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Anybody have any tips on how to show that the function $\frac{1}{2}\vert x_1 + x_2 + x_3 - x_1x_2x_3\vert^2$ is convex in $\mathbf{x}$, where $\vert x_i \vert \leq 1$?

This comes from the following expression, for general N:

\begin{equation} \frac{1}{2}\left\vert (\begin{array}{cc} 0 & 1 \end{array}) \left(\begin{array}{cc} 1 & -x_N \\\ x_N & 1 \end{array}\right) \cdots \left(\begin{array}{cc} 1 & -x_1 \\\ x_1 & 1 \end{array} \right) \left(\begin{array}{c} 1 \\\ 0\end{array}\right)\right\vert^2 \end{equation}

It is of course straightforward to calculate the Hessian of this function for N = 3, but it is not readily apparent to me that the Hessian is positive semidefinite. A Monte Carlo simulation over the range of the function does not reveal any $\mathbf{x}$ for which the Hessian has negative eigenvalues. So I believe the above N=3 function is convex. However, what I am hoping is to find a way to show that the squared norm of a similar function is convex for any N.

Thanks!

Anybody have any tips on how to show that the function $\frac{1}{2}\vert x_1 + x_2 + x_3 - x_1x_2x_3\vert^2$ is convex in $\mathbf{x}$, where $\vert x_i \vert \leq 1$?

This comes from the following expression, for general N:

\begin{equation} \frac{1}{2}\left\vert (\begin{array}{cc} 0 & 1 \end{array}) \left(\begin{array}{cc} 1 & -x_N \\\ x_N & 1 \end{array}\right) \cdots \left(\begin{array}{cc} 1 & -x_1 \\\ x_1 & 1 \end{array} \right) \left(\begin{array}{c} 1 \\\ 0\end{array}\right)\right\vert^2 \end{equation}

It is of course straightforward to calculate the Hessian of this function for N = 3, but it is not readily apparent to me that the Hessian is positive semidefinite. A Monte Carlo simulation over the range of the function does not reveal any $\mathbf{x}$ for which the Hessian has negative eigenvalues. So I believe the above N=3 function is convex. However, what I am hoping is to find a way to show that the function is convex for any N.

Thanks!

Anybody have any tips on how to show that the function $\frac{1}{2}\vert x_1 + x_2 + x_3 - x_1x_2x_3\vert^2$ is convex in $\mathbf{x}$, where $\vert x_i \vert \leq 1$?

This comes from the following expression, for general N:

\begin{equation} \frac{1}{2}\left\vert (\begin{array}{cc} 0 & 1 \end{array}) \left(\begin{array}{cc} 1 & -x_N \\ x_N & 1 \end{array}\right) \cdots \left(\begin{array}{cc} 1 & -x_1 \\ x_1 & 1 \end{array} \right) \left(\begin{array}{c} 1 \\ 0\end{array}\right)\right\vert^2 \end{equation}

It is of course straightforward to calculate the Hessian of this function for N = 3, but it is not readily apparent to me that the Hessian is positive semidefinite. A Monte Carlo simulation over the range of the function does not reveal any $\mathbf{x}$ for which the Hessian has negative eigenvalues. So I believe the above N=3 function is convex. However, what I am hoping is to find a way to show that the squared norm of a similar function is convex for any N.

Thanks!

Anybody have any tips on how to show that the function $\frac{1}{2}\vert x_1 + x_2 + x_3 - x_1x_2x_3\vert^2$ is convex in $\mathbf{x}$, where $\vert x_i \vert \leq 1$?

This comes from the following expression, for general N:

\begin{equation} \frac{1}{2}\left\vert (\begin{array}{cc} 0 & 1 \end{array}) \left(\begin{array}{cc} 1 & -x_N \\\ x_N & 1 \end{array}\right) \cdots \left(\begin{array}{cc} 1 & -x_1 \\\ x_1 & 1 \end{array} \right) \left(\begin{array}{c} 1 \\\ 0\end{array}\right)\right\vert^2 \end{equation}

It is of course straightforward to calculate the Hessian of this function for N = 3, but it is not readily apparent to me that the Hessian is positive semidefinite. A Monte Carlo simulation over the range of the function does not reveal any $\mathbf{x}$ for which the Hessian has negative eigenvalues. So I believe the above N=3 function is convex. However, what I am hoping is to find a way to show that the squared norm of a similar function is convex for any N.

Thanks!

Anybody have any tips on how to show that the function $\frac{1}{2}\vert x_1 + x_2 + x_3 - x_1x_2x_3\vert^2$ is convex in $\mathbf{x}$, where $\vert x_i \vert \leq 1$?

This comes from the following expression, for general N:

\begin{equation} \frac{1}{2}\left\vert (\begin{array}{cc} 0 & 1 \end{array}) \left(\begin{array}{cc} 1 & -x_N \\ x_N & 1 \end{array}\right) \cdots \left(\begin{array}{cc} 1 & -x_1 \\ x_1 & 1 \end{array} \right) \left(\begin{array}{c} 1 \\ 0\end{array}\right)\right\vert^2 \end{equation}

It is of course straightforward to calculate the Hessian of this function, but it is not readily apparent to me that the Hessian is positive semidefinite. A Monte Carlo simulation over the range of the function does not reveal any $\mathbf{x}$ for which the Hessian has negative eigenvalues. So I believe the above N=3 function is convex. However, what I am hoping is to find a way to show that the squared norm of a similar function is convex for any N.

Thanks!

Anybody have any tips on how to show that the function $\frac{1}{2}\vert x_1 + x_2 + x_3 - x_1x_2x_3\vert^2$ is convex in $\mathbf{x}$, where $\vert x_i \vert \leq 1$?

This comes from the following expression, for general N:

\begin{equation} \frac{1}{2}\left\vert (\begin{array}{cc} 0 & 1 \end{array}) \left(\begin{array}{cc} 1 & -x_N \\ x_N & 1 \end{array}\right) \cdots \left(\begin{array}{cc} 1 & -x_1 \\ x_1 & 1 \end{array} \right) \left(\begin{array}{c} 1 \\ 0\end{array}\right)\right\vert^2 \end{equation}

It is of course straightforward to calculate the Hessian of this function for N = 3, but it is not readily apparent to me that the Hessian is positive semidefinite. A Monte Carlo simulation over the range of the function does not reveal any $\mathbf{x}$ for which the Hessian has negative eigenvalues. So I believe the above N=3 function is convex. However, what I am hoping is to find a way to show that the squared norm of a similar function is convex for any N.

Thanks!