= Solution
Use the real <Hilbert space> $V=H_0^1(0,1)$, the <zero-boundary Sobolev space>, with <norm> $\|v\|_V=\|v'\|_{L^2}$. The <Poincaré inequality> makes this equivalent to its usual <Sobolev space> norm. Define the symmetric <bilinear form> and continuous <linear functional>
$$
a(u,v)=\int_0^1\bigl(u'v'+xuv\bigr)\,dx,\qquad
\ell(v)=\int_0^1fv\,dx.
$$
The <Cauchy-Schwarz inequality> and the sharp interval <Poincaré inequality> imply
$$
|a(u,v)|\leq(1+\pi^{-2})\|u\|_V\|v\|_V,\qquad
a(v,v)\geq\|v\|_V^2,\qquad
|\ell(v)|\leq\pi^{-1}\|f\|_{L^2}\|v\|_V.
$$
Thus $a$ is a <bounded bilinear form> and a <coercive bilinear form>, and $\ell$ is bounded because $f$ is <square-integrable>. The <Lax-Milgram theorem> supplies a unique <weak solution>:
$$
\boxed{y\in H_0^1(0,1),\qquad a(y,v)=\ell(v)\quad\text{for every }v\in H_0^1(0,1)}.
$$
This is obtained from the differential equation by <integration by parts>; the test functions have zero boundary trace. Conversely, the <weak solution> satisfies $y''=xy-f$ in the sense of <distributions>. Since $xy-f\in L^2$, it belongs to $H^2(0,1)\cap H_0^1(0,1)$ and solves the equation <almost everywhere>.
Equivalently, the <Ritz method> minimizes
$$
J(v)=\frac12\int_0^1\bigl((v')^2+xv^2\bigr)\,dx-\int_0^1fv\,dx
$$
over $V$. Indeed $J(y+w)-J(y)=a(w,w)/2$, since the mixed term is $a(y,w)-\ell(w)=0$. This proves existence and uniqueness of the minimizer directly from the <weak solution>, as well as its equivalence to the <variational problem>.
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