Suppose \(0 < a < b \leq \infty\), and find examples of functions \(f \in L^p((0, \infty ))\) if and only if:
\(a < p < b\)
\(a \leq p \leq b\)
\(p = a\)
Hint: consider functions of the following form: \[ f(x) \da x^{- \alpha} \abs{ \log(x) }^{ \beta} .\]
Solution
1. Case (i) (\(a < p < b\)): define \(f\) piecewise.
take, for fixed \(\delta > 0\) (chosen below), \[ f(x) \da \begin{cases} x^{-1/b}\,(-\log x)^{-1} & 0 < x \le 1/2,\\ 1 & 1/2 < x \le 2,\\ x^{-1/a}\,(\log x)^{-1} & x > 2 . \end{cases} \]
2. \(f \in L^p(0,\infty)\) iff \(a < p < b\).
near \(0\): \(\int_0^{1/2} x^{-p/b}(-\log x)^{-p}\,dx\) converges iff \(p/b < 1\), i.e. \(p < b\) (the power of \(x\) is integrable near \(0\) exactly when its exponent is \(> -1\); the log factor is irrelevant unless \(p = b\), where \(x^{-1}(-\log x)^{-1}\) is not integrable near \(0\): substituting \(u = -\log x\) gives \(\int du/u = \infty\)). Near \(\infty\): \(\int_2^\infty x^{-p/a}(\log x)^{-p}\,dx\) converges iff \(p/a > 1\), i.e. \(p > a\) (similarly, at \(p = a\) the integrand is \(x^{-1}(\log x)^{-1}\), not integrable). On \([1/2,2]\) the function is bounded. Hence \(f \in L^p\) exactly when \(a < p < b\).
3. Case (ii) (\(a \le p \le b\)): use \(\delta > 0\) with \((1+\delta)a > 1\) and\[ f(x) \da \begin{cases} x^{-1/b}\,(-\log x)^{-1-\delta} & 0 < x \le 1/2,\\ 1 & 1/2 < x \le 2,\\ x^{-1/a}\,(\log x)^{-1-\delta} & x > 2 . \end{cases} \]
4. \(f \in L^p\) iff \(a \le p \le b\).
near \(0\): \(\int_0^{1/2} x^{-p/b}(-\log x)^{-(1+\delta)p}\,dx\) converges iff \(p/b \le 1\), i.e. \(p \le b\): for \(p < b\) the power of \(x\) dominates, and at \(p = b\) we get \(\int_0^{1/2} x^{-1}(-\log x)^{-(1+\delta)b}\,dx < \infty\) since \((1+\delta)b > 1\) (substituting \(u = -\log x\) gives \(\int u^{-(1+\delta)b}\,du\)). Near \(\infty\): \(\int_2^\infty x^{-p/a}(\log x)^{-(1+\delta)p}\,dx\) converges iff \(p/a \ge 1\), i.e. \(p \ge a\). Hence \(f \in L^p\) exactly when \(a \le p \le b\).
5. Case (iii) (\(p = a\)): take \(\delta > 0\) with \((1+\delta)a > 1\) and\[ f(x) \da \begin{cases} x^{-1/a}\,(-\log x)^{-1-\delta} & 0 < x \le 1/2,\\ 1 & 1/2 < x \le 2,\\ x^{-1/a}\,(\log x)^{-1-\delta} & x > 2 . \end{cases} \]
6. \(f \in L^p\) iff \(p = a\).
at \(p = a\): near \(0\), \(\int_0^{1/2} x^{-1}(-\log x)^{-(1+\delta)a}\,dx < \infty\) (as \((1+\delta)a > 1\)), and near \(\infty\) similarly \(\int_2^\infty x^{-1}(\log x)^{-(1+\delta)a}\,dx < \infty\), so \(f \in L^a\). For \(p > a\): near \(0\), \(\int_0^{1/2} x^{-p/a}\,(-\log x)^{-(1+\delta)p}\,dx = \infty\) since \(p/a > 1\) makes the power of \(x\) non-integrable. For \(p < a\): near \(\infty\), \(\int_2^\infty x^{-p/a}(\log x)^{-(1+\delta)p}\,dx = \infty\) since \(p/a < 1\). Hence \(f \in L^p\) iff \(p = a\).
7. Q.E.D.