Let $I$ be the open interval $(0, 2)$. Consider the bilinear form on $H^1 (I)$ $$ a(u, v) = \int_0^2 u'v' + \left(\int_0^1 u\right) \left(\int_0^1 v\right) . $$
I'm trying to solve a problem in Brezis' Functional Analysis, i.e.,
Exercise 8.25
- Check that $a(u, v)$ is a continuous symmetric bilinear form, and that $a(u, u)=0$ implies $u=0$.
- Prove that $a$ is coercive.
- Deduce that for every $f \in L^2(I)$ there exists a unique $u \in H^1(I)$ satisfying $$ (1) \quad a(u, v)=\int_0^2 f v \quad \forall v \in H^1(I) . $$ What is the corresponding minimization problem?
- Show that the solution of (1) belongs to $H^2(I)$ and in particular $u \in C^1(\bar{I})$. Determine the equation and the boundary conditions satisfied by $u$.
- Assume that $f \in C(\bar{I})$, and let $u$ be the solution of (1). Prove that $u$ belongs to $W^{2, p}(I)$ for every $p<\infty$. Show that $u \in C^2(\bar{I})$ iff $\int_I f=0$.
In below attempt of (5.), I get $u \in W^{2, p}(I)$ even when $p = \infty$. Could you confirm if I made any subtle mistakes?
Clearly, $u \in H^2(I)$ implies $u \in C^1(\bar{I})$. Because $I$ is bounded, we get $u \in W^{1, p}(I)$ for any $p \in [1, \infty]$. Let $g (x) := (\int_0^1 u) 1_{(0, 1)} (x)$ for $x \in I$. From $(1)$, we get $$ (2) \quad \int_I [u'v' + gv] = \int_I f v \quad \forall v \in H^1(I). $$
By integration by parts, $u \in H^2(I)$ with $u'' = g-f$ and $u'(0)=u'(2)=0$. It follows from $f \in C(\bar{I})$ that $f$ is bounded. Clearly, $g$ is bounded, so $u''$ is bounded and thus $u \in W^{2, p}(I)$ for any $p \in [1, \infty]$.
We consider $u \in C^2(\bar{I})$. Then $u''$ is continuous. This implies $g$ is continuous and thus $g=\int_0^1 u=0$. We substitute $v=1$ into $(2)$ and get $\int_I f =0$.
We consider $\int_I f =0$. We substitute $v=1$ into $(2)$ and get $\int_I g =0$ and thus $g=0$. This implies $u''=-f \in C(\bar{I})$.