# 3.3 Constrained optimization  (Page 2/2)

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Example 2 Let $X={\mathbb{R}}^{2}$ ; solve

${x}_{0}=arg\underset{x={\left[{x}_{1}\phantom{\rule{3.33333pt}{0ex}}{x}_{2}\right]}^{T}}{max}f\left(x\right)={x}_{1}+{x}_{2}\phantom{\rule{3.33333pt}{0ex}}\mathrm{subjectto}\phantom{\rule{3.33333pt}{0ex}}{x}_{1}^{2}+{x}_{2}^{2}=1.$

In words, we look for the point in a unit circle in ${\mathbb{R}}^{2}$ that has the largest sum of its coordinates. It is easy to see by inspection that such point is given by ${x}_{0}={\left[\sqrt{2}\phantom{\rule{3.33333pt}{0ex}}\sqrt{2}\right]}^{T}$ . The feasible set can be given in terms of the single equality constrain

$g\left(x\right)={x}_{1}^{2}+{x}_{2}^{2}-1={x}^{T}x-1=⟨x,x⟩-1=⟨x,Ix⟩-1.$

From previous work, we know that the gradient is given by $\nabla g\left(x\right)=\left(I+{I}^{*}\right)x=2Ix=2x$ We can also write the objective function as

$f\left(x\right)=〈\left[\begin{array}{c}1\\ 1\end{array}\right],,,x〉,$

which has gradient $\nabla f\left(x\right)=\left[\begin{array}{c}1\\ 1\end{array}\right]$ . Thus, we can write the tangent space in this case as

${T}_{\Omega }\left({x}_{0}\right)=\left\{h\in X:⟨\nabla g\left({x}_{0}\right),h⟩=0\right\}=\left\{h\in X:⟨{x}_{0},h⟩=0\right\}=\left\{h:\sqrt{2}{h}_{1}+\sqrt{2}{h}_{2}=0\right\}.$

Therefore, we can write the tangent space as ${T}_{\Omega }\left({x}_{0}\right)=\left\{h\in X:{h}_{1}=-{h}_{2}\right\}$ . It is easy to see at this point that for any such $h\in {T}_{\Omega }\left({x}_{0}\right)$ we will have $⟨\nabla f\left({x}_{0}\right),h⟩=0$ , as stated by Theorem  [link] .

## Lagrangian multipliers

In this section, we will develop a method to solve optimization problems with linear objective functions and linear equality constraints.

Lemma 1 Let ${f}_{0},{f}_{1},{f}_{2},...,{f}_{n}$ be linear functionals on a Hilbert space X and suppose ${f}_{0}\left(x\right)=0$ for all $x\in X$ such that ${f}_{i}\left(x\right)=0,\phantom{\rule{3.33333pt}{0ex}}i=1,2,...,n$ . Then there exists constants ${\lambda }_{1},{\lambda }_{2},...,{\lambda }_{n}$ such that ${f}_{0}={\lambda }_{1}{f}_{1}+{\lambda }_{2}{f}_{2}+...+{\lambda }_{n}{f}_{n}$ .

Since our functionals are linear we have ${f}_{0},{f}_{1},...,{f}_{n}\in {X}^{*}$ . Define the subspace $M=\mathrm{span}\left(\left\{{f}_{1},{f}_{2},...,{f}_{n}\right\}\right)$ . Since $M$ is finite-dimensional, then $M$ is closed. We can therefore define its orthogonal complement:

${M}^{\perp }=\left\{f\in {X}^{*}:⟨f,{f}_{i}⟩=0,\phantom{\rule{3.33333pt}{0ex}}i=1,2,...,n\right\}.$

Since Hilbert spaces are self-dual, then for each function ${f}_{i}$ there exists ${w}_{i}\in X$ such that ${f}_{i}\left(x\right)=⟨x,{w}_{i}⟩$ ; therefore, we can rewrite the space above as its dual equivalent

${M}^{\perp }=\left\{w\in X:⟨w,{w}_{i}⟩=0,\phantom{\rule{3.33333pt}{0ex}}i=1,2,...,n\right\}.$

Now since $⟨w,{w}_{i}⟩={f}_{i}\left(w\right)$ , it follows that for all $w\in {M}^{\perp }$ we have that ${f}_{i}\left(w\right)=0$ , $w=1,...,n$ . Therefore, the Lemma implies that for all $w\in {M}^{\perp }$ we have that ${f}_{0}\left(w\right)=0=⟨w,{w}_{0}⟩$ . This implies that ${w}_{0}\in {\left({M}^{\perp }\right)}^{\perp }=M$ , due to $M$ being closed. Reversing to the dual space ${X}^{*}$ , this implies that ${f}_{0}\in M=\mathrm{span}\left(\left\{{f}_{1},...,{f}_{n}\right\}\right)$ , and so we can write $f={\lambda }_{1}{f}_{1}+{\lambda }_{2}{f}_{2}+...+{\lambda }_{n}{f}_{n}$ .

