Let $I$ be the open interval $(0, 1)$. I am trying to solve a problem in Brezis' Functional Analysis
Exercise 8.31 Let $V=\{v \in H^1 (I) : v(1)=0\}$. Let $H$ be the vector space of real-valued measurable functions on $I$ such that $x f(x) \in L^2 (I)$.
- Show that $H$ equipped with the scalar product $\langle \cdot, \cdot \rangle_H$ by $$ \langle f, g \rangle_H :=\int_I f(x)g(x)x^2 \, \mathrm d x. $$ is a Hilbert space.
- Given $f \in H$ and $\varepsilon>0$, check that there exists a unique $u \in V$ satisfying $$ \int_I u'(x) v'(x) (x^2+\varepsilon) \, \mathrm d x+ \int_I u(x) v(x) x^2 \, \mathrm d x = \int_I f(x) v(x) x^2 \, \mathrm d x \quad v \in V. $$ This $u$ is denoted by $u_{\varepsilon}$.
There are possibly subtle mistakes that I could not recognize in my below attempt of (2.). Could you please have a check on it?
Notice that $V$ is a closed subspace of $H^1(I)$, so $(V, \|\cdot\|_{H^1})$ is a Hilbert space. We define a continuous bilinear form $[\cdot, \cdot]$ on $V$ by $$ [u, v] := \int_I [ u'(x) v'(x) (x^2+\varepsilon) + u(x) v(x) x^2] \mathrm d x. $$
For $v \in V$, we have $v(x) = -\int_x^1 v'$ and thus $\| v \|_{\infty} \le \| v' \|_{L^2}$. Then for $v \in V$, $$ \begin{align*} [v, v] &= \int_I [ |v'(x)|^2 (x^2+\varepsilon) + |v(x)|^2 x^2] \mathrm d x \\ &\ge \varepsilon \int_I |v'|^2 \ge \frac{\varepsilon}{2} \|v\|_{H^1}. \end{align*} $$
Then $[\cdot, \cdot]$ is coercive on $V$. By Cauchy–Schwarz inequality, the linear map $$ V \to \mathbb R, \quad v \mapsto\int_I f(x) v(x) x^2 \, \mathrm d x $$ is continuous. The claim then follows from Lax-Milgram theorem.