To prove $v\ll\mu$ and $dv/d\mu$ $\in$ $L^q$

Question

I'm trying to prove the following:

Here's my attempt to prove it;

(b) To prove $dv/d\mu$ $\in$ $L^q$

By property (2), For any sequence $f_n$ $\subset$ $L^P(X, \mu)$, we have $\|f_n\|_p\to0$; and

$\lim_n\int_{X}f_ndv = 0$

Then, let $f_n = 1_{E_n}(x)$ and $E_n =\sum {1/2^n}$ and

$\|f_n\|_p = (\|1_{E_n}(x)\|)_p = \mu(E_n)^{1/p}$

$\implies$ $({\mu(\sum{1/2^n})})^{1/p}$

$\implies$ $(\lim \mu({1/2^n})^{1/p} \to 0$ as $n\to \infty$

Now since $X$ is $\sigma$-finite measure space and $v\ll\mu$ Then by Radon-Nikodym Theorem, $dv=fd\mu$. So, $\|dv/d\mu\|=(\int_{X}|f|^qd\mu)^{1/q}$

Since $\|f_n\|_p \to 0$ , we can assume that $f=0$ so $f_n \leq f =0$

Then, by dominated convergence theorem,

$\lim \int_{X} f_n dv = \lim \int_{X} f dv = 0$

$\implies \|dv/d\mu\|_q = \|f\|_q = 0 = (\lim \int_{X}f_n)^{1/q} = 0 < \infty$

Therefore, $dv/d\mu \in L^q$

(a) To prove $v\ll\mu$

Suppose \mu(E)=0, we need to prove that $v(E)=0$ Let $_ = _E$ Then $_ ∈ ^p(\mu) ⊂ ^1()$

$∫|_|_p d\mu=\mu(E) \to 0$

so, $∫_ = (E) \to 0$ as $n\to \infty$

that is (E)=0

$\implies v\ll\mu$

Professor's comment: Incorrectly assumes the inclusion of $L^p(X,\mu)$ into $L^1(X,\nu)$ implies $\|f\|_{L^1(X,\nu)} \le \|f\|_{L^p(X,\mu)}$. It is true if one scales on the right-hand side, but one must invoke property (ii). Proceeds to incorrectly assert that $\nu$ is dominated by $\mu$. Fails to show the Radon-Nikodym derivative is in $L^q(X,\mu)$.

I need help in correcting my solution. Thanks.

Answer 1

It seems that your professor is using the closed graph theorem. By assumption $L_p(\mu)\subset L_1(\nu)$. Consider the inclusion $\iota:L_p(\mu)\rightarrow L_1(\nu)$. If $(f_n,\iota(f_n))=(f_n,f_n)$ converges to some point $(f, g)\in L_p(\mu)\times L_1(\nu)$ then $\|f_n-f\|_{L_p(\mu)}\xrightarrow{n\rightarrow\infty}0$ and $\|f_n-g\|_{L_1(\nu)}\xrightarrow{n\rightarrow\infty}0$. Assumption (ii) implies that $\|f_n-f\|_{L_1(\nu)}\xrightarrow{n\rightarrow\infty}0$ (why?) whence one concludes that $f=g$. Hence, by $\iota$ is bounded and so, for some $B>0$, $$\|f\|_{L_1(\nu)}\leq B\|f\|_{L_p(\mu)}$$

Now, if $\mu(E)=0$, then $f=\mathbb{1}_{B}\in L_p(\mu)$ and $\|f\|_{L_p(\mu)}=(\mu(E))^{1/p}=0$; hence $\|f\|_{L_1(\nu)}=\nu(E)=0$. This shows that $\nu\ll \mu$. The $\sigma$-finiteness assumption allows one to apply the Radon-Nikodym theorem. This, there is a measurable function $g:=\frac{d\nu}{d\mu}$ such that $\nu=g\,d\mu$. It follows that for any $f\in L_p(\mu)(\subset L_1(\nu))$ $$\Big|\int f\,d\nu\Big|\leq\|f\|_{L_1(\nu)}=\int|f|g\,d\mu\leq B\|f\|_{L_p(\mu)}$$ This means that $f\mapsto \int fg\,d\mu$ is a bounded functional. It is easy to check (duality between $L_p$ spaces) that this implies that $g\in L_q(\mu)$.

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Here is an alternative solution that does not relay on the closed map theorem but uses the uniform boundedness principle.

Suppose $\mu(B)=0$. Define the sequence $f_n=\mathbb{1}_B$. Clearly $(f_n:n\in\mathbb{N})\subset L_p(\mu)$ and by assumptions (i) and (ii) we have that $\nu(E)=\int f_n\,d\nu\xrightarrow{n\rightarrow\infty}0$. This shows $\nu\ll\mu$.

The $\sigma$-finiteness assumption and the Radon-Nikodym theory implies that there an $\mu$-a.s. unique $g\in L^{loc}_1$, $g\geq0$, such that $\nu = g\cdot \mu$. Assumptions (i) and (ii) imply that for any $f\in L_p(\mu)$ $$\big|\int f\,d\nu\big|=\Big|\int fg\,d\mu\Big|<\infty$$ Suppose $A_n$ is an increasing sequence of measurable sets such that $X=\bigcup_nA_n$. For any $m\in\mathbb{N}$ define $g_m=\min(\max(-m,f),m)$. Notice that $|g_m|\leq \min(|g|,m)$ and $g_m\xrightarrow{m\rightarrow\infty}g$. Consider the linear functionals $\Lambda_{m, n}$ in $L_p(\mu)$ defined as $$\Lambda_{m, n}f=\int_{A_n}fg_m\,d\mu$$ Each $\Lambda_{m, n}$ is bounded: $|\Lambda_{m, n}f|\leq m\big(\mu(A_n)\big)^{1/q}\|f\|_p$. Also, for any $f\in L_p(\mu)$ $$|\Lambda_{m, n}f|\leq\int_{A_n}|f|g_m\,d\mu\leq\int_X|f|g\,d\mu=\|f\|_{L_1(\nu)}<\infty $$ Thus, by the uniform boundedness principle the $\Lambda_{m, n}$ are uniformly bounded; hence, for some $B>0$ $$|\int fg\,d\mu|\leq\int|f|g\,d\mu\leq B\|f\|_{L_p(\mu)}$$ Ir follows that $f\in L_q(\mu)$ by duality arguments.

Answer 2

(a) As proven above (b) By duality $L^p(X,\mu)^* = L^q(X,\nu)$

Now, we can define a linear function $l=L^p(X,\mu) \to \mathbb{C}$. Now, $l$ is $bounded \iff continous$ at zero

If we define, $f \in L^p(X,\mu) \subset L^1(X,\nu)$

$\implies l(f)=\int f d\nu \in \mathbb{C}$ which is a scalar, linear value because $\int f d\nu$ is linear

Also, if $||f_n||_p \to 0$ then from $(ii)$, we get

$l(f_n)=lim \int f_n d\nu \to o$

$\implies l$ is continuous at zero

$\implies l\in L^p(X, \mu)^*$

$\implies$ there exists $g \in L^q(X, \mu)$ such that $\int f d\nu = l(f)= \int fg d\mu$

By Radon-Nikodym Theorem,

$\int f{d\nu/d\mu} d\mu =\int fg d\mu$

$\implies d\nu/d\mu = g$

Now, $o = \int f(g - {d\nu/d\mu}) d\mu = f = 1_E = \mu(E) < \infty$

So, $E = ({x \in X: g(x) \neq d\nu/d\mu})$

$\implies \mu(E) = 0$

$\implies g(x) = d\nu/d\mu \in L^q(X, \mu)$