Let $X$ be an $n\times n$ symmetric matrix. Suppose $\lambda\_1(X)\geq \lambda\_2(X) \geq \cdots \geq \lambda\_n(X)$ are eigenvalues of $X$. Let $r$ be any integer with $1\leq r\leq n$. It is well-known that $\sum\_{i=1}^r \lambda\_i(X)$ is convex. Now, my question is: Is the following function convex?
$$\sum\_{i=1}^r \max(0,\lambda\_i(X))$$


Thanks!



A sum of eigenvalues
The answer is yes, because your function (let me call it $f$) is the maximum of convex functions. As such, it is convex. The formula :
$$f(X)=\max\left(0,\max\_{1\le r\le n}\sum\_{j=1}^r\lambda\_j(X)\right).$$


