= Solution
First scale the stated unit-ball approximation. For every nonzero residual $r\in F$, apply the hypothesis to $r/\|r\|$ and multiply the resulting vector by $\|r\|$. This gives $v\in E$ with
$$
\|v\|\le R\|r\|,\qquad \|Tv-r\|\le k\|r\|.
$$
For zero residual use $v=0$. Start with $r_0=y$, choose $v_j$ by this rule for $r_j$, and set $r_{j+1}=r_j-Tv_j$. Inductively,
$$
\|r_j\|\le k^j\|y\|,\qquad \|v_j\|\le Rk^j\|y\|.
$$
The <Banach space> $E$ therefore contains the sum $x=\sum_{j\ge0}v_j$, with
$$
\boxed{\|x\|\le\frac R{1-k}\|y\|.}
$$
The partial sums satisfy $T\sum_{j=0}^{n-1}v_j=y-r_n$. Boundedness of $T$ gives convergence of the left side to $Tx$, while $r_n\to0$ gives $Tx=y$. Thus \b[$T$ is <surjective> with the stated lifting bound]. This <geometric correction for approximate surjectivity> used <completeness> of $E$, not any unproved <completeness> of $F$.
Put $C=R/(1-k)$. Given a <Cauchy sequence> $(y_n)$ in $F$, choose a subsequence $(y_{n_j})$ such that $\|y_{n_{j+1}}-y_{n_j}\|\le2^{-j}$ for $j\ge1$. Lift each difference to $u_j\in E$ with $Tu_j=y_{n_{j+1}}-y_{n_j}$ and $\|u_j\|\le C2^{-j}$. Lift $y_{n_1}$ to $u_0$. The sum $u=u_0+\sum_{j\ge1}u_j$ exists in $E$, and its image is the limit of the subsequence. A <Cauchy sequence> with a convergent subsequence converges to the same limit: use the <triangle inequality> between an arbitrary late term, a later subsequence term, and the limit. This proves \b[$F$ is complete], the <completeness forced by uniformly bounded lifting> principle.
The <open mapping theorem> states that a bounded <surjective> linear operator between <Banach spaces> maps open sets to open sets. In particular a bounded linear bijection between <Banach spaces> has a bounded inverse. To deduce the <closed graph theorem>, let $S:E_1\to F_1$ be an everywhere-defined <linear map> between <Banach spaces> with closed graph. Its graph is a closed subspace of the Banach product $E_1\times F_1$, with <norm> $\|(x,y)\|=\|x\|+\|y\|$, so it is itself Banach. The projection
$$
\pi_1:\operatorname{graph}S\to E_1,\qquad(x,Sx)\mapsto x
$$
is a bounded linear bijection. The <open mapping theorem> makes its inverse bounded. Composing that inverse with the bounded second projection proves that $S$ is bounded. This is the required closed graph conclusion, rather than a continuity assumption on $S$.
Finally consider the identity map $I:(C(X),\|\cdot\|)\to(C(X),\|\cdot\|_\infty)$. If $f_n\to f$ in the new <norm> and $f_n\to g$ uniformly, each assumed continuous <point evaluation functional> gives $f_n(x)\to f(x)$, while <uniform convergence> gives $f_n(x)\to g(x)$. Hence $f=g$. Since both spaces are metric, this sequential argument proves the graph is closed. Both <norms> are complete, so the <closed graph theorem> makes $I$ bounded. Its inverse is bounded by the <open mapping theorem>. Consequently there are constants $c,C'>0$ such that
$$
\boxed{c\|f\|_\infty\le\|f\|\le C'\|f\|_\infty\qquad(f\in C(X)).}
$$
This proves <Banach norm rigidity from continuous point evaluations>. Pointwise continuity alone would not give a common bound on all evaluations; <completeness> supplies that through the closed graph argument.
Back to article page
