Suppose $f$ is of moderate decrease and that its Fourier transform $\hat{f}$ is supported in $I=[-1/2, 1/2]$, $\lambda > 1$. Show that: $$ f(x)=\sum_{n=-\infty}^{\infty}\frac{1}{\lambda}f(\frac{n}{\lambda})K_{\lambda}(x-\frac{n}{\lambda}) $$ where $$ K_{\lambda}(y)=\frac{\cos{\pi y} -\cos{\pi\lambda y}}{\pi^2(\lambda-1)y^2}. $$
My approach: Let $g(x)=\hat{f}(\lambda x)$, then by the Poisson summation formula we have $$ g(x)=\frac{1}{\lambda}\sum_{n=-\infty}^{\infty}f\left( \frac{n}{\lambda} \right)e^{ -2\pi inx } $$ by Fourier inversion we have $$ \begin{align} f(x)&=\int _{-\infty}^{\infty}\hat{f}(\xi)e^{ 2\pi i\xi x } \, d\xi=\int _{-\infty}^{\infty}g\left( \frac{\xi}{\lambda} \right)e^{ 2\pi i\xi x } \, d\xi \\ &=\int _{-\infty}^{\infty} \frac{1}{\lambda}\sum_{n=-\infty}^{\infty}f\left( \frac{n}{\lambda} \right)e^{ -2\pi in \frac{\xi}{\lambda} }e^{ 2\pi i\xi x } d\xi \\ &=\frac{1}{\lambda}\sum_{n=-\infty}^{\infty}f\left( \frac{n}{\lambda} \right) \frac{\sin\left( \pi\left( x-\frac{n}{\lambda} \right) \right)}{\pi\left( x-\frac{n}{\lambda} \right)} \end{align} $$ I'm not sure what's wrong with my calculations. The hint on page 167 in Stein's book Fourier Analysis is really confusing me, and I don't know how to "replace" the characteristic function so I didn't proceed with that approach.
I've also seen a similar problem on Function with product of sine kernel, but I don't understand the answer given because of my limited knowledge.
Any help would be appreciated!