemannian distance givesThe triangle inequality gives the reverse inequality. Therefore

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set, $d<\infty$, and $f:U\to {\mathbb{R}}^d$ a continuous function. Through every $(t_0,y_0)\in U$ there is a local solution of a differential equation: a continuously differentiable function $y:I\to {\mathbb{R}}^d$ on an open interval containing $t_0$, satisfying the displayed initial value problem pointwise. No Lipschitz condition is assumed.^[ref]

Continuity of a proposed rate law therefore guarantees at least one trajectory for a short time. For a physical model, this is a first consistency check: can the stated rule and initial condition actually be followed? Prediction also needs uniqueness. The delayed solutions of a square-root differential equation below show how the same rule and initial condition can leave a departure time undetermined. The Picard-Lindelöf theorem supplies a sufficient additional hypothesis for uniqueness; failure of that hypothesis alone does not prove nonuniqueness.

For the initial value problem $y^{\prime }=2\sqrt{|y|}$, $y(0)=0$, a forward solution of a differential equation can stay zero until any $c\ge 0$, then follow $y=(t-c{)}^2$. The delayed solutions of a square-root differential equation verify this freedom exactly.

Local existence gives neither a global solution of a differential equation nor a bound on all future values. The continuous function $f(t,y)=y^2$ already permits finite-time blowup; the finite-dimensional continuation criterion explains the additional issue of extending a trajectory within its domain.

Proof outline: choose a compact set $K=[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$. By the extreme value theorem, $M={\max }_K\Vert f\Vert <\infty$. Take $0<h\le a$ with $Mh\le r$. Forward Euler method polygons remain in the closed ball and satisfy $\Vert y_n(t)-y_n(s)\Vert \le M|t-s|$. The Arzelà-Ascoli theorem yields a convergent subsequence with uniform convergence. Uniform continuity of $f$ on $K$ makes their integral equation defect tend uniformly to zero, so the limit satisfiesThe fundamental theorem of calculus gives a continuously differentiable function. Repeat backward and join at $t_0$, where both derivatives equal $f(t_0,y_0)$.^[ref]

The compact set here uses a finite-dimensional vector space; bounded closed balls in an arbitrary Banach space need not be compact. When $M=0$, the constant solution of a differential equation $y=y_0$ works on the entire time interval in $K$. In general the proof selects a subsequence, without asserting that all polygons converge to one unique trajectory.

Selecting a subsequence is a regular matrix summability method: every row has one entry equal to one, every fixed column eventually vanishes, and all row sums equal one. Every bounded sequence of real or complex numbers has a convergent subsequence by the Bolzano-Weierstrass theorem, so each bounded sequence is summable by some sequence-dependent regular method. The order of the quantifiers matters: no single regular matrix sums all bounded sequences.