Based on a sample $X_1,...,X_n$ a biweight M-estimate of location is defined to be a solution, $ \hat{\theta_c}$, to $$\sum_{i=1}^{n} \chi_c (\frac{X_i - \theta}{\hat{\sigma}})=0 $$ where $\chi_c(x)=x(c^2 - x^2 )1_{(\mid x \mid \lt 1 )}$ and $\hat{\sigma}$ is an estimate of scale.
Show that $ \hat{\theta_c}$ is equal to a mode of the kernel density estimate based on $X_1,...,X_n$ with bandwidth $h=c\hat{\sigma}$ and the biweight kernel $K(x)=\frac{15}{16} (1-x^2)^2 1_{(\mid x \mid \lt 1 )} $
I am confused!!
The kernel density estimate is defined as
$$\hat{f}_h=\frac{1}{nh}\sum_{i=1}^n K\left(\frac{x-X_i}{h}\right)$$
Filling in the information you have on the kernel density estimator
$$\hat{f}_h=\frac{1}{nc\hat{\sigma}}\sum_{i=1}^n \frac{15}{16} \left(1-\left(\frac{x-X_i}{c\hat{\sigma}}\right)^2\right)^2 1_{(\mid x \mid \lt 1 )}$$
Taking the derivative with respect to $x$, you can show that this is
$$\hat{f}'_h=-\frac{15}{4}\frac{1}{nc^3\hat{\sigma}}\sum_{i=1}^n \chi_c\left(\frac{x-X_i}{c\hat{\sigma}}\right)=\frac{15}{4}\frac{1}{nc^3\hat{\sigma}}\sum_{i=1}^n \chi_c\left(\frac{X_i-x}{c\hat{\sigma}}\right)$$
The last step is because $\chi_c$ is an odd function. Since a mode is an extremum of the function, we know that the derivative has to be zero.