Bounded solvability for the Legendre operator ($((1-x^2)y^{\prime }{)}^{\prime }=f,{\int }_{-1}^{1}f=0$)

For $f\in C([-1,1])$, the equation $((1-x^2)y^{\prime }{)}^{\prime }=f$ has a bounded solution on $(-1,1)$ if and only if ${\int }_{-1}^{1}f=0$. All bounded solutions differ by a constant. To prove necessity, integrate the equation once. The quantity $(1-x^2)y^{\prime }$ 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 $F(x)={\int }_{-1}^{x}f(t)dt$ and $y(x)=C+{\int }_0^{x}F(s)/(1-s^2)ds$. Zero mean implies $|F(s)|\le \Vert f{\Vert }_{\infty }\min (1+s,1-s)$, so $y^{\prime }$ is bounded and $y$ extends continuously to the endpoints. The homogeneous solutions are a constant and $\frac{1}{2}\log ((1+x)/(1-x))$, 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 $y=C-{\sum }_{k\ge 1}{P}_k{\int }_{-1}^{1}f{P}_k/[k(k+1){\int }_{-1}^1{P}_k^2]$. 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 $f=F^{\prime }$, but $F/(1-x^2)$ has a divergent primitive. Thus zero mean alone is insufficient without endpoint control.