Gradient operator is it continuous?

Question

The following gradient operator is it continuous? \begin{array}{ccccc} \nabla :& H^1_0(\Omega) & \longrightarrow &(L^2(\Omega))^n\\ & y & \longmapsto & \displaystyle(\frac{\partial y(x)}{\partial x_1}, \frac{\partial y(x)}{\partial x_2}, ..., \frac{\partial y(x)}{\partial x_n}) \end{array} and how could i prove that it is continuous?

Answer 1

\begin{array}{ccccc} \nabla :& H^1_0(\Omega) & \longrightarrow &(L^2(\Omega))^n\\ & y & \longmapsto & \displaystyle(\frac{\partial y(x)}{\partial x_1}, \frac{\partial y(x)}{\partial x_2}, ..., \frac{\partial y(x)}{\partial x_n}) \end{array}

\begin{array}{ccccc} \|\nabla u\|^2_{(L^2(\Omega))^n} =\sum\limits_{i=1}^{n} \displaystyle\int_\Omega\Big|\frac{\partial u(x)}{\partial x_i}\Big|^2 dx\le \|u\|^2_{L^2(\Omega)}+\sum\limits_{i=1}^{n} \displaystyle\int_\Omega\Big|\frac{\partial u(x)}{\partial x_i}\Big|^2 dx = \|u\|^2_{H^1_0(\Omega)} \end{array}

That is \begin{array}{ccccc} \|\nabla u\|_{(L^2(\Omega))^n} \le\|u\|_{H^1_0(\Omega)} \end{array}