[EX] An operator of Hardy type

[gugo82]

Exercise:

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 \(T:L^{p^\prime} (0,\infty) \to C_0(]0,\infty[)\).

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 \(]0,\infty[\).

2. Is it possible to evaluate the operator norm \(\|T\|\) explicitly?

3. Is \(T\) one-to-one? Is \(T\) onto?

***

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.

[dissonance]

Some notes on notation, terminology, and all of that (almost useless) stuff.

In modern usage, a one-dimensional Hardy-type inequality is of the form

\[\tag{1} \left(\int_a^b \left(\int_a^x f(t)\, dt\right)^q u(x)\, dx\right)^{\frac{1}{q}} \le C \left( \int_a^b f^p(x)v(x)\, dx\right)^{\frac{1}{p}}, \qquad \forall f\ge 0\ \text{on}\ (a, b), \]

where \(-\infty \le aHardy operator \(Hf(x)=\int_a^xf(t)\, dt\):

\[\tag{2} \lVert Hf \rVert_{L^q \big( (a, b), u(x)dx\big)} \le C \lVert f \rVert_{L^p\big( (a, b), v(x)dx\big)}. \]

This last formulation makes sense for infinite \(p\) and/or infinite \(q\). It is the case of the exercise at hand, in which the conclusion "\(T\) is bounded" is equivalent to the Hardy-type inequality

\[\lVert Hf\rVert_{L^\infty \big( [0, \infty), x^{-1/p}dx\big)} \le C \lVert f\rVert_{L^{p'}\big( [0, \infty)\big)}.\]

See Kufner-Persson Weighted inequalities of Hardy type, §1.1:

http://books.google.it/books?id=4SqMH8D ... &q&f=false

To sum things up, I'd leave notation unchanged!!!