# If $u \notin L^p[0,1]$, can we find some $w \in L^{\frac{p}{p-1}}[0,1]$ such that $\int_0^1 u \cdot w = \infty$?

The question is as in the title.

For some fixed $$p \in (1,\infty)$$, let $$p' \in (1,\infty)$$ be such that $$\frac{1}{p}+\frac{1}{p'}=1$$.

Then, for any $$u \in L^1[0,1] - L^p[0,1]$$, I wonder if it is possible to find some $$w \in L^{p'}[0,1]$$ such that $$$$\int_0^1 u \cdot w = \infty$$$$ where we assume $$u$$ and $$w$$ to be $$\mathbb{R}^N$$-valued for some $$N \in \mathbb{N}$$.

I think this must be true, but cannot really prove myself..

• Its possible if $u\notin L^{p-r}$ for some $r > 0$ Jan 13 at 22:12
• @Jakobian How? Could you help me? Jan 13 at 22:18
• Are you asking if such a $w$ exists or if it is possible to build it explicitly? Jan 13 at 22:18
• @LL3.14 Just proof of exisetence will do.. Jan 13 at 22:20
• I'm not sure if it always exists though... my answer is still incomplete Jan 13 at 22:21

Thanks to Jakobian for pointing it out: some technical details require stronger hypothesis for this approach to work! It suffices to let $$u\in L^1[0,1]\setminus L^{p-r}[0,1]$$ for some $$r>0$$ small enough.

The main idea is to tweak the usual "dual" function so as to get the desired $$w$$.

Let $$u\in L^{1}[0,1] \setminus L^{p}[0,1]$$. A standard "dual" function in $$L^{p'}[0,1]$$ would be $$v = \operatorname{sign}(u)|u|^{p-1}$$ - if $$u \in L^p[0,1]$$. However, in this case $$v$$ is not in $$L^{p'}[0,1]$$, since $$\int_{0}^{1} |u(x)|^{p'(p-1)} dx = \int_{0}^{1} |u(x)|^p dx = \infty.$$

Now, lets fix this somehow. Let $$g_t = \frac{1}{|u|^t}\mathbf{1}_{\{|u|>1\}}$$, and let $$s=\inf_{t\in [0,\infty)}\{ug_t \in L^{p}[0,1]\};$$ note that since $$u\in L^{1}[0,1]\setminus L^{p}[0,1]$$ we have that the infimum exists, as $$t=\frac{1}{p'}$$ works and $$t=0$$ does not.

When $$u\notin L^{p-r}[0,1]$$ we can deduce that $$t=\frac{p-r}{p}>0$$ does not work either so furthermore $$s>0$$.

We have that then if $$t>s$$, $$w=v(g_t^{p-1})$$ is in $$L^{p'}$$, as $$\int_{0}^{1} |v(x)(g_t(x))^{p-1}|^{p'} dx = \int_{0}^{1} |u|^{(1-t)p} \mathbf{1}_{\{|u|>1\}} dx = \int_{0}^{1} |ug_t|^p < \infty.$$

However, $$\int_{0}^{1} |u(x)v(x)(g_t(x))^{p-1}| dx = \int_{0}^{1} |u(x)|^{1+(1-t)(p-1)} \mathbf{1}_{\{|u|>1\}}(x) dx = \int_{0}^{1} |u(x)|^{(1-t')p} \mathbf{1}_{\{|u|>1\}}(x) dx,$$ where $$t'=t-t/p$$, as $$(1-t')p = (1-t)p + t = 1 + (1-t)(p-1).$$

We now just need to take $$t$$ close enough to $$s$$ such that $$t'< s$$ and we get that for $$w=v(g_t)^{p-1}$$, $$uw\notin L^{1}[0,1]$$. This is possible when $$s>0$$ as $$t(1-\frac{1}{p})$$ can get arbitrarily close to $$0$$, but it is not clear how to proceed when $$s=0$$.

• In the definition of $s$, would it be better to have $t \in [0,\infty)$ for clarity? Jan 13 at 22:52
• In showing that $w \in L^{p'}$, I guess the inequality on the right-hand side is just "equality"? Jan 13 at 23:00
• What I mean is that $\lvert u \rvert^{(1-t)p} 1_{\lvert u \rvert >1} = \lvert u g_t \rvert^p$. Jan 13 at 23:06
• Yeah, sorry, you are right! I will change it for clarity. Thanks! Jan 13 at 23:09
• Note that such functions $u$ exist, for which $s = 0$, since for $u(x) = 1/\sqrt{x}, p = 2$ we have $s = 0$. It seems to be the same difficulty I was struggling with in my sketch of an answer @Keith Jan 13 at 23:27

Since my answer works for $$u\in L^1[0, 1]\setminus L^{p-\varepsilon}[0, 1]$$ for some $$\varepsilon > 0$$, I thought I'll include it as an alternative approach.

Let $$x_1 = 1$$ and $$x_{n+1} = (1/p')x_n+1$$

If $$u\in L^{x_n}[0, 1]$$, let $$w = \text{sgn}(u)|u|^{(1/p')x_n}$$, then $$\int wu = \int|u|^{x_{n+1}}$$.

If $$u\notin L^{x_{n+1}}[0, 1]$$ we are done. If not, we continue.

Since $$x_n = \sum_{k=0}^{n-1} (1/p')^k = \frac{(1/p')^n-1}{1/p'-1}$$, we see that $$x_n\to \frac{-1}{1/p'-1} = p$$.

If $$u\notin L^{p-\varepsilon}[0, 1]$$ for some $$\varepsilon > 0$$, we see that such $$w$$ exists, since we can take least $$n$$ with $$u\notin L^{x_{n+1}}$$.

If $$u\in L^s[0, 1]$$ for $$1\leq s < p$$, then its unclear how to proceed, but if $$w \in \bigcup_{r > 0} L^{p'+r}[0, 1]$$ then $$uw\in L^1$$. Note however that $$\bigcup_{r>0} L^{p'+r}[0, 1]\neq L^{p'}[0, 1]$$ since we can take one of the examples here, say $$f$$, then $$w = f^{p'}\in L^{p'}[0, 1]$$ but $$w\notin L^{p'+r}[0, 1]$$ for any $$r > 0$$.

If we are to look for a counter-example, we need to find it in $$L^{p'}[0, 1]\setminus \bigcup_{r > 0} L^{p'+r}[0, 1]$$.

• It is unclear to me why $\omega \in L^{p'}$. Jan 14 at 0:00
• @Keith It was an error, instead of taking $x_n$ with $p-\varepsilon < x_n < p$, simply take least $n$ for which $\int |u|^{x_{n+1}} = \infty$. Such least $n$ exists from assumption that $u\notin L^{p-\varepsilon}$ for some $\varepsilon > 0$ Jan 14 at 0:10
• So, $\omega \in L^{p'}$ since $\int \lvert u \rvert^{x_n} < \infty$? Jan 14 at 0:29
• @Keith yes, exactly. Jan 14 at 0:39
• I guess an argument similar to your answer works for $u \in L^p[0,1] - L^{p+\epsilon}[0,1]$ for any $\epsilon>0$? Jan 14 at 2:06