problem description
solution The following diagram illustrates epi \(J\) for \(n=1\). In words, it is the set of all points above the curve of the function \(J\left ( u\right ) \)
This is an iff proof, hence we need to show the following
Proof of first direction We pick any two arbitrary points in epi \(J\), such as \(p_{0}=(u^{0},y^{0})\) and \(p_{1}=(u^{1},y^{1})\). To show epi \(J\) is a convex set, we need now to show that any point on the line between \(p_{0},p_{1}\) is also in epi \(J\). The point between them is given by \(p_{\lambda }=\left ( u^{\lambda },y^{\lambda }\right ) \) where \(\lambda \in \left [ 0,1\right ] \). The following diagram helps illustrates this for \(n=1\).
The point \(p_{\lambda }\) is given by\begin{align*} \left ( u^{\lambda },y^{\lambda }\right ) & =\left ( 1-\lambda \right ) p_{0}+\lambda p_{1}\\ & =\left ( 1-\lambda \right ) \left ( u^{0},y^{0}\right ) +\lambda \left ( u^{1},y^{1}\right ) \\ & =\left ( \left ( 1-\lambda \right ) u^{0}+\lambda u^{1},\left ( 1-\lambda \right ) y^{0}+\lambda y^{1}\right ) \end{align*}
Therefore \(y^{\lambda }=\left ( 1-\lambda \right ) y^{0}+\lambda y^{1}\). Since \(p_{0},p_{1}\) are in epi \(J\), then by the definition of epi \(J\), we know that \(y^{0}\geq J\left ( u^{0}\right ) \) and \(y^{1}\geq J\left ( u^{1}\right ) \). Therefore we conclude that\begin{equation} y^{\lambda }\geq \left ( 1-\lambda \right ) J\left ( u^{0}\right ) +\lambda J\left ( u^{1}\right ) \tag{1} \end{equation} But since we assumed \(J\) is a convex function, then we also know that \(\left ( 1-\lambda \right ) J\left ( u^{0}\right ) +\lambda J\left ( u^{1}\right ) \geq J\left ( u^{\lambda }\right ) \) where \(u^{\lambda }=\left ( 1-\lambda \right ) u^{0}+\lambda u^{1}\). Therefore (1) becomes\[ y^{\lambda }\geq J\left ( u^{\lambda }\right ) \] This implies the arbitrary point \(p_{\lambda }\) is in epi \(J\).
We now need to proof the other direction. Given that epi \(J\) is a convex set, then show that \(J\) is a convex function. Since epi \(J\) is a convex set, we pick two arbitrary points in epi \(J\), such as \(p_{0},p_{1}\). We can choose \(p_{0}=\left ( u^{0},J\left ( u^{0}\right ) \right ) \) and \(p_{1}=\left ( u^{1},J\left ( u^{1}\right ) \right ) \). These are still in epi \(J\), but on the lower bound, on the edge with \(J\left ( u\right ) \) curve.
Since \(p_{0},p_{1}\) are two points in a convex set, then any point \(p^{\lambda }\) on a line between them is also in epi \(J\) (by definition of a convex set). And since \(p^{\lambda }=\left ( 1-\lambda \right ) p_{0}+\lambda p_{1}\) this implies\begin{align} p^{\lambda } & =\left ( u^{\lambda },y^{\lambda }\right ) \nonumber \\ & =\left ( \left ( 1-\lambda \right ) p_{0}+\lambda p_{1}\right ) \nonumber \\ & =\left ( \left ( 1-\lambda \right ) \left ( u^{0},J\left ( u^{0}\right ) \right ) +\lambda \left ( u^{1},J\left ( u^{1}\right ) \right ) \right ) \nonumber \\ & =\left ( \left ( 1-\lambda \right ) u^{0}+\lambda u^{1},\overset{y^{\lambda }}{\overbrace{\left ( 1-\lambda \right ) J\left ( u^{0}\right ) +J\left ( u^{1}\right ) }}\right ) \tag{1} \end{align}
Since \(p^{\lambda }\) is in epi \(J\) then by definition of epi \(J\) \begin{equation} y^{\lambda }\geq J\left ( u^{\lambda }\right ) \tag{2} \end{equation} But from (1) we see that \(y^{\lambda }=\left ( 1-\lambda \right ) J\left ( u^{0}\right ) +J\left ( u^{1}\right ) \), therefore (2) is the same as writing\begin{equation} \left ( 1-\lambda \right ) J\left ( u^{0}\right ) +J\left ( u^{1}\right ) \geq J\left ( u^{\lambda }\right ) \tag{3} \end{equation} But \(u^{\lambda }=\left ( 1-\lambda \right ) u^{0}+\lambda u^{1}\), hence the above becomes\[ \left ( 1-\lambda \right ) J\left ( u^{0}\right ) +J\left ( u^{1}\right ) \geq J\left ( \left ( 1-\lambda \right ) u^{0}+\lambda u^{1}\right ) \] But the above is the definition of a convex function. Therefore \(J\left ( u\right ) \) is a convex function. QED.
