# Weak convergence in $L^q$ but not in $L^p$, where $1\leq p<q<\infty$?

I've been trying to find a sequence $\{f_n\}$ of functions on $[0,1]$ that converges weakly in $L^q$ but does not converge weakly in $L^p$, where $1\leq p<q<\infty$. I'm stuck and any hints would be greatly appreciated.

(I find this to be tricky because the sequences are defined on $[0,1]$ and the functions cannot spread to infinity. Since we're working on $[0,1]$, we have $L^q\subset L^p$.)

The functions of the example below are not supported in $[0,1]$.

Sorry for overlooking the requirement that the functions are supported on $[0,1]$. If this condition is added, then the statement is true.

Straight from the definition, $f_n$ converges weakly to $f\in L^q$ iff for every $g\in L^{q^*}$,

$$\int f_ng\to \int fg,$$

where $1/q+1/q^*=1$. Since $p<q$, $p^*>q^*$, so $L^{p^*}([0,1])\subset L^{q^*}([0,1])$. Then the above convergence holds for every $g\in L^{p^*}$. This implies $f_n$ converges weakly to $f$ in $L^p$. As a sanity check, we see that $f$ is in $L^p$ because $L^q([0,1])\subset L^p([0,1])$.

More succinctly, this is because continuous linear maps between Banach spaces are also weakly continuous (See Theorem 1.1 of Chapter 6 of Conway, A Course in Functional Analysis).

Just for reference, the result is false if the functions are supported on $\mathbb R$.

Let $r\in(p,q)$ and $f_n=n^{-1/r}$ on $[0,n]$ and $f_n=0$ elsewhere. Then

$$\|f_n\|_{q}=(n\cdot n^{-q/r})^{1/q}=n^{1/q-1/r}\to 0$$

as $n\to\infty$, so $f_n\to0$ in $L^q$. On the other hand, we replace $q$ by $p$ in the above inequality to get

$$\|f_n\|_{p}=n^{1/p-1/r}\to\infty$$

as $n\to\infty$. By the uniform boundedness principle (among other approaches), $f_n$ does not converge weakly in $L^p$.

• The example seems actually impossible: Let p*, q* be the conjugate exponents of p, q resp. Since [0,1] contains sets of measure at most 1, if p<q (i.e., p*>q*) then L^q([0,1]) is a subset of L^p([0,1]) and L^{p^*}([0,1]) is a subset of L^{q^*}([0,1]). Therefore, if f_n are in L^q, hence also in L^p, then weak convergence in L^q implies weak convergence in L^p. – Aubrey Mar 10 '15 at 2:33
• Thank you! Perhaps you should just change the phrase "the statement is true" to "the statement is false". – Aubrey Mar 10 '15 at 2:35