Let $T:L^2([0,2\pi]) \to H^1([0,2\pi])$ be an integral operator given by
$$ T[\phi](x) = \int_0^{2\pi} \sin(\alpha |x-y|) \phi(y) dy. $$ I want to show that this operator can be asymptotically expanded with respect to $\alpha$ and that the expansion converges in the operator norm of $\mathcal{L}(L^2([0,2\pi]),H^1([0,2\pi]))$.
Regarding the asymptotic expansion, that is straightforward because I can just Taylor expand the kernel $k(x,y) = \sin(\alpha |x-y|)$ as
$$ \begin{align} k(x,y) & = \alpha|x-y| - \frac{1}{6}\alpha^3|x-y|^3 + \frac{1}{120}\alpha^5|x-y|^5 + \dots \end{align}. $$ So then I define $$ T_j[\phi](x) = \int_0^{2\pi}\frac{1}{j!}\frac{d^j}{d\alpha^j}\sin(\alpha|x-y|)\bigg|_{\alpha = 0}\phi(y)dy. $$
Now I can write $$ T[\phi](x) = \left(\sum_{j=0}^\infty \alpha^j T_j\right)[\phi](x) $$ So how do I then show that the series $\left(\sum_{j=0}^\infty \alpha^j T_j\right)$ converges in the operator norm of $\mathcal{L}(L^2([0,2\pi]),H^1([0,2\pi]))$?