maximizing the negative norm is it convex problem? I have an objective function with two decision variables x1 and x2
\begin{equation} 
\begin{array}{cl}
\underset{\alpha_{1},\alpha_{2}}{ \max}  a_1x_1+ a_2x_2 -c\sqrt{h_{1}(x) ^2 + h_{2}(x)^2}  \\ 
\text { s.t. }  x_{1}+x_{2}=1 
\end{array}
\end{equation}
I tried to introduce new variables
$b_1=h_1(x) ,b_2=h_2(x) $ so my objective function will be
\begin{array}{cl}
\underset{\alpha_{1},\alpha_{2}}{ \max}  a_1x_1+ a_2x_2 -cu  \\ 
\text { s.t. }  x_{1}+x_{2}=1 ,\\
 h_1(x)=b_1, \\ 
h_2(x)=b_2 ,\\
 \sqrt{b_1^2 + b_2^2 } \leq u 
\end{array}
my questions:

*

*does the reformulation with the new constraints correct ?


*because I am maximizing the negative norm does that mean it is a convex function ?


*how can I solve my problem in MATLAB
Thanks in advance

If I understand the problem correctly the optimization is over $(x_{1},x_{2})$ and not over $(a_{1},\,a_{2})$ because in the last case max=$+\infty$.
Just use the linear constraint and write $x_{2}=1-x$. Then we get
max $a_{1}x+a_{2}(1-x)-c\sqrt{H_{1}^{2}(x)+H_{2}^{2}(x)}$
where $H_{1}(x)=h_{1}(x,1-x)$ and $H_{2}(x)=h_{2}(x,1-x)$
which can be solved in the standard way by considering derivative of the function and finding stationary points, which is:
$a_{1}-a_{2}-c\dfrac{1}{2\sqrt{H_{1}^{2}(x)+H_{2}^{2}}(x)}(2H_{1}(x)H_{1}'(x)+2H_{2}(x)H_{2}'(x))=0$
Solving the last equality you get some stationary points and by considering second derivative you will decide which is the maximum! But it all depends on functions $\, h_{1},\,h_{2}$.
