= Solution
A <continuous embedding> $X\hookrightarrow Y$ is an injective continuous linear inclusion, so $\|x\|_Y\leq C\|x\|_X$ for a fixed constant. On a finite-measure domain, <Holder inequality> gives
$$
\boxed{\|u\|_q\leq|\Omega|^{1/q-1/p}\|u\|_p\quad(1\leq q\leq p\leq\infty),}
$$
with $1/\infty=0$. For finite $p>q$ apply <Holder inequality> to $|u|^q\cdot1$ with exponents $p/q$ and $p/(p-q)$; equality of exponents is immediate, and $p=\infty$ follows by integrating the supremum bound.
<Weak convergence> $u_j\rightharpoonup u$ means $\ell(u_j)\to\ell(u)$ for every <bounded linear functional> $\ell\in X^*$. The <Riemann-Lebesgue lemma> says that the <Fourier transform> of an integrable function tends to zero at high frequency. Since $L^2(0,1)\subset L^1(0,1)$, it makes $\cos(j\pi x)$ weakly null in $L^2$. Its absolute value is at most one, so $u_jv_j\in L^2$. For a test $\phi\in L^2$,
$$
\left|\int u_j(v_j-v)\phi\right|\leq\|v_j-v\|_2\|\phi\|_2\to0,
$$
while $\int u_jv\phi\to0$ by the <Riemann-Lebesgue lemma> applied to $v\phi\in L^1$. Hence \b[$u_jv_j\rightharpoonup0$]. Taking $w_j=u_j=\cos(j\pi x)$ instead gives $u_jw_j=(1+\cos(2j\pi x))/2\rightharpoonup1/2$, so two weakly null factors need not have a weakly null product.
For $\bar z_j=j^{1/p}\mathbf1_{(0,1/j)}$ and $p'=p/(p-1)$, <Holder inequality> gives
$$
\left|\int\bar z_jg\right|\leq\left(\int_0^{1/j}|g|^{p'}\right)^{1/p'}\to0\quad(g\in L^{p'}).
$$
Thus \b[$\bar z_j\rightharpoonup0$ in $L^p$]. In contrast, if $z_j$ had a weak $L^1$ limit $z$, testing on every interval strictly away from zero would give integral zero for $z$, hence $z=0$ <almost everywhere> by the permitted fact. Testing against the constant function one would give $\int z=1$, a contradiction. Therefore \b[$z_j$ has no weak $L^1$ limit].
Back to article page
