If $f_1,f_2,…$ are bounded continuous functions with $\sup_{B_n}|f_n|\to0$ and $T_n$ is a contraction on $C_b$, can we show $\sup_{B_n}|T_nf_n|\to0$?

Question

Let $E_n$ be a metric space, $B_n$ be a Borel subset of $E_n$ and $f_n:E_n\to\mathbb R$ be bounded and continuous for $n\in\mathbb N$ with $$\sup_{x\in B_n}\left|f_n(x)\right|\xrightarrow{n\to\infty}0.\tag1$$ Now assume $T_n$ is a bounded linear operator on the space $C_b(E_n)$ of bounded and continuous function from $E$ to $\mathbb R$ equipped with the supremum norm and assume the operator norm of $T_n$ is at most $1$ for $n\in\mathbb N$.

Are we able to show that $$\sup_{x\in B_n}\left|(T_nf_n)(x)\right|\xrightarrow{n\to\infty}0?\tag2$$

The problematic thing is that the contractivity only yields $$\sup_{x\in B_n}\left|(T_nf_n)(x)\right|\le\sup_{x\in E_n}\left|(T_nf_n)(x)\right|\le\sup_{x\in E_n}\left|f_n(x)\right|.\tag3$$

Answer 1

No.

Let $E=\Bbb R$, $B=[-\frac12,\frac12]$, $f_n(x)=\min\{1,x^{2n}\}$, $T_nf(x)=\frac12f(2x)$. Then $\sup_{x\in B}|f_n(x)|=2^{-n}\to 0$, $\|T_n\|=\frac12$, $\sup_{x\in B}|T_nf_n(x)|=1$.

Answer 2

No. Let $E=\{0,1\}$ as a subspace of $\mathbb{R}$, and let $f:E\to\mathbb{R}$ be the identity map. Let $B=\{0\}$. Define $T$ by the rule $(Tg)(0)=g(1)$ and $(Tg)(1)=g(0)$. Take $f_n=f$ and $T_n=T$ for all $n$.