# Math Help - Convergence in Lp, Lq

1. ## Convergence in Lp, Lq

Hi,
Reading through a proof in a book that the intersection of Lp and Lq spaces is complete, I'm wondering the following (which I think was used by the author):
In a measure space $(X, \mathcal{M}, \mu)$, with $1 \leq p, q \leq \infty$, is it true that if $(f_n)$ is a sequence in $L^p(\mu)\cap L^q(\mu)$ such that $f_n\to f$ in $L^p(\mu)$ and $f_n\to g$ in $L^q(\mu)$, then $f = g$ $\mu$-a.e.

Since there is no relationship between the two norms for arbitrary spaces X, I do not know how to go about proving it if it were indeed correct. Your insights would be most appreciated!

Thanks

2. Originally Posted by james123
Since there is no relationship between the two norms for arbitrary spaces X
The relation is that L^p convergence implies L^q convergence for all $q \leq p$.

Edit: You may need to assume that the measure is finite (may work with sigma finite).

3. Originally Posted by Focus
The relation is that L^p convergence implies L^q convergence for all $q \leq p$.

Edit: You may need to assume that the measure is finite (may work with sigma finite).
May I please have a reference?

4. If the measure is finite, you can assume wolog it is a probability measure (or else consider $\mu/\mu(X)$).

Use Jensen's inequality (p/q is greater than one),
$\left(\int |f|^q d\mu\right)^{\frac{p}{q}}\leq\left(\int |f|^{q\frac{p}{q}} d\mu\right)=\left (\int |f|^p d\mu\right)$.

Now take p-th root of both sides.

5. Originally Posted by james123
In a measure space $(X, \mathcal{M}, \mu)$, with $1 \leq p, q \leq \infty$, is it true that if $(f_n)$ is a sequence in $L^p(\mu)\cap L^q(\mu)$ such that $f_n\to f$ in $L^p(\mu)$ and $f_n\to g$ in $L^q(\mu)$, then $f = g$ $\mu$-a.e.
Assume $1\leq p,q<\infty$.
You can prove this by noting that both convergence in $L^p$ and $L^q$ imply convergence in measure: for every $\epsilon>0$, we have

$\mu(|f_n-f|>\epsilon)\leq\frac{\|f_n-f\|_p^p}{\epsilon^p}$

(in probability, this is called Markov inequality). Thus,
$\mu(|f_n-f|>\epsilon)\to_n 0$ and similarly $\mu(|f_n-g|>\epsilon)\to_n 0$. Since $\{|f-g|>\epsilon\}\subset \{|f-f_n|>\epsilon/2\}\cup\{|g-f_n|>\epsilon/2\}$, this implies that $\mu(|f-g|>\epsilon)=0$. Since this holds for every $\epsilon>0$, we have $\mu(f\neq g)=\lim_p \mu(|f-g|>\frac{1}{p})=0$, i.e. $f=g$ a.e..

If $p=\infty$, the convergence in measure is still true, in a stronger sense: for every $\epsilon>0$, $\mu(|f_n-f|>\epsilon)=0$ for large $n$. In particular, we can conclude like above.