 Solution

A continuous embedding $X\hookrightarrow Y$ is an injective continuous linear inclusion, so $\Vert x{\Vert }_Y\le C\Vert x{\Vert }_X$ for a fixed constant. On a finite-measure domain, Holder inequality giveswith $1/\infty =0$. For finite $p>q$ apply Holder inequality to $|u{|}^q\cdot 1$ 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 $\ell \in X^{\ast }$. 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 $u_j{v}_j\in L^2$. For a test $\varphi \in L^2$,while $\int u_{j}v\varphi \to 0$ by the Riemann-Lebesgue lemma applied to $v\varphi \in L^1$. Hence $u_j{v}_j\rightharpoonup 0$. Taking $w_j=u_j=\cos (j\pi x)$ instead gives $u_j{w}_j=(1+\cos (2j\pi x))/2\rightharpoonup 1/2$, so two weakly null factors need not have a weakly null product.

For ${\bar{z}}_j=j^{1/p}{1}_{(0,1/j)}$ and $p^{\prime }=p/(p-1)$, Holder inequality givesThus ${\bar{z}}_j\rightharpoonup 0$ 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 $z_j$ has no weak $L^1$ limit.