Since $I_{\nu}$ and $K_{\nu}$ satisfy

(5.34)		$\displaystyle\frac{d}{dy}(y\cdot I_{\nu}^{\prime}(y))-(y+\frac{\nu^{2}}{y})I_{\nu}(y)=0,$
(5.35)		$\displaystyle\frac{d}{dy}(y\cdot K_{\nu}^{\prime}(y))-(y+\frac{\nu^{2}}{y})K_{\nu}(y)=0.$

This implies that

(5.36)

$$K_{\nu}(y)\cdot\frac{d}{dy}(y\cdot I_{\nu}^{\prime}(y))-I_{\nu}(y)\cdot\frac{d}{dy}(y\cdot K_{\nu}^{\prime}(y))=0,$$

and hence

(5.37)

$$\frac{d}{dy}\Big(y\cdot\mathcal{W}(I_{\nu}(y),K_{\nu}(y))\Big)=0.$$

Therefore, $y\cdot\mathcal{W}(I_{\nu}(y),K_{\nu}(y))$ is a constant.

Next, we will compute this constant which equals the limit of $y\cdot\mathcal{W}(I_{\nu}(y),K_{\nu}(y))$ as $y\to 0$. By definition,

(5.38)

$$\lim\limits_{y\to 0}I_{\nu}(y)\Big/\Big(\frac{y^{\nu}}{\Gamma(\nu+1)\cdot 2^{\nu}}\Big)=1,\ \lim\limits_{y\to 0}K_{\nu}(y)\Big/\Big(\frac{\pi}{2\sin(\nu\pi)}\cdot\frac{2^{\nu}\cdot y^{-\nu}}{\Gamma(1-\nu)}\Big)=1.$$

Notice that

(5.39)

$$\Gamma(\nu+1)\Gamma(1-\nu)=\nu\Gamma(\nu)\Gamma(1-\nu)=\frac{\nu\pi}{\sin(\nu\pi)},$$

then it is straightforward that

(5.40)

$$\lim\limits_{y\to 0}y\cdot(I_{\nu}(y)K_{\nu}^{\prime}(y)-K_{\nu}(y)I_{\nu}^{\prime}(y))=-1.$$

This completes the proof. ∎