problem description
solution Let \(u_{0}^{\ast }\) and \(u_{1}^{\ast }\) be any two different minimizing elements in \(\Re ^{n}\) such that \(J\left ( u_{0}^{\ast }\right ) <J\left ( u_{1}^{\ast }\right ) \). We will show that this leads to contradiction. Since \(u_{0}^{\ast }\) is a minimizer, then there exists some \(R>0\), such that all points \(u\) that satisfy \(\left \Vert u^{\ast }-u\right \Vert \leq R\) also satisfy \[ J\left ( u_{0}^{\ast }\right ) \leq J\left ( u\right ) \]
We will consider all points along the line joining \(u_{0}^{\ast },u_{1}^{\ast }\), and pick one point \(u^{\lambda }\) that satisfies \(\left \Vert u^{\ast }-u^{\lambda }\right \Vert \leq R\), where \(\lambda \in \left [ 0,1\right ] \) is selected to make the convex mixture \(u^{\lambda }=\left ( 1-\lambda \right ) u_{0}^{\ast }+\lambda u_{1}^{\ast }\) satisfied. Therefore any \(\lambda \leq \frac{R}{\left \Vert u_{0}^{\ast }-u_{1}^{\ast }\right \Vert }\) will put \(u^{\lambda }\) inside the sphere of radius \(R\).
Hence now we can say that\begin{equation} J\left ( u_{0}^{\ast }\right ) \leq J\left ( u^{\lambda }\right ) \tag{1} \end{equation} But given that \(J\left ( u\right ) \) is a strict convex function, then\begin{equation} J( u^{\lambda }) <\left ( 1-\lambda \right ) J\left ( u_{0}^{\ast }\right ) +\lambda J\left ( u_{1}^{\ast }\right ) \tag{2} \end{equation} Since we assumed that \(J\left ( u_{0}^{\ast }\right ) <J\left ( u_{1}^{\ast }\right ) \), then if we replace \(J\left ( u_{1}^{\ast }\right ) \) by \(J\left ( u_{0}^{\ast }\right ) \) in the RHS of (2), it will change from \(<\) to \(\leq \) resulting in\begin{align} J ( u^{\lambda } ) & \leq \left ( 1-\lambda \right ) J\left ( u_{0}^{\ast }\right ) +\lambda J\left ( u_{0}^{\ast }\right ) \nonumber \\ J ( u^{\lambda } ) & \leq J\left ( u_{0}^{\ast }\right ) \tag{3} \end{align}
We see that equations (3) and (1) are a contradiction. Therefore our assumption is wrong and there can not be more than one minimizing element and \(u_{0}^{\ast }\) must be the same as \(u_{1}^{\ast }\).
