0

$M:[0,1]\times[0,1]\rightarrow \mathbb{R}$ be continuous. Let $L:C([0,1]) \rightarrow C([0,1])$ be defined by $$L(f)(t) = \int_{0}^{1}M(t,s)f(s) ds$$ and $C([0,1])$ has the norm $||f||_{\infty} = \sup\{|f(x)| \text{ such that } x \in [0,1]\}$,

$f \in C([0,1])$.

Questions:

1)Is $L:C([0,1]) \rightarrow C([0,1])$ continuous?

2)Show that $||L|| \le \sup \{\int_{0}^{1} | M(t,s)|\text{ }ds \text{ } | t \in[0,1]\}$

1) The first idea that came to my mind is to prove that $L$ is differentiable which automatically proves continuity. $M$ is continuous ,therefore Riemann differentiable.The question is if the integral is differentiable + I don't quite understand how $f$ and the norm of $C$ matters in this question. (infinite norm is just the maximal function value on $[0,1]$ (?))

2) As of two I would appreciate any help. Maybe it will follow from 1)

Update: This looks familiar...

Community
  • 1
John Lennon
  • 1,302
  • You have to be careful. Now $L: C([0,1])\to C([0,1])$ is a mapping between Banach space, do you really want to show that $L$ is differntiable? –  Mar 03 '15 at 10:09
  • What do you know about continuity of a linear map between normed linear spaces? If you can explain that, then perhaps someone here can help you better. – Prahlad Vaidyanathan Mar 03 '15 at 10:17
  • I just know that linear map is continuous iff it's bounded by a linear operator – John Lennon Mar 03 '15 at 10:21
  • 1
    Since $L$ is a linear operator, you have to show simply 2), and you will get that $L$ is automatically continuous. – Crostul Mar 03 '15 at 10:42

0 Answers0