Theorem  [link] shows that we can apply Lemma  [link] to the constrained optimization problem. Thus, at the extremum ${x}_{0}\in X$ of the constrained program there exist constants ${\lambda }_{1},...,{\lambda }_{n}$ such that for all $h\in X$ ,

$\begin{array}{cc}\hfill \delta f\left({x}_{0};h\right)& =\sum _{i=1}^{n}{\lambda }_{i}\delta {g}_{i}\left({x}_{0};h\right),\hfill \\ \hfill ⟨\nabla f\left({x}_{0}\right),h⟩& =\sum _{i=1}^{n}{\lambda }_{i}⟨\nabla {g}_{i}\left({x}_{0}\right),h⟩,\hfill \\ \hfill 〈\nabla ,f,\left({x}_{0}\right),-,\sum _{i=1}^{n},{\lambda }_{i},\nabla ,{g}_{i},\left({x}_{0}\right),,,h〉& =0,\hfill \end{array}$

which is equivalent to $\nabla f\left({x}_{0}\right)+{\sum }_{i=1}^{n}{g}_{i}\left({x}_{0}\right)=0$ . This can be written as the gradient of the Lagrangian function

$L\left(x,\lambda \right)=f\left(x\right)+\sum _{i=1}^{n}{\lambda }_{i}{g}_{i}\left(x\right).$

Thus, we say that the extremum must provide a zero-valued directional derivative of the Lagrangian in all directions or, equivalently, a zero-valued gradient for the Lagrangian. The results obtained in this section can be collected into a proof for the following theorem.

Theorem 2 If ${x}_{0}\in X$ is an extremum of a functional $f$ subject to constraints ${\left\{{g}_{i}\right\}}_{i=1}^{n}$ , then there exist scalars ${\lambda }_{1},\cdots ,{\lambda }_{n}$ , such that the Lagrangian $L\left(x,\lambda \right)=f\left(x\right)+{\sum }_{i=1}^{n}{\lambda }_{i}{g}_{i}\left(x\right)$ is stationary at ${x}_{0}$ , i.e., for all $h\in X$ we have that $\delta L\left(x,\lambda ;h\right)=0$ , i.e., $\nabla L\left(x,\lambda \right)=0$ .

The constants ${\lambda }_{1},{\lambda }_{2},\cdots ,{\lambda }_{n}$ are known as Lagrange multipliers.

Example 3 We want to minimize $f\left(x\right)={x}_{1}^{2}+{x}_{2}^{2}$ subject to $2{x}_{1}+{x}_{2}=3$ . The constraint function is

$g\left(x\right)=2{x}_{1}+{x}_{2}-3.$

From earlier we know that $\nabla f\left(x\right)=2x,$ while we can rewrite the constraint as

$g\left(x\right)=\left[\begin{array}{c}2\\ 1\end{array}\right]x+3,$

so that $\nabla g\left(x\right)=\left[\begin{array}{c}2\\ 1\end{array}\right]$ . Therefore, the extremum's condition on the gradient of the Lagrangian results in the equation

$\begin{array}{cc}\hfill \nabla f\left(x\right)+\lambda \nabla g\left(x\right)& =0,\hfill \\ \hfill 2x+\lambda \left[\begin{array}{c}2\\ 1\end{array}\right]& =\mathbf{0},\hfill \\ \hfill \left[\begin{array}{c}2{x}_{1}+2\lambda \\ 2{x}_{2}+\lambda \end{array}\right]& =\left[\begin{array}{c}0\\ 0\end{array}\right];\hfill \end{array}$

the solution to this equation is ${x}_{1}=-\lambda ,{x}_{2}=-\frac{1}{2}\lambda$ . To solve for the value of the Lagrangian multiplier $\lambda$ , we plug the solution into the constraint: Plug in the constraint function

$2\left(-\lambda \right)+\left(-,\frac{1}{2},\lambda \right)-3=0,$

which gives $\lambda =-\frac{6}{5}$ . Therefore, we end up with the solution ${x}_{1}=\frac{6}{5},{x}_{2}=\frac{3}{5}$ .

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