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Assume an exact solution of finite penalty exists and select the unique $J$-minimizing exact solution $u^{†}$. Minimality gives $\Vert K{u}_{\alpha }^{\delta }-f^{\delta }{\Vert }^2/2+\alpha J(u_{\alpha }^{\delta })\le {\delta }^2/2+\alpha J(u^{†})$. If $\alpha \to 0$ and ${\delta }^2/\alpha \to 0$, boundedness, weak sequential compactness of bounded sequences in a reflexive Banach space and lower semicontinuity identify every weak limit with $u^{†}$. The separate lower semicontinuity of $\Vert u{\Vert }_1$ and $\Vert u{\Vert }_2^2$, together with the upper bound on $J$, forces the Hilbert norms to converge. The Radon-Riesz theorem then gives norm convergence.

Use the standard inverse-problem setting of a bounded linear $K:{\ell }^2\to {\ell }^2$ and exact data having a finite-penalty solution. Let $u^{†}$ denote the unique $J$-minimizing exact solution for $J(u)=\Vert u{\Vert }_1+\beta \Vert u{\Vert }_2^2$, $\beta >0$. This is the limiting target of elastic net regularization in Hilbert sequence space; it need not be the Moore–Penrose minimum-norm solution unless additional hypotheses identify them, for example injectivity of $K$.

For each $\alpha >0$, the reconstruction objective is weakly lower semicontinuous, coercive by $\alpha \beta \Vert u{\Vert }_2^2$, and strictly convex. The direct method on the reflexive l2 sequence space gives existence, and strict convexity gives uniqueness. For two data sets, monotonicity of the one-norm subgradients and the quadratic penalty give, with $d=u_1-u_2$ and $g=f_1-f_2$,Completing the square yields $\Vert d\Vert \le \Vert g\Vert /(2\sqrt{2\alpha \beta })$, so each reconstruction map is continuous.

Minimality against $u^{†}$ givesChoose $\alpha \to 0$ and ${\delta }^2/\alpha \to 0$, for example $\alpha =\delta$. This bounds the Hilbert norms, drives the data residual to zero and gives $\mathrm{limsup}J(u_{\alpha }^{\delta })\le J(u^{†})$. Every weakly convergent subsequence has an exact solution as its limit, by boundedness of $K$; weak lower semicontinuity and uniqueness of the penalty minimizer identify that limit with $u^{†}$. Separately applying weak lower semicontinuity to the one-norm givesWeak convergence and convergence of Hilbert norms imply strong convergence by the Radon-Riesz theorem. This proves norm convergence of elastic net regularization:The argument applies to every sequence of noise levels and admissible data, hence gives the usual uniform noise-ball convergence for a fixed admissible exact datum.

For nonlinearity of elastic net reconstruction, zero is the minimizer exactly when $\Vert K^{\ast }{f}^{\delta }{\Vert }_{\infty }\le \alpha$. This includes a neighborhood of zero. If $K\ne 0$, choose $e_j$ with $K{e}_j\ne 0$ and $f^{\delta }=tK{e}_j$ for $t\Vert K{e}_j{\Vert }^2>\alpha$; zero then fails its optimality condition. A linear map vanishing on a neighborhood would vanish everywhere, so the elastic-net reconstruction is nonlinear whenever $K\ne 0$. For $K=I$ the explicit formula is $[R_{\alpha }f{]}_j=\mathrm{sign}(f_j)(|f_j|-\alpha {)}_{+}/(1+2\alpha \beta )$, exhibiting soft thresholding.

A concrete distinction from Moore–Penrose convergence is $K(u)=u_1+2u_2$ (embedded in the first output coordinate of ${\ell }^2$), exact datum one, and $\beta =1$. Its minimum-norm solution is $(1/5,2/5,0,\dots )$, with penalty $4/5$, while the unique penalty-minimizing exact solution is $(0,1/2,0,\dots )$, with penalty $3/4$. For $0<\alpha <2$ the exact-data elastic-net reconstruction is $(0,(2-\alpha )/(4+2\alpha ),0,\dots )$, converging to the latter. Thus it is not a Moore–Penrose-consistent regularization for every noninjective operator under the earlier definition.

