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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 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 \(-\infty \le a<b\le +\infty, 1\le p, q < \infty\). This is usually rewritten in terms of weighted Lebesgue spaces by means of the Hardy 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