There's that claim in the book I'm reading that I don't manage to prove.
We consider the function
$$ b(x)= \int_0^1 H(x(t))dt.$$
Where $x \in C^{\infty}(S^1,\mathbb{R}^n)$ and $H : \mathbb{R}^n \rightarrow \mathbb{R}$ is smooth.
We know that the function $H$ is such that
(1) $H(x) \equiv 0$ near $x=0$ and
(2) $H(x)= (\pi + \epsilon)q(x)$, for large values of $\mid x \mid$. Here $q(x)$ is a quadratic form.
Furthermore the following estimates hold for all $x \in \mathbb{R}$:
(3) $\mid H(x) \mid \leq M \mid x \mid^2$,
(4) $\mid d^2H(x) \mid \leq M.$
For a positive constant $M$.
The claim is that $b$ is defined for all $x \in L^2(S^1,\mathbb{R}^n)$. I am not sure what I should show here. This is what I've done
Let $x \in L^2 (S^1,\mathbb{R}^n)$, then by definition we have
$$ \int_0^1 \mid x(t) \mid^2 dt < \infty. $$
Now compute the value of $b(x)$ and use (3):
$$b(x)=\int_0^1 H(x(t)) dt \leq \int_0^1 M \mid x(t) \mid^2 dt = M \int_0^1 \mid x(t) \mid^2 dt < \infty.$$
I would argue that $b(x)$ cannot diverge to $- \infty$ because of (1) and (2).
Please let me know if I really missed something. Thank you very much