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Hello!Here is a solution for the second excercise:

Suppose \(T:X\rightarrow Y\) is bounded and \(f\in Y^{*}\).It follows that \(T\) is continuous on \(X\),so considering \(\{x_n,x\}\subset X\) such that \(x_n\rightarrow x \Rightarrow Tx_{n}\rightarrow Tx\).Since f is also continuous : \(f(Tx_n ...

Hello!

Let \(A=\{x_n,x\}\subset X\) and \(\{V_i\}_{i\in I}\) an open cover of \(A\),i.e. \(V_i\in\mathbb T\quad\forall i\in I\) and \(A\subset\bigcup_{i\in I}V_i\).

It follows that \(\forall n\in\mathbb N\quad\exists i_n\in I\) such that \(x_n\in V_{i_n}\) and also \(\exists i_x\in I\) such that ...

Hello!

Here is an answer to the second part.Let \(R\) be the ring of the 2X2 upper triangular real matrices (which is a subring of \(M_2(\mathbb R))\).

Consider a mapping \( \phi : R\rightarrow\mathbb R \) such that

\( \phi\left( \begin{array}{cc}a & b \\0 & c \end{array} \right)=a\)

Obviously ...

Hello!Let \(a_{k}=\sum_{n=0}^{k} x_{n}\) and \(l\geq m\in\mathbb N\).Since \(x_{n}\) are mutually orthogonal,applying Pythagoras' theorem gives us that:
$$ \|a_{l}-a_{m}\|^{2}=\|\sum_{n=m+1}^{l} x_{n}\|^{2}=\sum_{n=m+1}^{l}\|x_{n}\|^2=\sum_{n=1}^{l}\|x_{n}\|^2-\sum_{n=1}^{m}\|x_{n}\|^2 $$
It follows ...