Let $X$ be the function space defined in the following way. A function $x$ belongs to $X$ if and only if the two conditions below are satisfied:
• $x : R → R$ is continuous.
• there exists a compact set $I_x$ of $R$ such that $x(t) = 0$ for every $t ∈ R$ \ $I_x$
How can we show that, for every $x ∈ X$, there exists $c_x > 0$ such that $sup_{ n∈N}$ $|T_nx| ≤ c_x$
For any $x\in\mathbb{R}$, there are finitely many $n$ such that $n\in I_{x}$, but now for other $n$, we have $|T_{n}x|\leq\displaystyle\int_{I_{x}}|x(s)|ds$ since those other $n$ lie outside the interval $I_{x}$, so the integration can be taken only on $I_{x}$. But then $|T_{n}x|\leq|I_{x}|\|x\|$, in other words, $\sup_{n}|T_{n}x|=M_{x}<\infty$.