Preface
This is a problem in my Functional Analysis course. I'm really struggling to dissect part (b) any help/alternative solutions would be greatly appreciated on both parts of the question :)
Question
Let X be the space of continuous functions $x : [1,\infty)\rightarrow \mathbb R$ which have compact support, that is there exists a compact interval $I_x$ of $[1,\infty)$ such that $x(t) = 0 \quad \forall t ∈ [1,\infty) \backslash \space I_x$. Consider X with the norm $\left\lVert x \right\rVert = \max_{t\in[1,\infty)} |x(t)|$ and define the mapping $T : X \rightarrow X$ as $(Tx)(t) = \frac{1}{t^2}x(t)$,for every $t \in[1,\infty).\\$
(a) Show that T is linear, bounded, bijective and find $\left\lVert T \right\rVert$ and $T^{-1}$ and show that it is bijective. \begin{aligned} \forall x_{1}, x_{2} \in X \space \text{ and }\space \alpha, \beta \in \mathbb{R} \\ T\left(\alpha x_{1}+\beta x_{2}\right)(t) &=\frac{1}{t^{2}}\left(\alpha x_{1}+\beta x_{2}\right)=\alpha \frac{1}{t^{2}} x_{1}+\beta \frac{1}{t^{2}} x_{2} \\ & = \alpha\frac{1}{t^{2}} x_{1}(t)+\beta \frac{1}{t} x_{2}(t)\\ &=\alpha\left(T x_{1}\right)(t)+\beta\left(Tx_{2}\right)(t). \\ \text{Hence T is linear.} \end{aligned}
\begin{array}{l} \text { Note: } 0 \lt \frac{1}{t^{2}} \leqslant 1 \quad \forall t \in[1, \infty) \\ \Rightarrow \quad \frac{1}{t^{2}}|x(t)|=\left|\frac{1}{t^{2}} x(t)\right| \leqslant|x(t)| \quad\forall t \in[1, \infty) \\ \text { i.e }\|T x\|=\left\|\frac{1}{t^{2}} x(t)\right\|=\max _{t_{t}(1, \infty)}\left|\frac{1}{t^{2}} x(t)\right| \\ \leqslant \max _{t \in[1, \infty)}|x(t)|=\|x\| \\ \therefore T \text { is bounded } \end{array}
as for finding $\left\lVert T \right\rVert$ and showing that $T^{-1}$ it is bijective I am unsure how to do this :/
(b) The next problem I have is showing that $T^{-1}$ is not bounded and to explain why this not in contradiction with the Open Mapping Theorem and the Bounded Inverse Theorem