A sum of eigenvalues

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!

-

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 : $$f(X)=\max\left(0,\max_{1\le r\le n}\sum_{j=1}^r\lambda_j(X)\right).$$

-
Yes, your idea is correct. But the reason should be revised to be $$f(X) = \max\{0,\sum_{i=1}^1 \lambda_i(X), \cdots, \sum_{i=1}^r \lambda_i(X)\}.$$ –  user11870 May 30 '12 at 12:42
@wmmiao. I agree. –  Denis Serre May 30 '12 at 12:54

The matrix is assumed to be real, correct?

The answer seems to be trivially yes. Indeed, assume that $\lambda_{k} \geq 0$ and $\lambda_{k+1}<0$ and $r \geq k+1$. Then $\sum_{i=1}^{r}{\max(0,\lambda_{i})}=\sum_{i=1}^{k}{\lambda_{i}}$.

-
@Felix. Your argument does not work because your $k$ depends upon $X$. –  Denis Serre May 30 '12 at 12:21
Yes, assume to be real. But what we want to prove is for any 0≤α≤1, $$\alpha \sum_{i=1}^r \max(0, \lambda_i(X))+(1-\alpha)\sum_{i=1}^r \max(0,\lambda_i(Y)) \leq \sum_{i=1}^r \max(0,\lambda_i(\alpha X+(1-\alpha)Y)).$$It is not trivial. –  user11870 May 30 '12 at 12:24
Oh dear. Well, I had a feeling something was amiss. :) –  Felix Goldberg May 30 '12 at 13:41