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No. Let ${\lambda }_1>0$ be the First Dirichlet eigenvalue of $-\Delta$ on $B$, choose a corresponding nonzero eigenfunction $w$, and takeThen $L{u}_k=0$ and every $u_k$ vanishes on $\partial B$, so the boundary values converge uniformly to those of $v=0$. But $|u_k{|}_{0;B}\to \infty$, and hence no subsequence converges uniformly on $B$.

Suppose a nonzero solution $w$ did not change sign. Replacing it by $-w$ if necessary gives $w\ge 0$. The argument of part (b.ii) is local and invariant under translation and scaling: it applies on every ball whose concentric double lies in $B_2$. If the zero set of $w$ had positive measure, a density point and this local result would make $w$ vanish on one ball. Applying the same result successively on overlapping balls would then give $w\equiv 0$ throughout the connected ball $B_2$, a contradiction. Thus the zero set has measure zero in $B_2$. Applying part (a.i) to $u=w$ givesfor every compactly supported $C^1$ function $\zeta$; the omitted zero set has measure zero. But $\inf q>{\lambda }_1$, while the variational characterization of the First Dirichlet eigenvalue provides a $\zeta$ withThis contradiction proves that every nonzero solution takes both positive and negative values.

A weak solution belongs to $V=H_0^1(\Omega )$ and satisfieswith the nonlinear right-hand side defining a bounded functional on $V$. Under the stated global Lipschitz hypothesis this requirement is automatic: $|f(s)|\le |f(0)|+ϵ|s|$ and boundedness of $\Omega$ give $f(u)\in L^2$ whenever $u\in V$. Continuity of $f$ alone, without growth control, would not guarantee this for an arbitrary Sobolev function.

For the semilinear Poisson contraction threshold, use the gradient norm on $V$, and let $C_P$ satisfy the Poincaré inequality $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$. For $g\in L^2$, Lax-Milgram gives the zero-Dirichlet Poisson solution $Pg\in V$. Testing with $Pg$ yields $\Vert \nabla Pg{\Vert }_2\le C_P\Vert g{\Vert }_2$. Define $T(v)=P(f(v))$, a self-map of the complete space $V$. For $v,w\in V$,Therefore $ϵC_P^2<1$ is a sufficient condition for Banach's theorem to give a unique fixed point and hence a unique weak solution. There is no need to assume $f(0)=0$ or to restrict to a small ball: the Lipschitz bound makes the map defined on all of $V$. The same estimates giveWith the optimal Poincaré constant, the threshold is $ϵ<{\lambda }_1$ for the First Dirichlet eigenvalue; the stated sufficient condition does not depend on knowing that optimal constant.