Please explain the intuition behind the dual problem in optimization. I've studied convex optimization pretty carefully, but don't feel that I have yet &quot;grokked&quot; the dual problem.  Here are some questions I would like to understand more deeply/clearly/simply:

How would somebody think of the dual problem?  What thought process would lead someone to consider the dual problem and to recognize that it's valuable/interesting?
In the case of a convex optimization problem, is there any obvious reason to expect that strong duality should (usually) hold?
It often happens that the dual of the dual problem is the primal problem.  However, this seems like a complete surprise to me.  Is there any intuitive reason to expect that this should happen?
Does the use of the word &quot;dual&quot; or &quot;duality&quot; in optimization have anything to do with the dual space in linear algebra?  Or are they just different concepts that go by the same name.  What about the use of the word &quot;dual&quot; in projective geometry — is there a connection there?
You can define the dual problem and prove theorems about strong duality without ever mentioning the Fenchel conjugate.  For example, Boyd and Vandenberghe prove a strong duality theorem without mentioning the Fenchel conjugate in their proof.  And yet, people often talk as if the Fenchel conjugate is somehow the &quot;essence&quot; of duality, and make it sound as if the whole theory of duality is based on the Fenchel conjugate.  Why is the Fenchel conjugate considered to have such fundamental importance?

Note: I will now describe my current level of understanding of the intuition behind the dual problem.  Please tell me if you think I might be missing any basic insights.
I have read the excellent notes about convex optimization by Guilherme Freitas, and in particular the part about &quot;penalty intuition&quot;.  When we are trying to solve
\begin{align*}
\text{minimize} &amp;\quad f(x) \\
\text{such that} &amp; \quad h(x) \leq 0
\end{align*}
one might try to eliminate the constraints by introducing a penalty when constraints are violated.  This gives us the new unconstrained problem
\begin{equation}
\text{minimize} \quad f(x) + \langle \lambda ,h(x) \rangle
\end{equation}
where $\lambda \geq 0$. It's not hard to see that for a given $\lambda \geq 0$, the optimal value of this unconstrained problem is less than or equal to the optimal value for the constrained problem.  This gives us a new problem — find $\lambda$ so that the optimal value for the unconstrained problem is as large as possible.  That is one way to imagine how somebody might have thought of the dual problem.  Is this the best intuition for where the dual problem comes from?
Another viewpoint: the KKT conditions can be derived using what Freitas calls the &quot;geometric intuition&quot;.  Then, if we knew the value of the multipliers $\lambda$, it would be (often) much easier to find $x$.  So, a new problem is to find $\lambda$.  And if we can somehow recognize that $\lambda$ is a maximizer for the dual problem, then this suggests that we might try solving the dual problem.
Please explain or give references to any intuition that you think I might find interesting, even if it's not directly related to what I asked.
Here's what's really going on with the dual problem.  (This is my attempt to answer my own question, over a year after originally asking it.)

(A very nice presentation of this material is given in Ekeland and Temam.  These ideas are also in Rockafellar.)

Let $V$ be a finite dimensional normed vector space over $\mathbb R$.  (Working in an inner product space or just in $\mathbb R^N$ risks concealing the fundamental role that the dual space plays in duality in convex optimization.)

The basic idea behind duality in convex analysis is to think of a convex set in terms of its supporting hyperplanes.  (A closed
convex set $\Omega$ can be "recovered" from its supporting hyperplanes
by taking the intersection of all closed half spaces containing $\Omega$.
The set of all supporting hyperplanes to $\Omega$ is sort of a
"dual representation" of $\Omega$.)

For a convex function $f$ (whose epigraph is a convex set), this strategy leads
us to think about $f$ in terms of affine functions $\langle m^*, x \rangle - \alpha$
which are majorized by $f$.  (Here $m^* \in V^*$ and we are using the notation $\langle m^*, x \rangle = m^*(x)$.)

