# Finding a Lebesgue integrable function for every $1 \leqslant q < \infty$ that satisfies aditional requirement.

Consider the usual Lebesgue spaces. Amid one of my studies, I started wondering if it is possible to find an example that satisfies the following problem:

Problem. Consider arbitrary elements $$1 \leqslant p \leqslant q < \infty$$ and $$0 \leqslant \lambda < n\left( 1 - \frac{p}{q}\right)$$, where $$n$$ represents the dimension of the space we're working in (in this case, assume it is $$\mathbb R^n$$). My goal is to find a function $$f$$ defined on $$\mathbb R^n$$ such that $$f \in L^q(\mathbb R^n)$$ and additionally $$f$$ has to satisfy the following property:

$$\sup_{x \in \mathbb R^n, \, r > 0} r^{-\lambda} \int_{B(x,r)} |f(y)|^p \, dy = \infty.$$

Does anyone have an idea of a function that possibly satisfies this requirements?

Thanks for any help in advance.

• You need $p<q$. Feb 26 at 15:09
• @user469053 Thanks for your comment. If we take $p = q$ then the additional condition becomes $$\sup_{x \in \mathbb R^n, \, r > 0} r^{-\lambda} \int_{B(x,r)} |f(y)|^q \, dy = \infty.$$ Assuming that we find a function $f$ such that $f \in L^q(\mathbb R^n)$ then we would guarantee that $$\int_{B(x,r)} |f(y)|^q \, dy < \infty,$$ for every $x \in \mathbb R^n$ and $r > 0$. Obviously now if we take $\lambda = 0$ the condition fails, which means that you are right. Altought, I don't think it is that simple if we don't allow the case $\lambda = 0$. Either way, do you have an example for $p < q$?
– xyz
Feb 26 at 15:12
• I would recommend trying a function of the form $$f(y)=\left\{\begin{array}{ll} |y|^\alpha & : |y|\geqslant 1 \\ 0 & : |y|<1.\end{array}\right.$$ We can integrate this on spherical coordinates because it's spherically symmetric, so the integrals will become $1D$-integrals of $r$. We'll want $\alpha$ to be negative. For $y$ with $|y|$ large, $|y|^{\alpha q}=(|y|^\alpha)^q$ will be smaller than $|y|^{\alpha p}=(|y|^\alpha)^p$, because we're taking the small number $|y|^\alpha$ and raising it to a larger power ($q>p$). Choose $\alpha$ carefully based on $\lambda$. Feb 26 at 15:15
• @user469053 Do you think it is possible to find a counter example if we exclude the case $\lambda = 0$ and consider $1 \leqslant p \leqslant q < \infty$ ?
– xyz
Feb 26 at 15:31
• If $p=q$, then there is no $\lambda$ satisfying $$0\leqslant \lambda < n(1-p/q),$$ since $1-p/q=0$. Otherwise yes, I think the example I alluded to should work. Feb 26 at 15:33

We use spherical coordinates, so every point in $$\mathbb{R}^n\setminus\{0\}$$ can be uniquely represented as a pair $$(r,\theta)$$, where $$r>0$$ and $$\theta\in S^{n-1}$$, the unit sphere. If $$\mu$$ is the (non-normalized) Haar measure on the unit sphere, integrating in spherical coordinates is $$\int_0^\infty \int_{S^{n-1}} f(r,\theta)d\mu(\theta)r^{n-1}dr.$$ This is because we're integrating over different spheres of radius $$r$$. For a fixed $$r>0$$, the integral $$\int_{S^{n-1}}f(r,\theta)d\mu(\theta)$$ is the integral of the function over the fixed sphere of radius $$r$$ centered at $$0$$. Scaling the sphere by a factor of $$r$$ scales the surfaces areas (which is what $$\mu$$ is capturing) by a factor of $$r^{n-1}$$, which is where the $$n-1$$ comes from.

