= Solution
If $\lambda=0$, only a linear function can solve the equation, and zero endpoint values make it zero. If $\lambda<0$, the corresponding hyperbolic-sine solution also cannot vanish at both endpoints unless it is zero. For $\lambda=k^2>0$, the first endpoint gives $y=C\sin kx$, and the second requires $k=n\pi$ for a positive <integer> $n$. Thus $\boxed{\lambda_{\min}=\pi^2}$.
The admissible trial function $y=\sin\pi x$ has quotient
$$
\frac{\int_0^1\pi^2\cos^2\pi x\,dx}{\int_0^1\sin^2\pi x\,dx}=\pi^2.
$$
Therefore the minimum, or infimum before its attainment is established, is $\boxed{\leq\pi^2}$ as requested. The sharp <Poincaré inequality> supplies the reverse inequality on the finite-energy zero-trace class, so the exact minimum is also $\pi^2$ and is attained by this trial function. The one-sided assertion in the source does not require a full variational existence argument.
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