problem description
solution We are given that \(J\left ( u^{\ast }\right ) \leq J\left ( u\right ) \) for all \(u\) such that \(\left \Vert u^{\ast }-u\right \Vert <\delta \). Let us pick any arbitrary point \(u^{1}\), outside ball of radius \(\delta \). Then any point on the line between \(u^{\ast }\) and \(u^{1}\) is given by\[ u^{\lambda }=\left ( 1-\lambda \right ) u^{\ast }+\lambda u^{1}\] In picture, so far we have this setup
We now need to show that \(J\left ( u^{\ast }\right ) \leq J\left ( u^{1}\right ) \) even though \(u^{1}\) is outside the ball. Since \(J\) is a convex function, then\begin{equation} J\left ( u^{\lambda }\right ) \leq \left ( 1-\lambda \right ) J\left ( u^{\ast }\right ) +\lambda J\left ( u^{1}\right ) \tag{1} \end{equation} We can now select \(\lambda \) to push \(u^{\lambda }\) to be inside the ball. We are free to change \(\lambda \) as we want while keeping \(u^{1}\) fixed, outside the ball. If we do this we then we have\[ J\left ( u^{\ast }\right ) \leq J\left ( u^{\lambda }\right ) \]
Hence (1) becomes\begin{equation} J\left ( u^{\ast }\right ) \leq \left ( 1-\lambda \right ) J\left ( u^{\ast }\right ) +\lambda J\left ( u^{1}\right ) \tag{2} \end{equation} Where we replaced \(J\left ( u^{\lambda }\right ) \) by \(J\left ( u^{\ast }\right ) \) in (1) and since \(J\left ( u^{\ast }\right ) \leq J\left ( u^{\lambda }\right ) \) the \(\leq \) relation remained valid. Simplifying (2) gives\begin{align*} J\left ( u^{\ast }\right ) & \leq J\left ( u^{\ast }\right ) -\lambda J\left ( u^{\ast }\right ) +\lambda J\left ( u^{1}\right ) \\ \lambda J\left ( u^{\ast }\right ) & \leq \lambda J\left ( u^{1}\right ) \end{align*}
For non-zero \(\lambda \) this means \(J\left ( u^{\ast }\right ) \leq J\left ( u^{1}\right ) \). This completes the proof, since \(u^{1}\) was arbitrary point anywhere. Hence \(u^{\ast }\) is global minimum. QED
problem description
solution
We need to show that \(J\left ({\displaystyle \sum \limits _{i=1}^{N}} \lambda _{i}u^{i}\right ) \leq{\displaystyle \sum \limits _{i=1}^{N}} \lambda _{i}J\left ( u^{i}\right ) \) where \({\displaystyle \sum \limits _{i=1}^{N}} \lambda _{i}=1\). Proof by induction. For \(N=1\) and since \(\lambda _{1}=1\), then we have\[ J\left ( u^{1}\right ) =J\left ( u^{1}\right ) \] The case for \(N=2\) comes for free, from the definition of \(J\) being a convex function\begin{equation} J\left ( \left ( 1-\lambda \right ) u^{1}+\lambda u^{2}\right ) \leq \left ( 1-\lambda \right ) J\left ( u^{1}\right ) +\lambda J\left ( u^{2}\right ) \tag{A} \end{equation} By making \(\left ( 1-\lambda \right ) \equiv \lambda _{1},\lambda \equiv \lambda _{2}\), the above can be written as\[ J\left ( \lambda _{1}u^{1}+\lambda _{2}u^{2}\right ) \leq \lambda _{1}J\left ( u^{1}\right ) +\lambda _{2}J\left ( u^{2}\right ) \] We now assume it is true for \(N=k-1\). In other words, the inductive hypothesis below is given as true\begin{equation} J\left ({\displaystyle \sum \limits _{i=1}^{k-1}} \lambda _{i}u^{i}\right ) \leq{\displaystyle \sum \limits _{i=1}^{k-1}} \lambda _{i}J\left ( u^{i}\right ) \tag{*} \end{equation} Now we have to show it will also be true for \(N=k\), which is\begin{align}{\displaystyle \sum \limits _{i=1}^{k}} \lambda _{i}J\left ( u^{i}\right ) & =\lambda _{1}J\left ( u^{1}\right ) +\lambda _{1}J\left ( u^{1}\right ) +\cdots +\lambda _{k}J\left ( u^{k}\right ) \nonumber \\ & =\left ( 1-\lambda _{k}\right ) \left ( \frac{\lambda _{1}}{\left ( 1-\lambda _{k}\right ) }J\left ( u^{1}\right ) +\frac{\lambda _{1}}{\left ( 1-\lambda _{k}\right ) }J\left ( u^{1}\right ) +\cdots +\frac{\lambda _{k-1}}{\left ( 1-\lambda _{k}\right ) }J\left ( u^{k-1}\right ) +\frac{\lambda _{k}}{\left ( 1-\lambda _{k}\right ) }J\left ( u^{k}\right ) \right ) \nonumber \\ & =\left ( 1-\lambda _{k}\right ) \left ( \frac{\lambda _{1}}{\left ( 1-\lambda _{k}\right ) }J\left ( u^{1}\right ) +\frac{\lambda _{1}}{\left ( 1-\lambda _{k}\right ) }J\left ( u^{1}\right ) +\cdots +\frac{\lambda _{k-1}}{\left ( 1-\lambda _{k}\right ) }J\left ( u^{k-1}\right ) \right ) +\lambda _{k}J\left ( u^{k}\right ) \nonumber \\ & =\left ( 1-\lambda _{k}\right ) \left ({\displaystyle \sum \limits _{i=1}^{k-1}} \frac{\lambda _{i}}{\left ( 1-\lambda _{k}\right ) }J\left ( u^{i}\right ) \right ) +\lambda _{k}J\left ( u^{k}\right ) \tag{1} \end{align}
Now we take advantage of the inductive hypothesis Eq. (*) on \(k-1\), which says that \(J\left ({\displaystyle \sum \limits _{i=1}^{k-1}} \frac{\lambda _{i}}{\left ( 1-\lambda _{k}\right ) }u^{i}\right ) \leq \) \({\displaystyle \sum \limits _{i=1}^{k-1}} \frac{\lambda _{i}}{\left ( 1-\lambda _{k}\right ) }J\left ( u^{i}\right ) \). Using this in (1) changes it to \(\geq \) relation\begin{equation}{\displaystyle \sum \limits _{i=1}^{k}} \lambda _{i}J\left ( u^{i}\right ) \geq \left ( 1-\lambda _{k}\right ) J\left ({\displaystyle \sum \limits _{i=1}^{k-1}} \frac{\lambda _{i}}{\left ( 1-\lambda _{k}\right ) }u^{i}\right ) +\lambda _{k}J\left ( u^{k}\right ) \tag{2} \end{equation} We now take advantage of the case of \(N=2\) in (A) by viewing RHS of (2) as \(\left ( 1-\lambda _{k}\right ) J\left ( u^{1}\right ) +\lambda _{k}J\left ( u^{2}\right ) \), where we let \(u^{1}\equiv{\displaystyle \sum \limits _{i=1}^{k-1}} \frac{\lambda _{i}}{\left ( 1-\lambda _{k}\right ) }u^{i},u^{2}\equiv u^{k}\). Hence we conclude that\begin{equation} \left ( 1-\lambda _{k}\right ) J\left ({\displaystyle \sum \limits _{i=1}^{k-1}} \frac{\lambda _{i}}{\left ( 1-\lambda _{k}\right ) }u^{i}\right ) +\lambda _{k}J\left ( u^{k}\right ) \geq J\left ( \left ( 1-\lambda _{k}\right ){\displaystyle \sum \limits _{i=1}^{k-1}} \frac{\lambda _{i}}{\left ( 1-\lambda _{k}\right ) }u^{i}+\lambda _{k}u^{k}\right ) \tag{3} \end{equation} Using (3) in (2) gives (the \(\geq \) relation remains valid, even more now, since we replaced something in RHS of (2), by something smaller) \begin{align*}{\displaystyle \sum \limits _{i=1}^{k}} \lambda _{i}J\left ( u^{i}\right ) & \geq J\left ( \left ( 1-\lambda _{k}\right ){\displaystyle \sum \limits _{i=1}^{k-1}} \frac{\lambda _{i}}{\left ( 1-\lambda _{k}\right ) }u^{i}+\lambda _{k}u^{k}\right ) \\ & =J\left ( \left ({\displaystyle \sum \limits _{i=1}^{k-1}} \lambda _{i}u^{i}\right ) +\lambda _{k}u^{k}\right ) \end{align*}
Hence \[{\displaystyle \sum \limits _{i=1}^{k}} \lambda _{i}J\left ( u^{i}\right ) \geq J\left ({\displaystyle \sum \limits _{i=1}^{k}} \lambda _{i}u^{i}\right ) \] QED.