For a given slope $m^* \in V^*$, we only need to consider the "best" choice of $\alpha$ -- the other affine minorants with slope $m^*$ can be disregarded.
\begin{align*}
&amp; f(x) \geq \langle m^*,x \rangle - \alpha \quad \forall x \in V \\
\iff &amp; \alpha \geq  \langle m^*, x \rangle - f(x) \quad \forall x \in V \\
\iff &amp; \alpha \geq \sup_{x \in V} \quad \langle m^*, x \rangle - f(x)
\end{align*}
so the best choice of $\alpha$ is
\begin{equation}
f^*(m^*) = \sup_{x \in V} \quad \langle m^*, x \rangle - f(x).
\end{equation}
If this supremum is finite, then
$\langle m^*,x \rangle - f^*(m^*)$ is the best affine minorant of
$f$ with slope $m^*$.
If $f^*(m^*) = \infty$, then there is no affine minorant of $f$ with slope $m^*$.

The function $f^*$ is called the "conjugate" of $f$.  The definition and basic facts about $f^*$ are all highly intuitive.  For example, if $f$ is a proper closed convex function then $f$ can be recovered from $f^*$, because any closed convex set (in this case the epigraph of $f$) is the intersection of all the closed half spaces containing it.  (I still think the fact that the "inversion formula" $f = f^{**}$ is so simple is a surprising and mathematically beautiful fact, but not hard to derive or prove with this intuition.)

Because $f^*$ is defined on the dual space, we see already the fundamental role played by the dual space in duality in convex optimization.

Given an optimization problem, we don't obtain a dual problem until we specify how to perturb the optimization problem.  This is why equivalent formulations of an optimization problem can lead to different dual problems.  By reformulating it we have in fact specified a different way to perturb it.

As is typical in math, the ideas become clear when we work at an appropriate level of generality.  Assume that our optimization problem is
\begin{equation*}
\operatorname*{minimize}_{x}  \quad \phi(x,0).
\end{equation*}
Here $\phi:X \times Y \to \bar{\mathbb R}$ is convex.  Standard convex optimization problems can be written in this form with an appropriate choice of $\phi$.  The perturbed problems are
\begin{equation*}
\operatorname*{minimize}_{x}  \quad \phi(x,y)
\end{equation*}
for nonzero values of $y \in Y$.

Let $h(y) = \inf_x \phi(x,y)$.  Our optimization problem is simply to evaluate $h(0)$.

From our knowledge of conjugate functions, we know that
\begin{equation*}
h(0) \geq h^{**}(0)
\end{equation*}
and that typically we have equality.  For example, if $h$ is subdifferentiable at $0$ (which is typical for a convex function) then $h(0) = h^{**}(0)$.

The dual problem is simply to evaluate $h^{**}(0)$.

In other words, the dual problem is:
\begin{equation*}
\operatorname*{maximize}_{y^* \in Y^*} \quad - h^*(y^*).
\end{equation*}
We see again the fundamental role that the dual space plays here.

It is enlightening to express the dual problem in terms of $\phi$.  It's easy to show that the dual problem is
\begin{equation*}
\operatorname*{maximize}_{y^* \in Y^*} \quad - \phi^*(0,y^*).
\end{equation*}

So the primal problem is
\begin{equation*}
\operatorname*{minimize}_{x \in X} \quad \phi(x,0)
\end{equation*}
and the dual problem (slightly restated) is
\begin{equation*}
\operatorname*{minimize}_{y^* \in Y^*} \quad \phi^*(0,y^*).
\end{equation*}
The similarity between these two problems is mathematically beautiful, and we can see that if we perturb the dual problem in the obvious way, then the dual of the dual problem will be the primal problem (assuming $\phi = \phi^{**}$).  The natural isomorphism between $V$ and $V^{**}$ is of fundamental importance here.

The key facts about the dual problem -- strong duality, the optimality conditions, and the sensitivity interpretation of the optimal dual variables -- all become intuitively clear and even "obvious" from this viewpoint.

