# Why is this functional coercive?

Let $h\in L^2(0,1)$, such that $|\int h \mathrm{d}x|<1$. On the space $H^1(0,1)$, consider $$J(v)=\frac{1}{2}\int_0^1 v'^2\,\mathrm{d}x+\int_0^1\sqrt{1+v^2}\,\mathrm{d}x-\int_0^1hv \,\mathrm{d}x.$$
How to show that $J(u)\rightarrow +\infty$ when $\|v\|_{H^1}\rightarrow +\infty$?

I'm stuck. There is a hint: $$\forall v\in H^1(0,1), \|v-\int_0^1 v\,\mathrm{d}x\|_{L^\infty}\leqslant\int_0^1|v'|\,\mathrm{d}x,$$ but I don't see how it helps. Could anyone please help?

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The problematic term is $-\int_0^1hvdx$. So thanks to this hint we can try to bound it below. – Davide Giraudo Apr 25 '12 at 9:22
Did you show the assertion of the hint? Then use $v=v-\int v+\int v$. – Davide Giraudo Apr 25 '12 at 9:54
Well, the term $\int_0^1 h v$ is linear in $v$, while the norm of $W^{1,2}(0,1)$ is quadratic in $v$; hence it should not be very important for coercivity. Actually, I am in trouble even in the trivial case $h=0$. Indeed, is $v \mapsto \|v'\|_2^2 + \|v\|_1$ coercive in $W^{1,2}(0,1)$? – Siminore Apr 26 '12 at 8:11
In this case $||v'||_2^2+||v||_1\geq ||v'||_2^2$ and we can conclude by Poincaré inequality. – Davide Giraudo Apr 26 '12 at 9:37
Poincaré inequality is valid for $v\in H^1_0$. How to solve it if $v\in H^1$? – galath Apr 26 '12 at 9:46