Let $X$ be the function space defined in the following way. A function $x$ belongs to $X$ iff the two conditions are satisfied:

i) $x : \mathbb{R} \rightarrow \mathbb{R}$

ii) there exists a compact interval $I_x$ of $\mathbb{R}$ such that $x(t) = 0$ for every $t \in \mathbb{R} | I_x$.

$T_nx = \int^n_0 x(s) ds$ for every $x \in X$.

where the norm $||x|| = max_{t \in \mathbb{R}} |x(t)|$ and $n \in \mathbb{N}$.

I have already shown $X$ is a vector space over $\mathbb{R}$ and proved $T_n$ is linear and bounded. I now need to show that for every $x \in X$, there exists $c_x > 0$ such that $sup_{n \in \mathbb{N}}|T_nx| \leq c_x$. I'm not sure how to begin.


I assume you also want your functions in $X$ to be continuous, or at least bounded and integrable, so that $\|x\|$ is well-defined for $x\in X$. If this is the case, here's one way to obtain a bound:

\begin{align*} |T_nx|&=\left|\int_0^nx(s)\ ds\right|\\ &\leq\int_0^n|x(s)|\ ds\\ &=\int_{[0,n]\cap\operatorname{supp}(x)}|x(s)|\ ds\\ &\leq\|x\|m([0,n]\cap\operatorname{supp}(x))\\ &\leq\|x\|m(\operatorname{supp}(x)) \end{align*} Where $m$ denotes Lebesgue measure on $\mathbb R$ and $\operatorname{supp}(x)$ is the support of $x$.

You mention the uniform boundedness principle, but I should warn you that you can't use it here, since $X$ is not complete


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