Operator in funtional space

Question

Let $X = \{f :[0,1] \rightarrow \mathbb{R}/ f \hspace{1.0mm} is \hspace{1.0mm} continous\}$ and $T(f)_{(x)} = \int_{0}^{x}{f_{(s)}ds} $.

1. Prove that $\hspace{3.0mm}T^n(f)_{(x)} = \int_{0}^{x}{K_{n}(x,s)f(s)ds}\hspace{3.0mm}$ where $K_{n}$ is continous in $[0,1]$x$[0,1]$

2. Find $\|T\|$ and find $\sum_{n=1}^{\infty}{T^n}$

3. Solve the ecuation $(I-T)f = g$

Note: the norm we consider is the norm of uniform convergence in $X$

Answer 1

1. By induction, we can prove that $$\left( {{T^{\left( n \right)}}f} \right)\left( x \right) = \frac{1}{{\left( {n - 1} \right)!}}\int_0^x {{{\left( {x - s} \right)}^{n - 1}}u\left( s \right)ds} ,\,\forall n \geqslant 2,\,x \in \left[ {0,1} \right].$$

2. We have $$\left\| {Tf} \right\| = \mathop {\sup }\limits_{x \in \left[ {0,1} \right]} \left| {\int_0^x {f\left( s \right)ds} } \right| \leqslant \left\| f \right\|,$$ with $f \equiv 1$ we have $\left\| T \right\| = 1$.

3. You can use Fredholm alternative.