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Suppose $f,g\in L^q(\Omega)$ ($\Omega\subset \mathbb{R}^n$) for all $1\le q\le p$. Here, $L^p(\Omega)$ is defined with respect to some measure $\mu$ that is absolutely continuous wrt Lebesgue measure. Are there bounds on $\int|f-g|^p$ or $\int(f^p-g^p)$ in terms of $\int|f-g|^q$ for $q<p$?

Update. As pointed out in an answer below, without further assumptions this is false. At a high level, what I am trying to understand is the following: I wish to know the rate of convergence of $f_n\to f$ in $L^p$, but all I know is the rate of convergence in $L^q$ for some $q<p$. Can anything be said? It seems necessary to assume, at least, that $f_n,f\in L^r$ for some $r\ge p$.

For example, assuming sufficient regularity and additionally $L^2$ convergence of the gradients, Ladyzhenskaya's inequality is precisely such a bound for the case $q=2$ and $p=4$.

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  • $\begingroup$ The interpolation inequalities for a $p$-norm only work if you know the $q$-norm and the $r$-norm with $q\leq p \leq r$. I would guess that taking $h = f-g$ (it does not matter what $f$ and $g$ are) given by $h(x) = \big( \frac{d}{dx} (\frac{1}{\ln^{[k]} x} ) \big)^{1/p}$ will get you a counterexample (here $\ln^{[k]}$ is an $k$-times iterated logartihm). The integrand of the $p$-norm is logarithmic [or an interation of such], while the integrand of q norms will be dominated by the power of $x^{-q/p}$. But given the amount of downvote, I suspect there is a textbook example. $\endgroup$
    – ARG
    Commented May 4, 2020 at 13:00
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    $\begingroup$ PS: I think it's very unpolite that the downvoters and closers did not leave a comment... $\endgroup$
    – ARG
    Commented May 4, 2020 at 13:03
  • $\begingroup$ For your updated question: if you know that $f_n, f$ are bounded in $L^r$ for $r > p$, then you can directly interpolate to get convergence in $L^p$. (You don't need to assume $f_n \to f$ in $L^r$.) Iosif's example shows that this is sharp: with just $r = p$ this is not enough. $\endgroup$
    – Willie Wong
    Commented May 4, 2020 at 20:53
  • $\begingroup$ @WillieWong It seems I don't know enough about interpolation, then! What kind of interpolation inequalities give such explicit bounds? (If there is a standard reference, please feel free to share it.) $\endgroup$
    – JohnA
    Commented May 4, 2020 at 21:07
  • $\begingroup$ If $q < p < r$ there exists $\theta \in (0,1)$ such that $1/p = (1-\theta)/q + \theta/r$. Then $$ \int |f|^p = \int |f|^{(1-\theta)p} |f|^{\theta p} \leq \left( \int |f|^q \right)^{(1-\theta)p/q} \left( \int|f|^r\right)^{\theta p / r} $$ by Holder. So if $f_n \to f$ in $L^q$ and $f_n, f$ are uniformly bounded (say by $M$) in $L^r$, you have that $$ \int |f_n - f|^p \leq \left( \int |f_n - f|^q \right)^{(1-\theta)p/q} \left( 2M \right)^{\theta p} $$ using triangle inequality. $\endgroup$
    – Willie Wong
    Commented May 5, 2020 at 13:43

1 Answer 1

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Such a bound is impossible in general. E.g., suppose that $\Omega=[0,1]$, $g=0$, and $f=a^{-1/p}1_{[0,a]}$, where $p>0$ and $a\downarrow0$. Then for any $q\in(0,p)$ we have $\int\lvert f-g\rvert^q=a^{1-q/p}\to0$, whereas $\int\lvert f-g\rvert^p=\int(f^p-g^p)=1\not\to0$.

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  • $\begingroup$ This is a great counterexample. Do you have any idea if this extends to the case where either (a) $\int f^p-\int g^p \to 0$ or (b) $\int |f-g|^p \to 0$? $\endgroup$
    – JohnA
    Commented May 3, 2020 at 15:35
  • $\begingroup$ @JohnA : I am not sure how to understand your comment. Can you specify what kind of bound you want to prove or disprove? $\endgroup$
    – Iosif Pinelis
    Commented May 3, 2020 at 16:56
  • $\begingroup$ My comment was vague so my apologies; please see my edits to the "example" in my post. I'm curious if there is any relationship between the rate of convergence in $L^q$ vs $L^p$. Again, I suspect, the answer is no, but I am not sure. $\endgroup$
    – JohnA
    Commented May 3, 2020 at 17:33
  • $\begingroup$ @JohnA : I am still not sure about what specifically you mean by "arbitrarily bad". Can you just state it in formal terms? Also, I see no point in saying $c\alpha_n\to0$, because you can always rescale $f_n$ and $f$ (by replacing them, say, by $f_n/(c\alpha_n)^{1/q}$ and $f/(c\alpha_n)^{1/q}$) to get $c\alpha_n$ replaced by $1$. Alternatively, you can similarly rescale $f_n$ and $f$ to get $\int|f-g|^p=\int|(f^p-g^p)=1$, as was done in my example. $\endgroup$
    – Iosif Pinelis
    Commented May 3, 2020 at 18:08
  • $\begingroup$ It's probably better think of my question as asking "under what reasonable assumptions is such a bound possible". I edited my question to point out an example in Ladyzhenskaya's inequality, although this assumes a bit more than I wanted. (I am not sure about your point on $\alpha_n$, this is simply the rate of convergence.) $\endgroup$
    – JohnA
    Commented May 4, 2020 at 18:31

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