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let $(X,A,\mu)$ a measurable space.$1<p, q, s<\infty$

a.Prove that if $f\in L^p$ and $g\in L^q$ such that $1/p+1/q=1/s$ then $fg\in L^s$ and $\|fg\|_s\leq \|f\|_p\|g\|_q$.

b.Prove that if $f\in L^p$ , $g\in L^q$ and $h\in L^s$ such that $1/p+1/q+1/s=1$ then $fgh\in L^1(\mu)$ and $\|fgh\|_1\leq \|f\|_p\|g\|_q\|h\|_s$

My trial:

a. $\|fg\|_s= (\int_X|fg|^s)^{1/s}=(\int_X|fg|^s)^{1/p+1/q}= (\int_X |fg|^s)^{1/p} (\int_X |fg|^s)^{1/q}\leq$ (p,q>s) $(\int_X |fg|^p)^{1/p} (\int_X |fg|^q)^{1/q}= \|f\|_p\|g\|_q$ and since $\|f\|_p , \|g\|_q < \infty$ then $\|fg\|_s$ is finite too.so $fg \in L^s$

b. Denote $1/p+1/q=1/p'$ so by part a we get $fg\in L^{p'}$ and: $\|fg\|_{p'} \leq \|f\|_p \|g\|_q$, then we have $1/p'+1/s=1$ therefore by holder ineq1uality we get $\|fgh\|_1\leq \|fg\|_{p'} \|h\|_s \leq \|f\|_p\|g\|_q\|h\|_s$

Is what I solved right? Glad to her any notifications.

Calvin Khor
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user652838
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1 Answers1

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If I understand correctly, in part (a),

  1. you have used that $p>s$ implies that $$ \int_X |F|^s \le \int_X |F|^p$$ (and similarly for $q>s$). This is not true. For example consider $X=[1,\infty)$ with the usual Lebesgue measure, and $f(x) = 1/x$. Then $f\in L^2(X)$ but not $L^1(X)$, so $$\infty=\int_1^\infty\frac1{x} dx \not\le \int_1^\infty \frac1{x^2}dx<\infty. $$
    (a counterexample also in $L^1$ is $f(x)=\mathbf 1_{[1,N]}/x$, for some $N$ very large; $\mathbf 1_A$ is the indicator of $A$.)

  2. You then further went on to write $\|fg\|_p \|fg\|_q = \|f\|_p \|g\|_q$. This is also not true, and can be seen by scaling considerations: if you multiply $f$ by $2$, the left hand side is multiplied by $4$, but the right hand side only by $2$. So the equality is false.

Instead, here's a big hint: try use the normal Holder's inequality with the exponents $1=\frac1{p/s}+\frac1{q/s}$.

Some other minor points: you should write $fg\in L^s$, not $\|fg\|_s\in L^s$; and $p'$ is normally reserved for the Hölder conjugate; so in part (b), perhaps use $s'$ instead. (The proof for (b) is correct.)

Calvin Khor
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  • hey @Calvin Khor I guess i did manage to understand it! can I ask something in this context, if my measure space is $(N,P(N),\mu)$ with the counting measure then how would I show that if $L^p\subset L^q$ for $1\leq p<q$ then it is not true that $||f||_p<||f||_q$ ($f\in L^p$)? (I think that this is not true so I am trying to find a counterexample? – user652838 Nov 30 '20 at 16:25
  • @user726608 hi, I don’t understand. Saying ‘if $L^p\subset L^q$’ makes no sense. It’s just always true. And the inequality is not true because exactly the opposite inequality is true. Just take any sequence in $\ell^\infty$ that is not $\ell^1$. If this doesn’t help or I misunderstood, you can ask a new question – Calvin Khor Nov 30 '20 at 16:37
  • That example doesn’t belong to any $\ell^p$. Try one that is not bounded – Calvin Khor Nov 30 '20 at 16:56
  • @user726608 sorry, your messages woke me up so I was too tired. I obviously meant to say: try one that is bounded – Calvin Khor Dec 01 '20 at 00:54
  • Hey @Calvin Khor sorry for that, I took $f(n)=1/n^2$ p=1 and q=2 i know that the p and q norm which equals to the sum converge, but how can I show it formally? Can you help please – user652838 Dec 01 '20 at 08:35
  • How can you show what formally...? I dont understand @user726608 – Calvin Khor Dec 01 '20 at 10:01
  • I mean to show my counterexample formally, though i'm not sure bout it:( – user652838 Dec 01 '20 at 11:44
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    If you want to check your solution, you can post another question. I actually don't even know what your counterexample is, and its not convenient to look in the comments. You can link to the question here – Calvin Khor Dec 01 '20 at 12:07