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Suppose the continuous $f$ were not uniformly continuous. There would be an ${\epsilon }_0>0$ and pairs $x_n,y_n\in X$ with $d(x_n,y_n)<1/n$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0$. Sequential compactness gives a subsequence $x_{n_j}\to x\in X$. The triangle inequality gives $y_{n_j}\to x$ as well. Continuity at $x$ then gives $f(x_{n_j})-f(y_{n_j})\to 0$, a contradiction. Thus $f$ is uniformly continuous.

It need not be Lipschitz continuous. On the sequentially compact interval $[0,1]$, the continuous function $f(x)=\sqrt{x}$ hasNo finite Lipschitz constant exists.

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 any sequence in a complete, totally bounded metric space $X$. Cover $X$ by finitely many balls of radius $2^{-1}$; one contains infinitely many terms. From those terms choose an infinite subsequence lying in one ball of radius $2^{-2}$, and continue. The diagonal argument produces a subsequence $(x_{n_j})$ whose tail from its $j$th term onward lies in a ball of radius $2^{-j}$. Hencewhenever $k,\ell \ge j$, so the subsequence is Cauchy. Completeness makes it converge in $X$.

Every sequence in $X$ therefore has a convergent subsequence, so $X$ is sequentially compact. Sequential compactness is equivalent to compactness for metric spaces, and thus $X$ is compact. This proves that a complete totally bounded metric space is compact.

The assumptions prove convergence to the set of J-minimizing solutions, and subsequential convergence to a member of that set. Convergence to a single fixed member requires uniqueness or a suitable selection rule.

Let $u^{†}$ be any J-minimizing solution, whose existence is granted. It is feasible for every noisy-data constraint because $A{u}^{†}=f$ and $\Vert f-f_{\delta }\Vert \le \delta$. Minimality therefore gives $J(u_{\delta })\le J(u^{†})$. All reconstructions lie in this strongly sequentially compact space sublevel set. For any sequence ${\delta }_k\downarrow 0$, pass to a subsequence with $u_{{\delta }_k}\to u_{\ast }$. Boundedness of $A$ and the triangle inequality implyso $A{u}_{\ast }=f$. By sequential lower semicontinuity,Thus $u_{\ast }$ is a J-minimizing solution. The same compactness argument shows the distance to a set of exact minimizers tends to zero: otherwise a subsequence at a fixed positive distance would converge to an exact minimizer, a contradiction. If that minimizer is unique, every convergent subsequence has the same limit, giving the asserted full strong convergence.

For a counterexample to full convergence under the printed hypotheses alone, take $\mathcal{U}=\mathcal{V}=\mathbb{R}$, $A=0$, $f=f_{\delta }=0$, and $J(u)=\max \{|u|-1,0\}$. This is nonnegative and continuous, with compact sublevel sets $[-1-C,1+C]$ for $C\ge 0$. Every point of $[-1,1]$ is both an exact and a residual-method minimizer. Choosing $u_{1/k}=(-1{)}^k$ gives a valid sequence of minimizers that does not converge. The reusable corrected statement is compact-sublevel convergence of the residual method.

Assume for contradiction that $X$ is infinite-dimensional. Starting with $M_0=\{0\}$, apply Riesz lemma recursively to the proper closed finite-dimensional subspacesto choose unit vectors $x_n$ satisfyingFor $m<n$, the vector $x_m$ belongs to $M_{n-1}$, and thereforeThus $(x_n)$ lies in the closed unit ball but has no Cauchy subsequence, hence no convergent subsequence. A compact metric space is sequentially compact, contradicting compactness of the unit ball. Therefore $X$ is finite-dimensional, proving that a compact unit ball characterizes finite-dimensional normed spaces.

If $X$ is compact and $f:X\to \mathbb{R}$ is continuous, then $f(X)$ is compact by the continuous image of a compact space theorem. Every compact subset of $\mathbb{R}$ is bounded, so $f$ is bounded.

Conversely, suppose the metric space $X$ is not compact. For metric spaces, compactness is equivalent to sequential compactness, so there is a sequence of distinct points $(x_n)$ having no convergent subsequence. The setis closed: an accumulation point would supply a convergent subsequence. It is also discrete for the same reason. Consequently the functionis continuous. By part (a), $X$ is normal, and the Tietze extension theorem extends $g$ to a continuous $G:X\to \mathbb{R}$. Since $G(x_n)=n$, this extension is unbounded. Therefore, if every continuous real-valued function on $X$ is bounded, $X$ must be compact. This is the bounded-continuous-function characterization of compact metric spaces.