Maximizing a sum with ReLu functions.

Let $$f$$ be a vector valued function and $$W$$ a matrix. We denote the rows of the matrix by $$w_j$$. Suppose that $$\sum_{j=1}^h \lVert w_j\rVert^2\leq R^2$$. Show that the sum

$$\sum_{j=1}^h \lVert w_j\rVert^2\left(\sum_{i=1}^m \epsilon_i\sigma\left(\frac{w_j^T}{\lVert w_j\rVert}f(x_i)\right)\right)^2$$

achieves its supremum only if $$\lVert w_j\rVert=R$$ for some $$j$$ and $$\lVert w_i\rVert=0$$ for all $$i\neq j$$. Now I tried Lagrange multipliers but wasn't able to get the desired result. Is there a cleaner way to show this.

EDIT: $$\sigma$$ is the ReLu function and $$\epsilon_i\in \{-1,1\}$$ fixed. And $$x_i$$ are just fixed values.

• What are $\sigma, \epsilon_i, x_i,...$? – copper.hat Oct 27 '18 at 4:56
• I edited the post. – thegamer Oct 27 '18 at 5:17
• Nevermind I got it. – thegamer Oct 28 '18 at 1:33
• Just optimize the $w_j^T/||w_j||$ part first then bound the inequality by the max. – thegamer Oct 28 '18 at 1:34
• Assuming that $\sigma(x) = \max(0,x)$, it is not obvious to me... – copper.hat Oct 28 '18 at 4:34

Note that to maximize the sum $$\sum_{j=1}^h \lVert w_j\rVert^2\left(\sum_{i=1}^m \epsilon_i\sigma\left(\frac{w_j^T}{\lVert w_j\rVert}f(x_i)\right)\right)^2$$
we need to maximize each of the individual sums $$\left(\sum_{i=1}^m \epsilon_i\sigma\left(\frac{w_j^T}{\lVert w_j\rVert}f(x_i)\right)\right)^2$$ first. Since $$\frac{w_j^T}{||w_j||}f(x_i)$$ is a function over the unit sphere, which is compact, it attains a maximum at some $$\tilde w_j$$. Notice that this maximum is invariant under any positive scaling of $$\tilde{w}_j$$. Let $$M=\max_j\left(\sum_{i=1}^m \epsilon_i\sigma\left(\frac{\tilde{w}_j^T}{\lVert \tilde{w}_j\rVert}f(x_i)\right)\right)^2$$ then we have
$$\sum_{j=1}^h \lVert \tilde{w}_j\rVert^2\left(\sum_{i=1}^m \epsilon_i\sigma\left(\frac{\tilde{w}_j^T}{\lVert \tilde{w}_j\rVert}f(x_i)\right)\right)^2\leq \sum_{j=1}^h ||\tilde w_j|^2 M\leq R^2M$$
Note that there exists a $$k$$ such that $$M=\left(\sum_{i=1}^m \epsilon_i\sigma\left(\frac{\tilde{w}_k^T}{\lVert \tilde{w}_k\rVert}f(x_i)\right)\right)^2$$. Thus, if $$||\tilde{w}_k||^2=R^2$$ then our inequality become an equality. Therefore we have maximized the sum.