Computing the subgradient of an indicator function or the normal cone of a set

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I am reading a blog post by someone and they have the following example,

$K_2 = \{(x,y):y^2\geq x\}$, $I_{K_2}(x,y) = \begin{cases}0 & (x,y)\in K_2\\ +\infty & (x,y)\not\in K_2\end{cases}$.

When they calculate $\partial I_{K_2}(0,0)$ they get $\{(0,y):y\leq 0\}$. This does not seem correct to me, surely it would be $\{(x,0):x\geq 0\}$? Could it have been that they meant to define the set $K_2$ to be $K_2=\{(x,y):y\geq x^2\}$ instead of $y^2\geq x$?

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Since your tags all indicate convexity, I think your assumption is correct. The first $K_2$ does not define a convex set. With your second set, the solution is correct.