Let \(f\in L^1([0,1])\). Show that
- The limit \(\lim_{p\to 0^+} \| f \|_p\) exists.
- If \(m\{x:f(x)=0\}>0\), then this limit is zero.
Solution
1. The quantities \(\|f\|_p\) are finite for \(0<p\le1\).
For \(t\ge0\) and \(0<p\le1\), \[ t^p\le 1+t. \] Hence \[ \int_0^1 |f|^p\le 1+\|f\|_1<\infty. \] Thus \(\|f\|_p=(\int|f|^p)^{1/p}\) is well defined and finite for every \(0<p\le1\).
2. The logarithm of the \(p\)th moment is convex.
Assume first that \(f\) is not zero almost everywhere and define \[ A(p)=\int_0^1 |f(x)|^p\,dx, \qquad F(p)=\log A(p), \qquad 0<p\le1. \] For \(0<p,q\le1\) and \(0<\theta<1\), Hölder’s inequality gives \[ \begin{aligned} A((1-\theta)p+\theta q) &=\int |f|^{(1-\theta)p}|f|^{\theta q}\\ &\le A(p)^{1-\theta}A(q)^\theta. \end{aligned} \] Taking logarithms shows \[ F((1-\theta)p+\theta q) \le (1-\theta)F(p)+\theta F(q). \] Thus \(F\) is convex on \((0,1]\).
3. Compute the zeroth-moment limit.
Let \[ \alpha=m\{x:f(x)\ne0\}\in(0,1]. \] For almost every \(x\), \[ |f(x)|^p\longrightarrow \mathbf1_{\{f\ne0\}}(x) \qquad(p\downarrow0). \] For \(0<p\le1\) we have \(|f|^p\le1+|f|\), and \(1+|f|\) is integrable on \([0,1]\). Dominated convergence therefore gives \[ A(p)\longrightarrow\alpha. \] Hence \[ F(p)\longrightarrow\log\alpha. \]
4. If the zero set has positive measure, the limit is zero.
If \(m\{f=0\}>0\), then \(\alpha<1\). By step 3, \[ F(p)\longrightarrow\log\alpha<0. \] Therefore \[ \log\|f\|_p=\frac{F(p)}p\longrightarrow-\infty, \] and so \[ \boxed{\|f\|_p\longrightarrow0.} \] If \(f=0\) almost everywhere, this conclusion is immediate because every \(\|f\|_p=0\).
5. If \(f\ne0\) almost everywhere, the limit still exists.
Now suppose \(\alpha=1\). Step 3 gives \(F(p)\to0\), so define \(F(0)=0\). The resulting function is convex on \([0,1]\).
For \(0<p<q\le1\), convexity at \[ p=\frac pq q+\left(1-\frac pq\right)0 \] gives \[ F(p)\le \frac pq F(q). \] Thus \[ \frac{F(p)}p\le\frac{F(q)}q. \] Consequently the function \[ p\longmapsto \frac{F(p)}p=\log\|f\|_p \] is nondecreasing on \((0,1]\). Hence it has a limit as \(p\downarrow0\), equal to its infimum, possibly \(-\infty\) but never \(+\infty\) because \[ \frac{F(p)}p\le F(1)=\log\|f\|_1. \] Exponentiating, \(\|f\|_p\) therefore has a limit in \([0,\infty)\).
6. Conclude both assertions.
If \(f=0\) almost everywhere, the limit is \(0\). If the zero set has positive but not full measure, step 4 gives limit \(0\). If \(f\ne0\) almost everywhere, step 5 proves existence of the limit. These three cases cover every \(f\in L^1([0,1])\), which proves both claims.