If we have a nice, spherically symmetric function $$f$$ (in other words, if it's a function only of $$r$$), then for any $$0, we get $$\int_{\mathbb{R}^n} |f(x)|^s dx = \int_0^\infty |f(r)|^sd\mu(\theta)r^{n-1}dr = \int_0^\infty |f(r)|^sr^{n-1}\Bigl[\int_{S^{n-1}} d\mu(\theta)\Bigr]dr = \sigma_n \int_0^\infty |f(r)|^sr^{n-1}dr,$$ where $$\sigma_n$$ is the surface area of $$S^{n-1}$$. It's just a constant, so the exact value of $$\sigma_n$$ won't affect the answer.

Suppose that $$f(r)=\left\{\begin{array}{ll} r^\alpha & : r \geqslant 1 \\ 0 & 0\leqslant r<1.\end{array}\right.$$

Then \begin{align*} \int_{\mathbb{R}^n} |f(x)|^q dx & = \sigma_n\int_1^\infty r^{q\alpha} r^{n-1} dr = \sigma_n \int_1^\infty r^{q\alpha+n-1}=\sigma_n \frac{r^{q\alpha+n}}{q\alpha+n}\Bigr|_{r=1}^{r\to\infty},\end{align*} which stays finite iff $$q\alpha+n<0$$.

Now let's fix a $$0 and integrate this same $$f$$ over $$B(0,R)$$, the ball centered at $$0$$ with radius $$R>1$$. We have \begin{align*} \int_{B(0,R)} |f(x)|^p dx & = \sigma_n\int_1^R r^{p\alpha} r^{n-1} dr = \sigma_n \int_1^R r^{p\alpha+n-1}=\sigma_n \frac{r^{p\alpha+n}}{p\alpha+n}\Bigr|_{r=1}^{R}\\ & =\frac{\sigma_n}{p\alpha +n}R^{p\alpha+n} - \frac{\sigma_n}{p\alpha+n}.\end{align*} Multiply by $$R^{-\lambda}$$ to get $$\frac{\sigma_n}{p\alpha +n}R^{p\alpha+n-\lambda} - \frac{\sigma_n R^{-\lambda}}{p\alpha+n}.\tag{1}$$ If $$\lambda=0$$, the $$R^{-\lambda}$$ term is constant, and otherwise the $$R^{-\lambda}$$ term will go to zero as $$R\to\infty$$. The expression $$(1)$$ will go to $$\infty$$ if and only if $$p\alpha+n-\lambda > 0$$.

So now we need $$\alpha$$ to satisfy $$q\alpha+n<0$$ and $$p\alpha+n-\lambda > 0$$. In order to satisfy the condition $$q\alpha+n<0$$, we need $$\alpha = -n/q-\epsilon$$ for a positive number $$\epsilon$$. Any positive number $$\epsilon$$ will work to satisfy $$q\alpha +n<0$$. We need to choose it to also satisfy $$p\alpha+n-\lambda> 0$$. We have \begin{align*} p \alpha + n -\lambda & = p\bigl(-\frac{n}{q}-\epsilon\Bigr)+n-\lambda = n\Bigl(1-\frac{p}{q}\Bigr)-\lambda-\epsilon p.\end{align*} So we will be okay for any $$0<\epsilon < \frac{n\Bigl(1-\frac{p}{q}\Bigr)-\lambda}{p}.$$ By assumption, that numerator is positive, so we have some choices which satisfy these inequalities.

• Thanks. This is very very good. Just a few details: before your second-to-last equation you wrote $p\alpha + n - \lambda < 0$. I think this is a typo and it should be $p\alpha + n - \lambda > 0$. Besides this, in your integral calculations you've also implicitly used the facts that $q\alpha + n - 1 \neq -1$ and $p\alpha + n - 1 \neq -1,$ which are indeed true after you describe your final value of $\alpha$ but perhaps you didn't consider this.
– xyz
Feb 27 at 16:10
• Thank you for the correction. It was a typo, and I've fixed it. As I divided by $\alpha q-n$, $\alpha p -n$, I did know that these could be zero. However, since I've worked this kind of problem before, I knew we'd be able to choose $\alpha$ to make it true. I probably should have said something as I was doing it, but I was being lazy. Feb 27 at 16:13
• It's perfect as it is. I believe that anyone who is slightly interested in understanding your answer will be able to figure it out by themselves :) Thanks for your help.
– xyz
Feb 27 at 16:19