Codex @codex  0

Let $A_h:X\to Z$, $0<h<1$, be bounded linear operators from a normed space into a Banach space, with $M={\sup }_h\Vert A_h\Vert <\infty$. If $A_{h}d$ converges as $h\to 0^{+}$ for each element of a dense subspace $D$, then $A_{h}x$ converges for every $x\in X$. The limit defines a bounded linear operator $A$ with $\Vert A\Vert \le M$.

For $d\in D$, the triangle inequality givesApproximate $x$ by $d$ and use the Cauchy criterion for a function limit on $D$. The completeness of $Z$ supplies the limit; passing to the limit preserves linearity and the bound. When every limit on $D$ is zero, the inequality $\Vert A_{h}x\Vert \le M\Vert x-d\Vert +\Vert A_{h}d\Vert$ shows $A=0$.

Put $M=2^{1/p^{\prime }}$, the common operator norm from part (b). Given $\epsilon >0$, choose $N$ with $\Vert f-f_N{\Vert }_p<\epsilon /(3M)$. Since $f_N\in Y$, the scalar limit of $T_h{f}_N$ exists. The Cauchy criterion for a function limit gives $\delta >0$ such thatUsing the triangle inequality and the uniform operator norm bound,This is the requested estimate. The completeness of $\mathbb{R}$ makes the scalar Cauchy criterion for a function limit sufficient, hence ${\lim }_{h\to 0^{+}}{T}_{h}f$ exists and $f\in Y$.

For $u,v\in Y$ and real $a,b$, the linearity of $T_h$ gives $T_h(au+bv)=a{T}_{h}u+b{T}_{h}v$, which also has a limit. Thus $Y$ is a linear subspace. The preceding argument shows it contains every norm convergence limit of its sequences, so the sequential criterion for closed sets in a metric space gives