1

here we have two cases to study

$(1)$ let us fix any $f \in C^{1}[ [0,1] \times [0,1]]$ ($k \neq 0$). Set $$[T(v)](x) := x^{-1}\int_0^xf(x,y)v(y)dy $$ for any $x \neq 0$ otherwise $[T(v)](0) := k(0,0)v(0)$

(i) is T a bounded linear operator $C^{0}([0,1]) \to C^0([0,1])$ ?

(ii) in the case (i) is true, is a compact operator between the two spaces?

my attempt: just tried to obtain some upper bound given an element of unitary norm but the fact that there is an "$x$" at the denominator gives me trouble. Tried to apply the integral mean value theorem (i feel it very similar to my case) but the fact the integral goes from $0$ to $x$ doesn't allow me to proceed any further. assuming it is bounded my idea was to try using Ascoli-Arzelà. Is a good idea?

$(2)$ let us fix any $f \in L^{\infty}(\mathbb{R}^2)$ and set Set $$[T(v)](x) := x^{-1}\int_0^xf(x,y)v(y)dy $$ for any $x \geq 0$

(i) is T a bounded linear operator $L^1([0,1]) \to L^1([0,1])$ ?

(ii) in the case (i) is true, is a compact operator between the two spaces?

my attempts: here my attempts are very poor. I could do something in the case $L^2$ (approximate $f$ restricted to $[0,1] \times [0,1]$ with an orthonormal bases) and then try to prove that the sequence of finite rank operator converge in operator norm to T. But here in $L^1$ i don't know how to proceed.

Thanks in advance for the help :)

Riccardo
  • 7,401
  • 1
    This could be helpful – Norbert Jan 12 '14 at 20:51
  • @Norbert Hardy inequality applies for $f \in L^p$ with $p \in (1,\infty)$. in my case $p=\infty$,but i think that restricting $f$ to $[0,1]^2$ is not a problem. in this case, which constant $p$ i have to take in the inequality? – Riccardo Jan 12 '14 at 21:12
  • I didn't read carefully your question. I think it is better for you google compact integral operators thann that – Norbert Jan 12 '14 at 22:14

0 Answers0