The PDF supplies neither a nonzero assumption on $K$ nor a target/domain qualification here. Its unconditional nonlinearity claim is false for $K=0$, whose reconstruction is the zero linear map. The convergence proof above states the finite-penalty exact-solution hypotheses explicitly; it establishes convergence to the penalty-minimizing solution, not an automatically identical minimum-norm least-squares solution.

The Radon-Riesz property of a norm is the implicationIn a Hilbert space, expand the square using its inner product:Weak convergence gives $⟨u_k,u⟩\to \Vert u{\Vert }^2$, and convergence of norms gives $\Vert u_k{\Vert }^2\to \Vert u{\Vert }^2$. Hence the right side tends to zero. This proves the Hilbert-space Radon-Riesz theorem.

Assume (ii). Weak convergence givesUsing the inner product identity,Hence (ii) implies (i); this is the Radon-Riesz theorem in a Hilbert space.

Strong convergence immediately implies convergence of norms by the reverse triangle inequality.

Conversely, suppose $x_n\rightharpoonup x$ and $\Vert x_n\Vert \to \Vert x\Vert$. ThenThus $x_n\to x$ strongly. This is the Radon-Riesz theorem for Hilbert spaces.

Part (c)(i) already gives weak continuity of $t\mapsto \theta (t)$ in $L^2$. In the assumed energy equality, the dissipation integralis absolutely continuous. The forcing integrand is in $L^1(0,T)$ because $u\in L^{\infty }(0,T;L^2)$ and $\theta \in L^{\infty }(0,T;L^2)$. The equality therefore makes $t\mapsto \Vert \theta (t){\Vert }_2^2$ continuous.

Whenever $t_n\to t$, weak continuity gives $\theta (t_n)\rightharpoonup \theta (t)$ and the energy equality gives convergence of their norms. The Radon-Riesz theorem, or directly the strong continuity from weak continuity and an energy equality, now gives $\theta (t_n)\to \theta (t)$ in $L^2$. Thus

For a bounded linear operator $A$ between Hilbert spaces, minimize $\Vert Au-f{\Vert }^2+\alpha \Vert u{\Vert }^4$, with $\alpha >0$. The objective is coercive and weakly lower semicontinuous, and its strictly convex norm penalty gives a unique minimizer. Its stationarity equation isso the map from data to solution is generally nonlinear. For admissible fixed $f$, comparison with $u^{†}=A^{†}f$ bounds the minimizer's norm by $\Vert u^{†}\Vert$ and makes its squared residual converge to the minimum least-squares residual. Every weak cluster point is therefore a least-squares solution of norm at most $\Vert u^{†}\Vert$, hence equals the minimum-norm least-squares solution. Weak lower semicontinuity then forces convergence of norms, and the Radon-Riesz theorem gives strong convergence. This supplies a nonlinear regularization of an inverse problem.

For completeness, the fixed-$\alpha$ data-to-solution map is continuous. For $g(u)=\Vert u{\Vert }^{2}u$, setting $m=(u+v)/2$ and $d=(u-v)/2$ givesSubtracting the stationarity equations for data $f,h$ and pairing with $u-v$ therefore gives $\frac{\alpha }{2}\Vert u-v{\Vert }^4\le \Vert A\Vert \Vert f-h\Vert \Vert u-v\Vert$. Thuswhich proves continuity directly.

A normed vector space has the Radon-Riesz property when weak convergence $u_k\rightharpoonup u$ together with $\Vert u_k\Vert \to \Vert u\Vert$ implies strong convergence $u_k\to u$. The Hilbert-space case is the Radon-Riesz theorem. The condition prevents norm from remaining in weakly invisible directions.

If $u:[0,T]\to H$ is weakly continuous and an energy equality makes $t\mapsto \Vert u(t){\Vert }_H$ continuous, then $u$ is strongly continuous. This follows from the Radon-Riesz theorem applied whenever $t_n\to t$.