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I have this example and I don't understand its resolution:

Let $\Omega \subset \mathbb{R}^n , n\geq3$ be a bounded open set (with smooth boundary), let $f\colon\mathbb{R}\rightarrow \mathbb{R}$ be a continuous function, and assume that there exist $a,b>0$ such that: $$|f(t)|\leq a+b|t|^{{2^*}-1}\tag{*}$$ for all $t\in \mathbb{R}$. Recall that $2^*=\frac{2n}{n-2}$ is the critical exponent for the embedding of $H^1(\Omega)$ into $L^p(\Omega)$. Define: $F(t)=\int_0^1 f(s) ds$, and consider the functional $J:H^1(\Omega) \rightarrow \mathbb{R}$ given by: $J(u)=\int_{\Omega} F(u(x))dx$. Then $J$ is differentiable on $H^1(\Omega)$ et $J'(u)v=\int_{\Omega}f(u(x))v(x)dx$ for all $u,v \in H^1(\Omega)$.

To prove that they process as follows: first we prove that $J$ is Gâteaux differentiable and then we show that $J'_G$ is continuous. From the growth assumption $(*)$ one easily derives that $J(u)$ is well defined via the Sobolev inequalities.

My question is:

Why should we check that $ J $ is well defined on using the fact that $H^1$ injects into $L^p$? And as I am in the specialty "ode" we have a single variable, what should I change for this example is in line with my specialty ?

Does the critical exponent have a meaning in ordinary differential equations?

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up vote 0 down vote accepted

For your first question: You have to use this fact, because of the kind of inequality you have on $f$. In general, nonlinear differential problems where $f$ has critical growth, i.e. $f$ is comaparable to a polynomious of degree the Sobolev exponent (or critical exponent) are called critical problems.

You also have sub-critial ($p<2^\star$) and super-critical ($p>2^\star$) problems. You can google this term and find a lot on literature on it.

For your second question: If $\Omega\subset\mathbb{R}$, the space $H_0^1(\Omega)$ is continuous embedded in $C(\overline{\Omega})$, and hence $H_0^1$ is continuous embedded in $L^p$ for all $p$, so my "guess" is that the critical exponent is meaningless in this case.

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After that they calculate $\displaystyle\lim_{t\rightarrow0} \frac{F(u(x)+tv(x))-F(u(x))}{t}$ and they say bye the Lahrange theorem there exists a real number $|\theta| \leq |t|$ and $|\displaystyle\frac{F(u(x)+tv(x))-F(u(x))}{t}|=|f(u(x)+\theta u(x))v(x)|\leq (a+b|u(x)+\theta v(x)|^{2^*-1})|v(x)| \leq C(|v(x)+|u(x)|^{2^*-1}|v(x)|+|v(x)|^{2^*})$ i dont understand what is the Lagrange theorem ? please – Vrouvrou Feb 10 '13 at 14:46
Sorry, but I cant understand your comment. I think that it is missing something... – Tomás Feb 10 '13 at 17:41

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