Let \(p\in [1,\infty[\) and, as usual, let \(p^\prime \in ]1,\infty]\) be the Hölder conjugate exponent of \(p\), i.e. \(\frac{1}{p} + \frac{1}{p^\prime} = 1\).
1. Prove that equation:
\[
\tag{1} Tu(x) := \frac{1}{x^{1/p}}\ \int_0^x f(t)\ \text{d} t
\]
defines a bounded linear operator \(T:L^{p^\prime} (0,\infty) \to C_0(]0,\infty[)\).
Testo nascosto, fai click qui per vederlo
Hints: Linearity and boundedness are trivial. The difficult part is proving that \(Tu\in C_0(]0,\infty[)\); to this end, use an approximation argument.
In particular, approximate \(u\) with \(u_n:=u\ \chi_{]0,n[}\) in \(L^{p^\prime}\) and prove that a) \(Tu_n \in C_0(]0,\infty[)\) and b) \(Tu_n\) converges uniformly to \(Tu\) in \(]0,\infty[\).
In particular, approximate \(u\) with \(u_n:=u\ \chi_{]0,n[}\) in \(L^{p^\prime}\) and prove that a) \(Tu_n \in C_0(]0,\infty[)\) and b) \(Tu_n\) converges uniformly to \(Tu\) in \(]0,\infty[\).
2. Is it possible to evaluate the operator norm \(\|T\|\) explicitly?
3. Is \(T\) one-to-one? Is \(T\) onto?
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I don't know if "Hardy-type operator" can be correctly attached to the operator \(T\).
Neverthless, such a name came to my mind because the righthand side of (1) remembered to me the lefthand side of the classical Hardy inequality...
So, if anyone has a better name for the operator \(T\), just let me know and I'll modify the thread's title.