Conditioning changes the interval on a triangular support

A random vector $(X,Y)$ has joint density $$ f_{X,Y}(x,y)= \begin{cases} C,&0<x<y<1,\\ 0,&\text{otherwise}. \end{cases} $$ Find $C$. Derive the marginal densities $f_X$ and $f_Y$, including their supports, and then derive $f_{X\mid Y}(x\mid y)$ and $f_{Y\mid X}(y\mid x)$ for conditioning values where the relevant marginal density is positive. Verify directly that each conditional density integrates to one over its own support. Compute $$ \mathbb P(X<Y/2\mid Y=y),\qquad \mathbb P\!\left(Y<\frac{1+X}{2}\,\middle|\,X=x\right), $$ for $0<y<1$ and $0<x<1$. Find $\mathbb E[X\mid Y=y]$ and $\mathbb E[Y\mid X=x]$, and verify the tower identity $$ \mathbb E[\mathbb E[X\mid Y]]=\mathbb E[X] $$ by explicit integration. Finally compute $\operatorname{Cov}(X,Y)$ and prove that $X$ and $Y$ are not independent in two distinct ways: once from the support or density factorization, and once from a conditional law or covariance. Explain why a joint density being constant on its support does not imply independence. Values assigned to densities on boundary sets of Lebesgue measure zero are irrelevant.

Draw $0<x<y<1$. For fixed $x$, the allowed $y$-interval is $(x,1)$; for fixed $y$, the allowed $x$-interval is $(0,y)$.