= Bounded solvability for the Legendre operator
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{title2=$((1-x^2)y')'=f,\quad\int_{-1}^1f=0$}
= Zero-mean compatibility for the Legendre operator
{c}
{synonym}
For $f\in C([-1,1])$, the equation $((1-x^2)y')'=f$ has a bounded solution on $(-1,1)$ if and only if $\int_{-1}^1f=0$. All bounded solutions differ by a constant. To prove necessity, integrate the equation once. The quantity $(1-x^2)y'$ has finite endpoint limits, and any nonzero endpoint limit would force logarithmic growth in $y$. Both limits must therefore vanish, yielding the zero-mean condition. For sufficiency set $F(x)=\int_{-1}^xf(t)dt$ and $y(x)=C+\int_0^x F(s)/(1-s^2)ds$. Zero mean implies $|F(s)|\leq\|f\|_\infty\min(1+s,1-s)$, so $y'$ is bounded and $y$ extends continuously to the endpoints. The homogeneous solutions are a constant and $\tfrac12\log((1+x)/(1-x))$, proving uniqueness up to a constant. This is the <Fredholm alternative> for the zero <eigenvalue> of this singular <Sturm-Liouville problem>.
The <eigenfunction expansion> in <Legendre polynomials> is $y=C-\sum_{k\geq1}P_k\int_{-1}^1fP_k/[k(k+1)\int_{-1}^1P_k^2]$. It follows by projecting onto the <eigenfunctions> with <eigenvalues> $-k(k+1)$. Completeness follows because the <Legendre polynomials> span the <polynomials>, which are dense by the <Weierstrass approximation theorem>.
Continuity only in the open interval requires more care. Even if $f$ is integrable and has zero mean, the primitive of $F/(1-x^2)$ must still be bounded at both ends. For example, $F(x)=1/\log(e/(1-x^2))$ has zero endpoint limits and absolutely integrable derivative $f=F'$, but $F/(1-x^2)$ has a divergent primitive. Thus zero mean alone is insufficient without endpoint control.
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