Show that the set of $x$ such that $x$ majorizes $y$ is convex I have this question involving Majorization :
show that $x$ majorizes $y$ if and only if
$$\sum_{j=1}^{d}\max(x_j-t,0)\leq\sum_{j=1}^{d}\max(y_j-t,0)$$
and
$$\sum_{j=1}^{d}x_j=\sum_{j=1}^{d}y_j$$
for all real values of $t$, where $d$ is the dimension of $x$ and $y$.
I can do this bit just fine by picking values of $t$ and showing that one implies the other. But the next part is:
Use the previous exercise to show that the set of $x$ such that $x$ majorizes $y$ is convex.
From which I'm not entirely sure how to proceed. Do I have to define a specific form of y?
Exercise was taken from Nielsen and Chuang "Quantum Computation and Information" section 12.5.1

The definition of a convex set $S$ is that for any two points inside it, $x^{1}$ and $x^{2},$ and $\alpha\in[0,1],$ $\alpha x^{1}+(1-\alpha)x^{2}\in S.$ We want to show that the set $S=\{x\text{ majorizes }y\}$ is convex using the characterization of majorization above. Noting that $\max(x-t,0)$ is convex in $x$ for each fixed $t,$ and since $x^{1}$ and $x^{2}$ majorize $y,$ we see that \begin{align*}\sum_{j=1}^{d}\max(\alpha x^{1}_{j}+(1-\alpha)x^{2}_{j}-t,0)&\leq \sum_{j=1}^{d}\alpha\max(x^{1}_{j}-t,0)+(1-\alpha)\max(x^{2}_{j}-t,0)\\&\leq \sum_{j=1}^{d}\alpha\max(y_{j}-t,0)+(1-\alpha)\max(y_{j}-t,0)\\&=\sum_{j=1}^{d}\max(y_{j}-t,0).\end{align*} Lastly, again since $x^{1},x^{2}$ majorize $y$, $\sum_{j=1}^{d}\alpha x_{j}^{1}+(1-\alpha)x_{j}^{2}=\alpha\left(\sum_{j=1}^{d}y_{j}\right)+(1-\alpha)\left(\sum_{j=1}^{d}y_{j}\right)=\sum_{j=1}^{d}y_{j}.$ This proves that $\alpha x^{1}+(1-\alpha)x^{2}$ majorizes $y,$ hence belongs to $S$, which completes the proof.