problem description
solution
Assuming \(J\left ( u\right ) \) is twice continuously differentiable (\(C^{2}\)) in \(u_{1},u_{2},\cdots ,u_{n}\), then if we can show that the Hessian \(\nabla ^{2}J\left ( u\right ) \) is positive semi-definite on \(u_{i}>0\), then this implies \(J\left ( u\right ) \) is convex. The first step is to determined \(\nabla ^{2}J\left ( u\right ) \). \begin{align*} \frac{\partial J}{\partial u_{i}} & =-\frac{1}{n}\left ( u_{1}u_{2}\cdots u_{n}\right ) ^{\frac{1}{n}-1}{\displaystyle \prod \limits _{k=1,k\neq i}^{n}} u_{k}=\frac{1}{n}\frac{J\left ( u\right ) }{\left ( u_{1}u_{2}\cdots u_{n}\right ) }{\displaystyle \prod \limits _{k=1,k\neq i}^{n}} u_{k}=\frac{1}{n}\frac{J\left ( u\right ) }{{\displaystyle \prod \limits _{k=1}^{n}} u_{k}}{\displaystyle \prod \limits _{k=1,k\neq i}^{n}} u_{k}\\ & =\frac{1}{n}\frac{J\left ( u\right ) }{u_{i}} \end{align*}
And\begin{align*} \frac{\partial ^{2}J}{\partial u_{i}^{2}} & =\frac{1}{n}\frac{\left ( \frac{1}{n}\frac{J\left ( u\right ) }{u_{i}}\right ) }{u_{i}}-\frac{1}{n}\frac{J\left ( u\right ) }{u_{i}^{2}}\\ & =\frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{i}^{2}}-\frac{1}{n}\frac{J\left ( u\right ) }{u_{i}^{2}}\\ & =\frac{1}{n}\frac{J\left ( u\right ) }{u_{i}^{2}}\left ( \frac{1}{n}-1\right ) \end{align*}
And the cross derivatives are\begin{align*} \frac{\partial ^{2}J}{\partial u_{i}\partial u_{j}} & =\frac{\partial }{\partial u_{j}}\left ( \frac{1}{n}\frac{J\left ( u\right ) }{u_{i}}\right ) \\ & =\frac{1}{n}\frac{\frac{1}{n}\frac{J\left ( u\right ) }{u_{j}}}{u_{i}}\\ & =\frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{i}u_{j}} \end{align*}
Therefore \[ \boxed{ \nabla ^{2}J\left ( u\right ) =\begin{pmatrix} \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{1}^{2}}\left ( 1-n\right ) & \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{1}u_{2}} & \cdots & \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{1}u_{n}}\\ \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{2}u_{1}} & \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{2}^{2}}\left ( 1-n\right ) & \cdots & \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{2}u_{n}}\\ \vdots & \cdots & \ddots & \vdots \\ \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{n}u_{1}} & \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{n}u_{2}} & \cdots & \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{n}^{2}}\left ( 1-n\right ) \end{pmatrix} } \] Now we need to show that \(\nabla ^{2}J\left ( u\right ) \) is positive semi-definite. For \(n=1\), the above reduces to\[ \nabla ^{2}J\left ( u\right ) =\frac{J\left ( u\right ) }{u_{1}^{2}}\left ( 1-1\right ) =0 \] Hence non-negative. This is the same as saying the second