# consequences of the uniform boundedness theorem

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