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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 \begin{equation} \int_0^1 u \cdot w = \infty \end{equation} 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..

Could anyone please help me?

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    $\begingroup$ Its possible if $u\notin L^{p-r}$ for some $r > 0$ $\endgroup$
    – Jakobian
    Jan 13 at 22:12
  • $\begingroup$ @Jakobian How? Could you help me? $\endgroup$
    – Keith
    Jan 13 at 22:18
  • $\begingroup$ Are you asking if such a $w$ exists or if it is possible to build it explicitly? $\endgroup$
    – LL 3.14
    Jan 13 at 22:18
  • $\begingroup$ @LL3.14 Just proof of exisetence will do.. $\endgroup$
    – Keith
    Jan 13 at 22:20
  • $\begingroup$ I'm not sure if it always exists though... my answer is still incomplete $\endgroup$
    – Jakobian
    Jan 13 at 22:21

2 Answers 2

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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 $s=0$.

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    $\begingroup$ In the definition of $s$, would it be better to have $t \in [0,\infty)$ for clarity? $\endgroup$
    – Keith
    Jan 13 at 22:52
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    $\begingroup$ In showing that $w \in L^{p'}$, I guess the inequality on the right-hand side is just "equality"? $\endgroup$
    – Keith
    Jan 13 at 23:00
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    $\begingroup$ What I mean is that $\lvert u \rvert^{(1-t)p} 1_{\lvert u \rvert >1} = \lvert u g_t \rvert^p$. $\endgroup$
    – Keith
    Jan 13 at 23:06
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    $\begingroup$ Yeah, sorry, you are right! I will change it for clarity. Thanks! $\endgroup$ Jan 13 at 23:09
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    $\begingroup$ 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 $\endgroup$
    – Jakobian
    Jan 13 at 23:27
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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 $L^{p'}[0, 1]\setminus \bigcup_{r > 0} L^{p'+r}[0, 1]$.

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  • $\begingroup$ It is unclear to me why $\omega \in L^{p'}$. $\endgroup$
    – Keith
    Jan 14 at 0:00
  • $\begingroup$ @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 $\varepsilon > 0$ $\endgroup$
    – Jakobian
    Jan 14 at 0:10
  • $\begingroup$ So, $\omega \in L^{p'}$ since $\int \lvert u \rvert^{x_n} < \infty$? $\endgroup$
    – Keith
    Jan 14 at 0:29
  • $\begingroup$ @Keith yes, exactly. $\endgroup$
    – Jakobian
    Jan 14 at 0:39
  • $\begingroup$ I guess an argument similar to your answer works for $u \in L^p[0,1] - L^{p+\epsilon}[0,1]$ for any $\epsilon>0$? $\endgroup$
    – Keith
    Jan 14 at 2:06

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