derivative is zero. For \(n=2\)\[ \nabla ^{2}J\left ( u\right ) =\begin{pmatrix} \frac{1}{4}J\left ( u\right ) \frac{1-2}{u_{1}^{2}} & \frac{1}{u_{1}u_{2}}\frac{1}{4}J\left ( u\right ) \\ \frac{1}{u_{2}u_{1}}\frac{1}{4}J\left ( u\right ) & \frac{1}{4}J\left ( u\right ) \frac{1-2}{u_{2}^{2}}\end{pmatrix} =\begin{pmatrix} \frac{-1}{u_{1}^{2}}\frac{1}{4}J\left ( u\right ) & \frac{1}{u_{1}u_{2}}\frac{1}{4}J\left ( u\right ) \\ \frac{1}{u_{2}u_{1}}\frac{1}{4}J\left ( u\right ) & \frac{-1}{u_{2}^{2}}\frac{1}{4}J\left ( u\right ) \end{pmatrix} \] The first leading minor is \(\frac{-1}{4u_{1}^{2}}J\left ( u\right ) \), which is positive, since \(J\left ( u\right ) <0\) and \(u_{i}>0\) (given). The second leading minor is \[ \Delta _{2}=\begin{vmatrix} \frac{-1}{u_{1}^{2}}\frac{1}{4}J\left ( u\right ) & \frac{1}{u_{1}u_{2}}\frac{1}{4}J\left ( u\right ) \\ \frac{1}{u_{2}u_{1}}\frac{1}{4}J\left ( u\right ) & \frac{-1}{u_{2}^{2}}\frac{1}{4}J\left ( u\right ) \end{vmatrix} =0 \] Hence all the leasing minors are non-negative. Which means \(\nabla ^{2}J\left ( u\right ) \) is semi-definite. We will look at \(n=3\)\[ \nabla ^{2}J\left ( u\right ) =\begin{pmatrix} \frac{-2}{u_{1}^{2}}\frac{1}{9}J\left ( u\right ) & \frac{1}{u_{1}u_{2}}\frac{1}{9}J\left ( u\right ) & \frac{1}{u_{1}u_{3}}\frac{1}{9}J\left ( u\right ) \\ \frac{1}{u_{2}u_{1}}\frac{1}{9}J\left ( u\right ) & \frac{-2}{u_{2}^{2}}\frac{1}{9}J\left ( u\right ) & \frac{1}{u_{2}u_{3}}\frac{1}{9}J\left ( u\right ) \\ \frac{1}{u_{3}u_{1}}\frac{1}{9}J\left ( u\right ) & \frac{1}{u_{3}u_{2}}\frac{1}{9}J\left ( u\right ) & \frac{-2}{u_{3}^{2}}\frac{1}{9}J\left ( u\right ) \end{pmatrix} \] The first leading minor is \(\frac{-2}{9u_{1}^{2}}J\left ( u\right ) \), which is positive again, since \(J\left ( u\right ) <0\) for \(u_{i}>0\) (given). And the second leading minor is \(\frac{1}{27}J^{2}\frac{u^{2}}{u_{1}^{2}u_{2}^{2}}\)
which is positive, since all terms are positive. The third leading minor is \[ \Delta _{3}=\begin{vmatrix} \frac{-2}{u_{1}^{2}}\frac{1}{9}J\left ( u\right ) & \frac{1}{u_{1}u_{2}}\frac{1}{9}J\left ( u\right ) & \frac{1}{u_{1}u_{3}}\frac{1}{9}J\left ( u\right ) \\ \frac{1}{u_{2}u_{1}}\frac{1}{9}J\left ( u\right ) & \frac{-2}{u_{2}^{2}}\frac{1}{9}J\left ( u\right ) & \frac{1}{u_{2}u_{3}}\frac{1}{9}J\left ( u\right ) \\ \frac{1}{u_{3}u_{1}}\frac{1}{9}J\left ( u\right ) & \frac{1}{u_{3}u_{2}}\frac{1}{9}J\left ( u\right ) & \frac{-2}{u_{3}^{2}}\frac{1}{9}J\left ( u\right ) \end{vmatrix} =0 \] Hence non-of the leading minors are negative. Therefore \(\nabla ^{2}J\left ( u\right ) \) is semi-definite. The same pattern repeats for higher values of \(n.\) All leading minors are positive, except the last leading minor will be zero.
