If $u=0$ in the border can I say $\frac{\partial u}{\partial \vec n}=0$ in the border?

Question

$$u(x) = \int_{\Omega} T_y(x)\Delta u(y) dy - \int_{\partial \Omega} u(y)\frac{\partial T_y(x)}{\partial \vec n} - T_y(x)\frac{\partial u}{\partial \vec n} d\sigma(y)$$

If I have that $u=0$ in $\partial \Omega$ then can I say that

$$\frac{\partial u}{\partial \vec n}=0$$

in $\partial \Omega$?