这个地方验证 ∫−11yn(x)ym(x)dx=0\int_{-1}^1y_n(x)y_m(x)dx=0∫−11yn(x)ym(x)dx=0 可以用分部积分搞，假设 m<nm<nm<n：
∫−11dndxn(x2−1)ndmdxm(x2−1)mdx=∫−11ddxdn−1dxn−1(x2−1)ndmdxm(x2−1)mdx=dn−1dxn−1(x2−1)ndmdxm(x2−1)m∣−11−∫−11dn−1dxn−1(x2−1)ndm+1dxm+1(x2−1)mdx=−∫−11dn−1dxn−1(x2−1)ndm+1dxm+1(x2−1)mdx=(−1)n∫−11(x2−1)ndm+ndxm+n(x2−1)mdx \begin{aligned} \int_{-1}^1\frac{d^n}{dx^n}(x^2-1)^n\frac{d^m}{dx^m}(x^2-1)^mdx&=\int_{-1}^1\frac{d}{dx}\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n\frac{d^m}{dx^m}(x^2-1)^mdx\\ &=\left.\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n\frac{d^m}{dx^m}(x^2-1)^m\right|_{-1}^1-\int_{-1}^1\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n\frac{d^{m+1}}{dx^{m+1}}(x^2-1)^mdx\\ &=-\int_{-1}^1\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n\frac{d^{m+1}}{dx^{m+1}}(x^2-1)^mdx\\ &=(-1)^{n}\int_{-1}^1(x^2-1)^n\frac{d^{m+n}}{dx^{m+n}}(x^2-1)^mdx\\ \end{aligned} ∫−11dxndn(x2−1)ndxmdm(x2−1)mdx=∫−11dxddxn−1dn−1(x2−1)ndxmdm(x2−1)mdx=dxn−1dn−1(x2−1)ndxmdm(x2−1)m∣∣∣∣−11−∫−11dxn−1dn−1(x2−1)ndxm+1dm+1(x2−1)mdx=−∫−11dxn−1dn−1(x2−1)ndxm+1dm+1(x2−1)mdx=(−1)n∫−11(x2−1)ndxm+ndm+n(x2−1)mdx
此时由于 m<nm<nm<n，因此 dm+ndxm+n(x2−1)m=0\frac{d^{m+n}}{dx^{m+n}}(x^2-1)^m=0dxm+ndm+n(x2−1)m=0 （因为被求导的多项式是 2m2m2m 次的，求了 m+nm+nm+n 次导之后是 000） 规律是零点越来越多了 ，特征函数有非常好的震动性。
微分比积分多一个条件才能交换，多要求了微分之后的级数一致收敛。
希尔伯特空间的共轭空间和共轭算子就是满足类似于 (Ax,y)=(x,Ay)(Ax,y)=(x,Ay)(Ax,y)=(x,Ay)