An optimization problem in the form
\begin{align*}
\operatorname*{minimize}_x &amp; \quad f(x) \\
\text{subject to} &amp; \quad g(x) \leq 0,
\end{align*}
can be perturbed as follows:
\begin{align*}
\operatorname*{minimize}_x &amp; \quad f(x) \\
\text{subject to} &amp; \quad g(x) + y \leq 0.
\end{align*}

This perturbed problem has the form given above with
\begin{equation*}
\phi(x,y) = 
\begin{cases}
f(x) \quad \text{if } g(x) + y \leq 0 \\
\infty \quad \text{otherwise}.
\end{cases}
\end{equation*}
To find the dual problem, we need to evaluate $-\phi^*(0,y^*)$, which is a relatively straightforward calculation.
\begin{align*}
-\phi^*(0,y^*) &amp;= -\sup_{g(x) + y \leq 0} \quad \langle y^*,y \rangle - f(x) \\
&amp;= -\sup_{\substack{ x \\ q \geq 0 }} \quad \langle y^*, -g(x) - q \rangle - f(x) \\
&amp;= \inf_{\substack{ x \\ q \geq 0 }} \quad f(x) + \langle y^*, g(x) \rangle + \langle y^*, q \rangle.
\end{align*}
We can minimize first with respect to $q$, and we will get $-\infty$ unless $\langle y^*, q \rangle \geq 0$ for all $q \geq 0$.  In other words, we will get $-\infty$ unless $y^* \geq 0$.

The dual function is
\begin{equation*}
-\phi^*(0,y^*) =
\begin{cases}
\inf_x \quad f(x) + \langle y^*, g(x) \rangle \quad \text{if } y^* \geq 0 \\
-\infty \quad \text{otherwise}.
\end{cases}
\end{equation*}

