# Bounded linear functions on Lp spaces

• Dec 26th 2009, 04:22 PM
GTO
Bounded linear functions on Lp spaces
Let g be an integrable functions on [0,1] and $\displaystyle 1/p+1/q=1$ with $\displaystyle 1<p<\infty$. Suppose that there is a constant $\displaystyle M>0$ such that $\displaystyle |\int fg| \leq M||f||_p$ for all bounded measurable functions $\displaystyle f$. Prove that $\displaystyle g \in L^q$ and $\displaystyle ||g||_q \leq M$.

• Dec 27th 2009, 11:10 AM
Opalg
Quote:

Originally Posted by GTO
Let g be an integrable functions on [0,1] and $\displaystyle 1/p+1/q=1$ with $\displaystyle 1<p<\infty$. Suppose that there is a constant $\displaystyle M>0$ such that $\displaystyle |\int fg| \leq M||f||_p$ for all bounded measurable functions $\displaystyle f$. Prove that $\displaystyle g \in L^q$ and $\displaystyle ||g||_q \leq M$.

Here's the classic proof as given in Dunford and Schwartz. First, we may assume that $\displaystyle g(t)\geqslant0\; (0\leqslant t\leqslant1)$. Reason: let $\displaystyle \alpha (t) = \overline{g(t)}/|g(t)|$ (with $\displaystyle \alpha(t)=1$ if $\displaystyle g(t)=0$). In other words, $\displaystyle \alpha(t)$ is the complex number of absolute value 1 such that $\displaystyle \alpha(t)g(t)\geqslant0$. Replace $\displaystyle g$ by $\displaystyle \alpha g$, which has the same norm as $\displaystyle g$ in any of the $\displaystyle L^p$ spaces, and is always real and non-negative. We can then assume that
$\displaystyle \textstyle\int fg \leqslant M\|f\|_p\qquad ({\color{blue}^*})$
for all non-negative $\displaystyle f\in L^p$. (We are only told that this holds for bounded $\displaystyle f$, but you can easily see from the Monotone Convergence theorem that it holds for all $\displaystyle f\in L^p$.)

Now comes the clever part. It follows by putting $\displaystyle f\equiv1$ in (*) that $\displaystyle \|g\|_1\leqslant M$. Also, $\displaystyle g^{1/p}\in L^p$, with $\displaystyle \|g^{1/p}\|_p = \|g\|_1^{1/p} \leqslant M^{1/p}$. Therefore, putting $\displaystyle f=g^{1/p}$ in (*), we get $\displaystyle \textstyle\int g^{1+\frac1p} \leqslant M\|g^{1/p}\|_p \leqslant M^{1+\frac1p}$.

Now let $\displaystyle g_2 = g^{1+\tfrac1p}$. We have shown that $\displaystyle g_2\in L^1$, with $\displaystyle \|g_2\|_1\leqslant M^{1+\tfrac1p}$. So $\displaystyle g_{2}^{1/p} \in L^p$, and we can repeat the argument in the previous paragraph to get $\displaystyle \textstyle\int g^{1+\tfrac1p + \tfrac1{p^2}}\leqslant M^{1+\tfrac1p + \tfrac1{p^2}}$. Thus $\displaystyle g_3 = g^{1+\tfrac1p + \tfrac1{p^2}} \in L^1$, with $\displaystyle \|g_3\|_1 \leqslant M^{1+\tfrac1p + \tfrac1{p^2}}$.

Continue in that way to see that $\displaystyle \textstyle\int g^{1+\tfrac1p + \ldots + \tfrac1{p^n}}\leqslant M^{1+\tfrac1p + \ldots + \tfrac1{p^n}}$ for n=1,2,3,... . But $\displaystyle \sum_{n=0}^\infty1/p^n = \frac1{1-\tfrac1p} = q$. By Fatou's lemma, we can let $\displaystyle n\to\infty$ and get $\displaystyle \textstyle\int g^q \leqslant M^q$, from which $\displaystyle g\in L^q$ with $\displaystyle \|g\|_q \leqslant M$.