Let $N\geq 2$ and $\Omega$ be an open bounded subset of $\mathbb{R}^N$ and consider the Trudinger-Moser inequality $$ \sup_{\substack{u\in W_0^{1, N}(\Omega)\\\|u\|_{W_0^{1, N}}\leq 1}}\int_{\Omega} e^{\alpha_N |u|^{N^{\prime}}} dx <+\infty, $$ where $\|u\|_{W_0^{1, N}}^N =\int_{\Omega}|\nabla u|^N dx$ and $\alpha_N =N\omega_{N-1}^{\frac{1}{N-1}}$.
The inequality is clear, but my question is: if I only know that $u\in W_0^{1, N}(\Omega)$ and $\|u\|_{W_0^{1, N}}$ is bounded (I don't know if $\|u\|_{W_0^{1, N}}\leq 1$), I can say, anyway, that that supremum is finite? Intuitively I would say "yes", because it seems a bit strange to have the supremum finite in the unitary ball and not, e.g., in the ball of radius 2.
If yes, why? Alternatively, there is a version of Trudinger-Moser inequality which does not ask for the norm to be less or equal than $1$ but only bounded?
Thank you in advance!
Let $v=u/\|u\|_{W_0^{1,N}},$ with $\|u\|_{W_0^{1,N}}<\infty.$ Since $v$ has $W^{1,N}_0$ norm $1$, we know that
$$\int_{\Omega} e^{\alpha_N |v|^{N^{\prime}}}\, dx<\infty.$$ Now,
$$\int_{\Omega} e^{\alpha_N |v|^{N^{\prime}}}\, dx=e^{\alpha_N \|u\|_{W_0^{1,N}}^{-N^{\prime}}}\int_{\Omega} e^{\alpha_N |u|^{N^{\prime}}}\, dx\implies \int_{\Omega} e^{\alpha_N |u|^{N^{\prime}}}\, dx=e^{\alpha_N \|u\|_{W_0^{1,N}}^{N^{\prime}}}\int_{\Omega} e^{\alpha_N |v|^{N^{\prime}}}\, dx<\infty.$$ So, it is true whenever you take a sup over elements of $W_0^{1,N}(\Omega)$ with fixed, finite norm $M$.