This is the expected result.
Here are some counter-examples to help you understand KKT conditions and strong duality. The answer is from my other post: https://math.stackexchange.com/a/4154563/273731
${\bf counter-example 1}$
If one drops the convexity condition on objective function, then strong duality could fails even with relative interior condition.
The counter-example is the same as the following one.
${\bf counter-example 2}$
For non-convex problem where strong duality does not hold, primal-dual optimal pairs may not satisfy KKT condition.
Consider the optimization problem
\begin{align}
 \operatorname{minimize} &amp; \quad e^{-x_1x_2} \\
 \text{subject to} &amp; \quad x_1\le 0.
\end{align}
The domain for the problem is $D = \{ (x_1,x_2) \ge 0 \}$. The problem is not convex by calculating the Hessian matrix. Clearly, any $x_1 = 0, x_2 \in\mathbb R_+$ is a primal optimal solution with primal optimal value $1$ .
The Lagrangian is
$$
L(x_1,x_2,\lambda) = e^{-x_1x_2} + \lambda  x_1.
$$
The dual function is
\begin{align}
 G(\lambda) &amp;= \inf L(x_1,x_2,\lambda) = 
 \begin{cases}
  0&amp;  \lambda\ge 0\\
  -\infty&amp; \lambda &lt; 0
 \end{cases}
\end{align}
Thus, $\lambda \geq 0$ is dual optimal solution with dual optimal value $0$, so dual gap is $1$, strong duality fails. As for the KKT conditions, remember the domain is $D = \{ (x_1,x_2) \ge 0 \}$
\begin{align*}
 &amp;\lambda-x_2e^{-x_1x_2}=0\\
 &amp;x_1\le 0\\
 &amp;\lambda\ge 0\\
 &amp;\lambda x_1=0\\
\end{align*}
Pick any primal-dual pair satisfying $x_1 = 0, x_2\ge 0, \lambda\ge0, \lambda\ne x_2$, the KKT conditions fail.
${\bf counter-example 3}$
For a non-convex problem, even strong duality holds, solutions for KKT conditions may not be primal-dual optimal solution.
Consider the optimization problem on $\mathbb R$
\begin{align}
 \operatorname{minimize} &amp; \quad x^3 \\
 \text{subject to} &amp; -x^3-1\le 0.
\end{align}
The objective function is not convex by calculating the Hessian matrix. Clearly, $x=-1$ is the unique primal optimal solution with primal optimal value $-1$.
The Lagrangian is
$$
L(x,\lambda) = x^3 - \lambda  (x^3+1).
$$
The dual function is
\begin{align}
 G(\lambda) &amp;= \inf L(x,\lambda) = 
 \begin{cases}
  -1&amp;  \lambda=1\\
  -\infty&amp; otherwise
 \end{cases}
\end{align}
Thus, $\lambda = 1 $ is dual optimal solution with dual optimal value $-1$, so dual gap is $0$, strong duality holds. While the KKT conditions are
\begin{align*}
 &amp;3x^2(1-\lambda)=0\\
 &amp;-x^3-1\le 0\\
 &amp;\lambda\ge 0\\
 &amp;\lambda (-x^3-1)=0\\
\end{align*}
Solutions for KKT conditions are $x=-1, \lambda=1$ or $x=0,\lambda=0$. Notice that $x=0,\lambda=0$ satisfies KKT conditions but has nothing to do with primal-dual optimal solutions.
The discussion indicates for non-convex problem, KKT conditions may be neither necessary nor sufficient conditions for primal-dual optimal solutions.
${\bf counter-example4}$
For a convex problem, even strong duality holds, there could be no solution for the KKT condition, thus no solution for Lagrangian multipliers.
Consider the optimization problem on domain $\mathbb R$
\begin{align}
 \operatorname{minimize} &amp; \quad x \\
 \text{subject to} &amp; \quad x^2\le 0.
\end{align}
Obviously, the problem is convex with unique primal optimal solution $x=0$ and optimal value $0$; Feasible set is $\{0\}$, therefore Slater's condition fails.
The Lagrangian is
$$
L(x,\lambda) = x + \lambda  x^2.
$$
The dual function is
\begin{align}
 G(\lambda) &amp;= \inf L(x,\lambda) = 
 \begin{cases}
  -\infty&amp;  \lambda\le 0\\
  -\frac{1}{4\lambda} &amp;\lambda &gt;0
 \end{cases}
\end{align}
Thus, dual optimal value is $0$, so dual gap is $0$, strong duality holds. However, there are no solution for dual optimal solution because the optimal value is attained as $\lambda\rightarrow \infty$. As for the KKT conditions
\begin{align*}
 &amp;1+2\lambda x=0\\
 &amp;x^2\le 0\\
 &amp;\lambda\ge 0\\
 &amp;\lambda x^2=0\\
\end{align*}
No solution for KKT conditions.
This is the convex problem where the dual problem has no feasible solution and KKT conditions have no solution but the primal problem is simple to solve.
${\bf counter-example 5}$
For a differentiable convex problem, there could be no solution for the KKT conditions, even if the primal-dual pair exists. In that case, the strong duality fails.
Consider the optimization problem on domain $D:=\{(x,y):y&gt;0\}$
\begin{align}
 \operatorname{minimize} &amp; \quad e^{-x} \\
 \text{subject to} &amp; \quad \frac{x^2}{y}\le 0.
\end{align}
The problem can be proved to be convex with primal optimal solution $x=0, y&gt;0$ and optimal value $1$; Feasible set is $\{(0,y): y &gt; 0\}$, therefore Slater's condition fails.
The Lagrangian is
$$
L(x,y,\lambda) = e^{-x} + \lambda  \frac{x^2}{y}.
$$
After some careful calculation, the dual function is
\begin{align}
 G(\lambda) &amp;= \inf L(x,y,\lambda) = 
 \begin{cases}
  0&amp;  \lambda\ge 0\\
  -\infty &amp;\lambda &lt;0
 \end{cases}
\end{align}
Thus, dual optimal value is $0$, so dual gap is $1$, strong duality fails. We can pick primal-dual pair to be $x=0, y=1, \lambda=2$. As for the KKT conditions
\begin{align*}
 &amp;-e^{-x}+\frac{2\lambda x}{y}=0\\
 &amp;\frac{x^2}{y}\le0\\
 &amp;\lambda\ge 0\\
 &amp;\lambda \frac{x^2}{y}=0\\
\end{align*}
with $y&gt;0$, thus no solution for KKT conditions.
This counter-example warns us that we have to be careful about the strong duality condition even for differentiable convex problems.