$J$ is not coercive in $W^{1,2}$ For that to happen you need to show that $\Vert u_n\Vert_{1,2}\to \infty$ implies $J(u_n)\to \infty$. Take for example the function $u_n$ which is identically $0$ for $x>\frac{1}{n}$ and equal to $1-nx$ on the interval $[0,1/n]$.

Then

$$J(u_n)<\frac{1}{2n} $$

and

$$\Vert u_n\Vert_{1,2}^2\geq n. $$

$J$ is not coercive in $W^{1,2}$ For that to happen you need to show that $\Vert u_n\Vert_{1,2}\to \infty$ implies $J(u_n)\to \infty$. Take for example the function $u_n$ which is identically $0$ for $x>\frac{1}{n}$ and equal to $1-nx$ on the interval $[0,1/n]$.

Then

$$J(u_n)<\frac{1}{2n} $$

and

$$\Vert u_n\Vert_{1,2}^2\geq n. $$

Next observe that

$$ \frac{1}{\sqrt{2}}(|a|+|b|)\leq \sqrt{a^2+b^2}\leq (|a|+|b|) $$

which shows that

$$ I(u_n)\to \infty\iff \Vert u_n\Vert_{1,1}\to\infty. $$

Choose a sequence of smooth functions $u_n$ which converge in $W^{1,1}$ to a function $u\in W^{1,1}\setminus W^{1,2}$. Then $\Vert u_n\Vert_{1,1}\to \Vert u\Vert_{1,1}$ but $\Vert u_n\Vert_{1,2}\to\infty$. This proves that $I$ is not coercive either.