I'm having difficulty in proving the solution of this problem:
The given vectors $\mathbf{a}_1$, $\mathbf{a}_2 \in \mathbb{R}^M$ and the variables $\mathbf{b}_{1},\mathbf{b}_2\in\mathbb{R}^M$. Assume that $l<\frac{1}{2}\|\mathbf{a}_1-\mathbf{a}_2\|_2$, $\mathbf{c}$ is an unit vector orthogonal to $(\mathbf{a}_1-\mathbf{a}_2)$, and $0<p<1$.
How to prove $$[\mathbf{a}_1,\mathbf{a}_2]=\arg\min_{[\mathbf{b}_1,\mathbf{b}_2]}\sum_{n=1}^{99}\Bigg\|\frac{n}{100}\mathbf{a}_1 + (1-\frac{n}{100})\mathbf{a}_2 - \big(\frac{n}{100}\mathbf{b}_1 + (1-\frac{n}{100})\mathbf{b}_2\big)\Bigg\|_2^p \\+ \Bigg\|0.5\mathbf{a}_1+0.5\mathbf{a}_2-\big(0.5\mathbf{b}_1+0.5\mathbf{b}_2\big)+l\mathbf{c}\Bigg\|_2^p$$
Any help would be highly appreciated.
Update:
I have already tried to use the KKT condition for solving this problem. The KKT condition of the unconstrained problem is only derivative equal to zeros.
Let the notation of the function in the problem is $J(\mathbf{b}_{1},\mathbf{b}_2)$,$$ u_n(\mathbf{b}_1,\mathbf{b}_2) = \Bigg\|\frac{n}{100}\mathbf{a}_1 + (1-\frac{n}{100})\mathbf{a}_2 - \big(\frac{n}{100}\mathbf{b}_1 + (1-\frac{n}{100})\mathbf{b}_2\big)\Bigg\|_2, n = 1,\dots,99, $$ and $$ u_{100}(\mathbf{b}_1,\mathbf{b}_2) = \Bigg\|0.5\mathbf{a}_1+0.5\mathbf{a}_2-\big(0.5\mathbf{b}_1+0.5\mathbf{b}_2\big)+l\mathbf{c}\Bigg\|_2. $$
The partial derivative are derived as following chain rule $$ \frac{\partial J(\mathbf{b}_1,\mathbf{b}_2)}{\partial \mathbf{b}_1}\triangleq \frac{\sum_{n=1}^{100}\partial \big(u_n(\mathbf{b}_1,\mathbf{b}_2)\big)^p}{\partial \mathbf{b}_1} =\sum_{n=1}^{100}\frac{\partial \big(u_n(\mathbf{b}_1,\mathbf{b}_2)\big)^p}{\partial u_n(\mathbf{b}_1,\mathbf{b}_2)}\frac{\partial u_n(\mathbf{b}_1,\mathbf{b}_2)}{\partial \mathbf{b}_1}. $$ Then, the partial derivative of this problem can be obtained as follows: $$ \mathbf{0} = \frac{\partial J(\mathbf{b}_1,\mathbf{b}_2)}{\partial \mathbf{b}_1} \\=\sum_{n=1}^{99}\frac{-np}{100}\big(u_n(\mathbf{b}_1,\mathbf{b}_2)\big)^{p-2}\Bigg(\frac{n}{100}\mathbf{a}_1 + (1-\frac{n}{100})\mathbf{a}_2 - \big(\frac{n}{100}\mathbf{b}_1 + (1-\frac{n}{100})\mathbf{b}_2\big)\Bigg) \\+ (-0.5p)\big(u_{100}(\mathbf{b}_1,\mathbf{b}_2)\big)^{p-2}\Bigg(0.5\mathbf{a}_1+0.5\mathbf{a}_2-\big(0.5\mathbf{b}_1+0.5\mathbf{b}_2\big)+l\mathbf{c}\Bigg). $$ Can any one help me to check the partial derivative of $\mathbf{b}_1$ in this problem is correct or not?