We have $f \in L^p$. The goal is to show that $\exists \psi \in C(\mathbb{R^+}, \mathbb{R^+})$ such that $$ \lim_{s \to +\infty} \frac{\phi(s)}{s}=+ \infty \text{ and } \phi(|f|) \in L^p$$

I neeed some pointer on how to begin.

Edit: I went to my professor for a hint and he said that there exists a continous function $\psi$ such that $\lim_{s \to \infty}\psi(s)=+\infty$ with $\sum \psi(n)a_n < \infty$ for a convergent series of poistive terms. Use $\psi$ to construct $\phi$.


1 Answer 1


There are several approaches, though fundamentally the idea is the same: when a series converges, it can be made bigger while preserving convergence.

Consider the sequence $$f_n(x) = \begin{cases} |f(x)|, \quad &|f(x)|\ge n, \\ 0 \quad & |f(x)|<n \end{cases} \tag{1}$$ Observe that $\|f_n\|_{L^p}\to 0$ as $n\to\infty$. Therefore, there is a stricly increasing sequence $n_k$ such that $\|f_{n_k}\|_{L^p}\le 2^{-k}$. Define $g(x) = \sum_k f_{n_k}$, which is in $L^p$.

By construction, $g$ is of the form $\psi(|f|)$ with $$\psi = \sum_{k} \chi_{[n_k,\infty)} $$ Hence, $\psi(s)/s\to\infty $ as $s\to\infty$.

However there is the annoying requirement of continuity. One way around it is to replace (1) with $$f_n(x) = \begin{cases} |f(x)|, \quad &|f(x)|\ge n+1, \\ |f(x)|\left(|f(x)|-n\right), \quad &n\le |f(x)|\le n+1, \\ 0 \quad & |f(x)|<n \end{cases} \tag{2}$$ and repeat. The function $\psi$ will now be the sum of "tapered" characteristic functions. Continuity of the sum is easy to check because only one summand changes at a time.

  • $\begingroup$ Isn't $\psi$ just a sum of 1's (the indicator functions). I do not see how that is equal to $\sum_k f_{n_k}$. $\endgroup$
    – Iconoclast
    May 9, 2014 at 15:09
  • $\begingroup$ How can $\sum_k \chi_{[n_k, \infty)}(|f(x)|)$ take values other than integers? Maybe I'm not getting your point. $\endgroup$
    – Iconoclast
    May 9, 2014 at 18:12

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