1
$\begingroup$

Let $f$ be a function such that $\|fg\|_1<\infty$ whenever $\|g\|_2<\infty$. I would like to show that $\|f\|_2<\infty$. It seems that I should use some kind of Hölder inequalities, since we have $\|fg\|_1\leq \|f\|_2\|g\|_2$, but I don't know how. Any help would be appreciated. Thanks!

2 Answers 2

1

You have to assume that

$$M := \sup \{ \|f \cdot g\|_1; \|g\|_2 \leq 1\}<\infty$$

... otherwise it won't work. (Assume $M=\infty$. Then for all $n \in \mathbb{N}$ there exists $g_n \in L^2$, $\|g_n\|_2 \leq 1$, such that $\|f \cdot g_n\|_1 \geq n$. And this means that there cannot exist a constant $c$ such that $\|f \cdot g\|_1 \leq c \cdot \|g\|_2$, in particular $f \notin L^2$ (by Hölder inequality).)

  • 0
    I see. And if $M<\infty$, do you necessarily have that $f\in L^2$?2012-12-09
  • 0
    Yes. In this case you can use Zarrax' proof, because $\|T\| = \sup\{\|f \cdot g\|_1; \|g\|_2 \leq 1\}=M<\infty$ which means that $T$ is a bounded operator. You could also use some approximating sequences, see e.g. J. Yeh: "Real Analysis: Theory of Measure And Integration", Theorem 18.4. Maybe this [link](http://books.google.de/books?id=0IZg1kjfRIMC&pg=PA434&lpg=PA434&dq=converse+h%C3%B6lder-inequality&source=bl&ots=1sqo4YthHK&sig=lMgd2Cf86Kx8iy9rnF7F0KV_sJM&hl=de&sa=X&ei=cffEULiIGMjGtAbEv4GwAQ&ved=0CEMQ6AEwAg) works.2012-12-09
  • 0
    Thank you for the reference!2012-12-09
  • 0
    Now I see what you mean. I remember having seen this formula before: $\|f\|_2 = sup \{\|fg\|_1 ; \|g\|_2 = 1\}$. Then $f\in L^2$ if and only if the $sup$ is finite.2012-12-09
0

Do you have Hilbert space theory? Because then you can use that $T(g) = \int fg = $ is a bounded linear functional on $L^2$, in which case there must be an $h \in L^2$ for which $ = $ for all $g \in L^2$. Since $\int g(\bar{f} - h) = 0$ for all $g \in L^2$, it is not hard to show that $\bar{f} - h = 0$ a.e., so that $f = \bar{h}$ is in $L^2$.

  • 1
    How do you know that $T$ is bounded?2012-12-09
  • 0
    Thanks! Unfortunately, I am not familiar with Hilbert space theory, so I don't really understand.2012-12-09