Let $(X, \mathscr{M}, \mu)$ be a measure space. Let $f \in L^p.$ Verify that there exists $g \in L^q$ such that $||g||_q = 1$ and $$\int_Xfgd\mu = ||f||_p$$ where $1 < p < \infty$ and $\frac{1}{p} + \frac{1}{q} = 1.$

$\textbf{Proof}$ Assume $X = \mathbb{R}, \mathscr{M} = \ \mbox{collection of Lebesgue measurable sets}$ and $\mu = m \ (\mbox{Lebesgue measure}).$

Assume first that $f = 0$ almost everywhere. Set $g = \chi_{[0,1]}.$ Then $$||g||_q = \int_\mathbb{R} \chi_{[0,1]}(x) dx = 1$$ and $$\int_\mathbb{R}fg = 0 = 0||g||_q = ||f||_p||g||_q.$$ Now assume that $f \neq 0$ almost everywhere. Set $g$ such that $g = |f|^{\frac{p}{q}}$ where $f$ is not zero and $g = 1$ where $f$ is zero. Then $$||g||_q = \int_\mathbb{R}(|f|^{p/q})^q dx < \infty.$$ So $g \in L^q.$ The problem is it is possible that $||g||_q \neq 1.$ So I set $$h = \frac{g}{||g||_q}.$$ It is valid since $g$ is positive almost everywhere so $||g||_q > 0.$ Now I try to apply the special case of young 's inequality $$ab \leq a^p/p + b^q/q$$ which it is actually equlity if and only if $a^p = b^q$. However, I can apply it only to $g$ since $|g|^q = |f|^p$ (cant apply to $h$ since $|h|^q \neq |f|^p$).

So I am not sure how to proceed.

Moreover, this is just a very specific space $(\mathbb{R}, \mathscr{L},m)$. I am not sure how to do the proof in abstract space $(X, \mathscr{M}, \mu)$. Actually, I can choose $X$ to be any nonempty set, $\mathscr{M} = 2^X$ and $\mu(A) := 0$ for all $A \subseteq X.$ Then it is a measure space, and I think that no integrable function $g$ on this space has $L^q$ norm equals $1$. Also, if I define $\mu(\phi) = 0$ and $\mu(A) = \infty$ for any nonempty subset $A$ of $X$. Then again this is a measure space with I think that no $g$ has $L^q$ norm equals $1$.

So I really doubt if the statement is really ture. Any suggestion ?

  • $\begingroup$ Hint: Solve it first for $\lVert f \rVert_p = 1$, then scale your solution by linearity. $\endgroup$ – B. Mehta Nov 27 '17 at 5:16
  • $\begingroup$ Is the result also holds on general measure space ? Or the proof is able to go through only when it is Lebesgue measure ? $\endgroup$ – Both Htob Nov 27 '17 at 5:20
  • $\begingroup$ Nothing in your proof used specifically that the measure in question was Lebesgue $\endgroup$ – B. Mehta Nov 27 '17 at 5:21
  • $\begingroup$ Like when $f = 0 $ a.e. I define $g$ specifically as interval which use property of $\mathbb{R}$. The general measure space $X$, as I stated above, I am not sure how to find $g$. $\endgroup$ – Both Htob Nov 27 '17 at 5:24
  • $\begingroup$ Ah, true. Instead of using $[0,1]$, can you see why finding a subset of finite measure will help construct such a $g$? $\endgroup$ – B. Mehta Nov 27 '17 at 5:27

If $f=0$ a.e., then $fg=0$ and $|f|^p=0$ a.e., thus $$\int_Xfg d\mu=\|f\|_p=0,\quad \forall g\in L^q.$$

Now assume $f$ is not $0$ a.e., then we consider two cases

  • If $p=1$, the define $$g(x)=\begin{cases}\frac{|f(x)|}{f(x)}, &f(x)\neq 0\\ 0,&\text{otherwise}\end{cases}$$

Then it's straight forward to verify that $\|g\|_\infty=1$ and $$\int_Xfg d\mu=\int_X|f|d\mu=\|f\|_1,$$

  • If $1<p<\infty$, then define $$g(x)=\begin{cases}\frac{|f(x)|^p}{f(x)}, &f(x)\neq 0\\ 0,&\text{otherwise}\end{cases}$$ and $$\tilde{g}(x)=\frac{g(x)}{\|f\|_p^{p-1}}.$$ Then note that $$|g|^q=|f|^{pq-q}=|f|^p,$$ and thus $g\in L_q$ and so is $\tilde{g}.$ Furthermore by direct calculation $$\|\tilde{g}\|_q^q=\|f\|_p^{(1-p)q}\int_X|f|^pd\mu=\|f\|_p^{-p}\|f\|_p^p=1,$$ and $$\int_Xf\tilde{g}d\mu=\|f\|_p^{1-p}\int_X|f|^pd\mu=\|f\|_p^{1-p}\|f\|_p^p=\|f\|_p.$$
  • $\begingroup$ In the $f=0$ ae case, how can you be sure there is some $\lVert g \rVert =1$? $\endgroup$ – B. Mehta Nov 27 '17 at 5:53
  • 1
    $\begingroup$ We can construct such a function from the indicator function of some set with positive finite measure. Note that if every set in the measure space has measure either zero or infinity, then all funcitons in $L^p$ have zero norm. This can be seen from the fact that $$\{|f|>0\}=\bigcup_{n\in\mathbb{N}}\{|f|\geq\frac{1}{n}\},$$ and it follows that $$\|f\|_p^p\geq n^{-p}\mu(\{|f|\geq\frac{1}{n}\}).$$ Thus the extreme case is not appetizing. So I think if the question is stated in this way, it implicitly excluded the extreme case, i.e. $L^p\neq\{0\}$ and there exists an element with positive norm. $\endgroup$ – Frank Lu Nov 27 '17 at 6:44

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