# Suppose $f(\beta)$ is strictly convex. Show that $f(\hat{\beta}_k^*) < f(\hat{\beta}_k)$

Suppose $f(\beta)$ is strictly convex. Show that $f(\hat{\beta}_k^*) < f(\hat{\beta}_k)$, where $\hat{\beta}, \hat{\beta}^* \in \mathbb{R}^n$ and

$\hat{\beta}_k^* = \begin{cases} \hfill \frac{1}{2}(\hat{\beta}_i+\hat{\beta}_j) \hfill & \text{ if$k \in \{i,j\}$} \\ \hfill \hat{\beta}_k \hfill & \text{ otherwise.} \\ \end{cases}$

This is a part of the proof to Lemma 2 (a) in [1]. The proof to this Lemma can be found in the Appendix of [1], but it includes this step, which I unfortunately do not understand. If I understand it correctly, it is also given that $f(\beta) >0$ $\forall \beta \neq 0$.

UPDATE: Any help would be appreciated. I might need to provide you with more information, but if you tell me what other information you need, then let me know. I attempted to solve this question several times and only am able to start with $f(\hat{\beta}_k^*)$ and separate this into $f(\frac{1}{2} \hat{\beta}+\frac{1}{2}\tilde{\beta})$, where $\tilde{\beta}$ is a vector, whose elements are $\{\hat{\beta}_1, \dotsc, \hat{\beta}_j, \dotsc, \hat{\beta}_i, \dotsc, \hat{\beta}_n\}$, where $n$ is the last element and I simply assume $i<j$. Then I can make use of the strict convexity, but that's how far I get.

[1] Zou, Hui, and Trevor Hastie. "Regularization and variable selection via the elastic net." Journal of the Royal Statistical Society: Series B (Statistical Methodology) 67.2 (2005): 301-320.

• just added my new information inside the post. – pthesling Aug 31 '15 at 15:20

I think that the claim is false. Consider $X = \mathrm{diag}(1, 0, 0)$, $y = 0$ and the strictly convex function $J(x) = ||x - (0, 1, 2)^t||_2^2$. Then the minimizer of $|y - X\beta|^2 + \lambda J(\beta)$ is clearly $\hat{\beta} = (0, 1, 2)^t$.