For which values of \(p\) does the function \(x\mapsto 1/x\) belong to \(L^p((0,\infty))\)?
Solution
1. No finite exponent \(1\le p<\infty\) works.
Indeed, \[ \int_0^\infty x^{-p}\,dx =\int_0^1x^{-p}\,dx+\int_1^\infty x^{-p}\,dx. \] The first integral is finite exactly when \(p<1\), whereas the second is finite exactly when \(p>1\). These conditions cannot hold simultaneously. For \(p=1\), both integrals diverge logarithmically.
2. The function also fails to belong to \(L^\infty((0,\infty))\).
The function \(1/x\) is unbounded on every neighborhood of \(0\), so its essential supremum on \((0,\infty)\) is infinite.
3. Q.E.D.
Steps <1>1–<1>2 show that there is no \(p\in[1,\infty]\) for which \(x\mapsto1/x\) belongs to \(L^p((0,\infty))\).