A counter example I have this two spaces $$C_{\theta}=\{u\in C(\overline{\Omega}), \sup (|x|^{\theta} |u(x)|)<\infty\}$$ with the norm $\displaystyle\|u
\|_{\theta}=\sup_{\Omega}(|x|^{\theta} |u(x)|)$
and $$L_{1}^{p^*}=\{u ~~\text{measurable};~~\int_{\Omega} (|x|\cdot|u(x)|)^{p^*} dx<\infty\}$$ with the norm $\|u\|_{L^{p^*}_{1}}^{p^*}=\int_{\Omega} (|x|\cdot|u(x)|)^{p^*}dx$
Where $\Omega\subset\mathbb{R}^N$ is bounded with $0\in \Omega$ and $\theta>\frac{N}{p}>1,$ $p^*=\frac{pN}{N-p}$ and $N>p.$
I need a counter example to say that  $C_{\theta}$ is not continuously embeded in $L^{p^*}_{1}$, so i'm searching a function which is in $C_{\theta}$ but not in $L^{p^*}_1$
Or a sequence $u_n$ which converge to $u$ un $C_{\theta}$ but not in $L^{p^*}_{1}$
Please help me thank you
 A: Let's try a canonical test function, $u(x)=|x|^\alpha$ and see if some value of $\alpha$ works. Suppose $\Omega=B(0,1)$. Then
$$
\sup_{x\in\Omega}|x|^\theta|u(x)|=1\tag{1}
$$
as long as $\alpha+\theta\ge0$.
$$
\int_\Omega|xu(x)|^{p^*}\,\mathrm{d}x
=\omega_{N-1}\int_0^1r^{(1+\alpha)p^*}r^{N-1}\,\mathrm{d}r\tag{2}
$$
which is finite when $0\lt(1+\alpha)p^*+N=N\left[\frac{(1+\alpha)p}{N-p}+1\right]$. This is equivalent to
$$
\frac{(1+\alpha)p}{N-p}\gt-1\iff\alpha\gt-\frac Np\tag{3}
$$
For a counterexample, we would need
$$
-\theta\le\alpha\le-\frac Np\tag{4}
$$
Since $\frac Np\lt\theta$ by hypothesis, we can find an $\alpha$ strictly between $-\theta$ and $-\frac Np$. Thus, for $\alpha$ satisfying $(4)$, $u=|x|^\alpha$ and $\Omega=B(0,1)$ satisfy $(1)$, yet fail to have $(2)$ converge.

Since the question also wants $u\in C(\overline{\Omega})$, $C_\theta$ is not closed. We have to limit our functions to be in $C(\overline{\Omega})$. So, define
$$
u_n(x)=\left\{\begin{array}{}
|x|^\alpha&\text{if }|x|\ge\frac1n\\
\frac1{n^\alpha}&\text{if }|x|\lt\frac1n
\end{array}\right.
$$
So each $u_n$ is in $C(\overline{\Omega})$ and satisfies $(1)$. However, $u_n\to u$ monotonically as $n\to\infty$; thus, by Monotone Convergence, we have that $\|u_n\|_{\large L_{p^\ast}^1}\to\infty$.