Another way to show that \(\nabla ^{2}J\left ( u\right ) \) is positive semi-definite is to show that \(x^{T}\left ( \nabla ^{2}J\left ( u\right ) \right ) x\geq 0\) for any vector \(x.\) (since \(\nabla ^{2}J\left ( u\right ) \) is symmetric).\[ x^{T}\left ( \nabla ^{2}J\left ( u\right ) \right ) x=\begin{pmatrix} x_{1} & x_{2} & \cdots & x_{n}\end{pmatrix}\begin{pmatrix} \frac{1}{n}\frac{J\left ( u\right ) }{u_{1}^{2}}\left ( \frac{1}{n}-1\right ) & \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{1}u_{2}} & \cdots & \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{1}u_{n}}\\ \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{2}u_{1}} & \frac{1}{n}\frac{J\left ( u\right ) }{u_{2}^{2}}\left ( \frac{1}{n}-1\right ) & \cdots & \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{2}u_{n}}\\ \vdots & \cdots & \ddots & \vdots \\ \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{n}u_{1}} & \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{n}u_{2}} & \cdots & \frac{1}{n}\frac{J\left ( u\right ) }{u_{n}^{2}}\left ( \frac{1}{n}-1\right ) \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ \vdots \\ x_{n}\end{pmatrix} \] Now the idea is to set \(n=1,2,3,\cdots \) and show that the resulting values \(\geq 0\) always. For \(n=1\), we obtain \(0\) as before. For \(n=2\), let
\(\Delta =\begin{pmatrix} x_{1} & x_{2}\end{pmatrix}\begin{pmatrix} \frac{1}{n}\frac{J\left ( u\right ) }{u_{1}^{2}}\left ( \frac{1}{n}-1\right ) & \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{1}u_{2}}\\ \frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{2}u_{1}} & \frac{1}{n}\frac{J\left ( u\right ) }{u_{2}^{2}}\left ( \frac{1}{n}-1\right ) \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} \). Expanding gives\begin{align*} \Delta & =\begin{pmatrix} x_{1}\frac{1}{n}\frac{J\left ( u\right ) }{u_{1}^{2}}\left ( \frac{1}{n}-1\right ) +x_{2}\frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{2}u_{1}} & x_{1}\frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{1}u_{2}}+x_{2}\frac{1}{n}\frac{J\left ( u\right ) }{u_{2}^{2}}\left ( \frac{1}{n}-1\right ) \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\end{pmatrix} \\ & =x_{1}^{2}\frac{1}{n}\frac{J\left ( u\right ) }{u_{1}^{2}}\left ( \frac{1}{n}-1\right ) +x_{1}x_{2}\frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{2}u_{1}}+x_{2}x_{1}\frac{1}{n^{2}}\frac{J\left ( u\right ) }{u_{1}u_{2}}+x_{2}^{2}\frac{1}{n}\frac{J\left ( u\right ) }{u_{2}^{2}}\left ( \frac{1}{n}-1\right ) \\ & =x_{1}^{2}\frac{1}{2}\frac{J\left ( u\right ) }{u_{1}^{2}}\left ( \frac{1}{2}-1\right ) +x_{1}x_{2}\frac{1}{4}\frac{J\left ( u\right ) }{u_{2}u_{1}}+x_{2}x_{1}\frac{1}{4}\frac{J\left ( u\right ) }{u_{1}u_{2}}+x_{2}^{2}\frac{1}{2}\frac{J\left ( u\right ) }{u_{2}^{2}}\left ( \frac{1}{2}-1\right ) \end{align*}
The RHS above becomes, and by replacing \(J\left ( u\right ) =-\sqrt{u_{1}u_{2}}\) for \(n=2\)\begin{align*} -\frac{1}{4}x_{1}^{2}\frac{J\left ( u\right ) }{u_{1}^{2}}+x_{1}x_{2}\frac{1}{2}\frac{J\left ( u\right ) }{u_{2}u_{1}}-\frac{1}{4}x_{2}^{2}\frac{J\left ( u\right ) }{u_{2}^{2}} & =\frac{1}{4}x_{1}^{2}\frac{\sqrt{u_{1}u_{2}}}{u_{1}^{2}}-x_{1}x_{2}\frac{1}{2}\frac{\sqrt{u_{1}u_{2}}}{u_{2}u_{1}}+\frac{1}{4}x_{2}^{2}\frac{\sqrt{u_{1}u_{2}}}{u_{2}^{2}}\\ & =\left ( \frac{1}{\sqrt{4}}\frac{\left ( u_{1}u_{2}\right ) ^{\frac{1}{4}}}{u_{1}}x_{1}-\frac{1}{\sqrt{4}}\frac{\left ( u_{1}u_{2}\right ) ^{\frac{1}{4}}}{u_{2}}x_{2}\right ) ^{2} \end{align*}
Where we completed the square in the last step above. Hence \(x^{T}\left ( \nabla ^{2}J\left ( u\right ) \right ) x\geq 0\). The same process can be continued for \(n\) higher. Hence \(\nabla ^{2}J\left ( u\right ) \) is positive semi-definite.
