Prove or disprove a claim related to $L^p$ space

Solution 1:

We want to mimic $f(x) = x^{-1/2}$ on $[0,1]$, but then cut $[0,1]$ into lots of intervals, and then on each interval remake the function so that its integral over that interval remains the same, but the support of the function is much smaller.

So let $x_n \to 0$ be a decreasing sequence with $x_0 = 1$. Write $x_{n} = (1+\epsilon_n) x_{n+1}$, and suppose that $1 > \epsilon_n \to 0$ is a decreasing sequence. Then define $f$ on $[x_{n+1},x_n)$ to be $$ f(x) = \cases{\frac1{\epsilon_n} x_{n}^{-1/2} & if $x_{n+1} \le x \le (1 + \epsilon_n^2) x_{n+1}$ \cr 0 & if $ (1 + \epsilon_n^2) x_{n+1} \le x < (1 + \epsilon_n) x_{n+1}$\cr} $$ Check that $\int_a^b f(x) \, dx \le \sqrt{b-a}$.

Now set $E_N = \bigcup_{n=N}^\infty [x_{n+1},(1 + \epsilon_n^2) x_{n+1})$. Then $\int_{E_N} f(x) \, dx = \int_0^{x_N} f(x) \, dx \approx \sqrt{x_N}$, whereas $|E_N| \le \epsilon_N x_N$.

Suppose $x_n = 1/n^\alpha$ for $0<\alpha \le 1$. Then $\epsilon_n \approx \alpha/n$. Hence $$ \int_{E_N} f(x) \, dx \gtrsim (\alpha^{-1}|E_N|)^{\alpha/(2(\alpha+1))} .$$

Still lots of details to be checked. But I think this will provide a counterexample.

Solution 2:

Through out the following, I use $P(E)$ to denote the perimeter of the set $E$. (See the definition of "sets of finite perimeter" in Giusti, minimal surfaces and functions of bounded variation. One may think of it as surface area.)

The following example shows $h \in \tilde{M}_p$, but $h \notin M_q,\forall$ $\frac{1}{1-t}\le q<p$

Let $Q_k=[x_{k+1},x_k) \times [x_{k+1},x_k)$, and define $h$ on $Q_k$ to be $h_k=k^{1+\alpha}$, where $\alpha$ is to be specified later, and zero otherwise. Let $E_K=\bigcup_{k=K}^{\infty}Q_k$.

Since $x_k=k^{-\alpha}$, $x_k-x_{k+1}\approx k^{-\alpha-1}$, then we have the following:

$|Q_k| \approx k^{-2\alpha-2}, P(Q_k) \approx k^{-1-\alpha}, \int_{Q_k} h\approx k^{-1-\alpha}, |E_K| \approx K^{-1-2\alpha}$ and$\int_{E_K}h \approx K^{-\alpha}$

For any $t \in (0,1/2)$, choose $0<\alpha < \frac{1}{1-2t}$, then $$\frac{\int_{E_K}h}{|E_K|^t} \approx \frac{K^{-\alpha}}{(K^{-1-2\alpha})^t}=K^{1-\alpha(1-2t)}\rightarrow \infty$$ Therefore, since $M_p(\Omega) \subset M_q(\Omega), \forall 1 \le q \le p$ and let $p_0 = p_0(t)=\frac{1}{1-t}$, $h \notin M_p(\Omega)$ for any $p \ge p_0$.

However, for any $E \subset \Omega$ and $E$ has finite perimeter, since $\mathcal{H}^1(Q_k \cap Q_{k-1})=\mathcal{H}^1(\{(x_k,y_k)\})=0$,we have $$P(E) \ge P(E \bigcap (\cup_{k=1}^{\infty} Q_k) = \Sigma_{k=1}^{\infty} P(E\cap Q_k)$$ Also, since $h$ is supported in $\cup_{k=1}^{\infty} Q_k$, we have $$\int_E h=\Sigma_{k=1}^{\infty}\int_{E\cup Q_k}h$$ Therefore, $$\frac{\int_E h}{P(E)} \le \frac{\int_E h}{\Sigma_{k=1}^{\infty} P(E\cap Q_k)} \le \sup\{\frac{\int_F h}{P(F)}: F \subset Q_k, k=1,2, \dots\}$$For any $F \subset Q_k$, we have by the isoperimetric inequality that $$\frac{\int_F h}{P(F)} = h_k |F|/P(F) \le C h_k P(F) \le C h_k P(Q_k) \approx k^{1+\alpha}k^{-1-\alpha}=1 \quad (1)$$where C in the estimate above is the isoperimetric constant in $\mathbb{R}^2$, thus $$\sup\{\frac{\int_F h}{P(F)}: F \subset Q_k, k=1,2, \dots\} < \infty \quad \quad $$ and thus $$\sup\{\frac{\int_E h}{P(E)}: E \subset \Omega\} < \infty$$

By (1), we conclude that for any cube $Q \subset \Omega$, $\frac{\int_Q h}{P(Q)} \le C $. Since $Q$ is a cube, $P(Q) \approx |Q|^{1/2}$, hence $\frac{\int_Q h}{|Q|^{1/2}} \le C $. Therefore, we showed that $$ \sup\{\frac{\int_Q h}{|Q|^{1/2}}: Q \subset \Omega\} \le C,$$ thus $h \in \tilde{M}_2$

Remark: I did plan to summarize what I had done, and share some experiences, but I'm lack of time. So I just put another example here, whose proof is very sneaky. Maybe it's better to know this idea. Thanks to @Stephen Montgomery-Smith's inspiration, I found a whole bunch of examples by myself, and I only choose this one to post because it is quite interesting.