I'm looking for an example of an harmonic function in the unit disk $D$, say $u$, such that the family $$(t\mapsto u(re^{it}))_{r\in[0,1)}$$ is uniformly integrable on the torus $\mathbb{T}$ and such that the radial maximal function $$u^*:\mathbb{T}\rightarrow[0,+\infty], t\mapsto \sup_{r\in[0,1)}|u(re^{it})|$$ is not integrable.
Knowing that:
- the Poisson integral of an integrable function on the torus gives us the uniform integrability condition;
- if an harmonic function is the real part of a function in the Hardy space $H^1(D)$, then the radial maximal function is still integrable;
I tried to get a function that falls in the first but not in the second category, and so I caught the classical example of such a function, i.e. the Poisson integral of the integrable function: $$\varphi(t):=\sum_{n=2}^{+\infty}\frac{\cos(nt)}{\log(n)},$$ where this sum can be interpreted, for example, in the sense of Cesaro means. So, writing down explicitly the Poisson integral of this function, we get: $$u:D\rightarrow\mathbb{R}, re^{it}\mapsto\sum_{n=-\infty\\|n|\ge2}^{+\infty}\frac{\operatorname{sgn}(n)}{2\log(|n|)}r^{|n|}e^{int}.$$ However, now I completely got lost in evaluating $u^*$.
So the question: is this an example of what I'm looking for? If it is, can someone help me in showing how to compute $u^*$ and in showing that it isn't an integrable function? If it isn't, can someone give me a working example? Thanks in advance.
I don't know if your attempted example works, but here is another one: define $f:\mathbb{T}\to\mathbb{R}$ by $$f(e^{it}) = \frac{1}{|t| \log^2(4/|t|)},\quad -\pi \le t\le \pi \tag1$$ This is an integrable function on $\mathbb{T}$, hence its harmonic extension $u$ to the disk is uniformly integrable on circles. It is also positive. I claim that $$ u^*(e^{it}) \ge \frac{C}{|t| \log(4/|t|)} \tag2 $$ for some constant $C>0$, which implies $u^*$ is not integrable. The idea is that the right hand side of (2) represents the size of the Hardy-Littlewood maximal function of $f$, and the radial maximal function is comparable to that for nonnegative functions.
It suffices to consider $t$ with $0<|t|<1$. Let $r = 1-|t|$. At the point $re^{it}$, the Poisson integral is estimated from below as $$ u(re^{it}) = \frac{1}{2\pi}\int_{-\pi}^{\pi} \frac{1-r^2}{|re^{it}-e^{is}|^2} f(s)\,ds \ge \frac{1}{2\pi} \int_{-|t|}^{|t|} \frac{1-r^2}{|re^{it}-e^{is}|^2} f(s) \,ds \tag3$$ By the triangle inequality, $$|re^{it}-e^{is}|\le 1-r + |e^{it}-e^{is}| \le |t| + 2|t| = 3|t| \tag4 $$ which in view of $1-r^2\ge 1-r = |t|$ means that the coefficient of $f(s)$ in the last integral in (3) is bounded from below by a positive constant, namely $1/3$. This is the key point of the proof: the values of $f$ on the interval $[-|t|, |t|]$ strongly influence the harmonic extension at $(1-|t|)e^{it}$. Specifically, $$ u(re^{it}) \ge \frac{1}{6\pi} \int_{-|t|}^{|t|} \frac{1}{|s| \log^2(4/|s|)} \,ds \ge \frac{C}{|t|\log (4/|t|)} \tag5$$ the last step being an explicit computation of the integral.
A more general form of the above is a theorem that appears in books on harmonic analysis: a nonnegative function is in $H^1$ if and only if it is in $L\log L$. Thus, any nonnegative function in $L^1\setminus (L\log L)$ works.