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We have the following two conditions:

$C1.$ $l(x) \geq 0$ for all $x\in \mathbb{R}^n$,
$C2.$ $l(x) \geq 1$ for all $x\in \mathbb{R}^n$ such that $a+b^Tx \leq 0.$

Here $l(x) = \begin{bmatrix} x^T 1 \end{bmatrix} M \begin{bmatrix} x \\ 1 \end{bmatrix}.$

We have to show that $C2.$ is equivalent to
$C3.$: there exists $\tau \geq 0$ such that for every $x$, $l(x) \geq 1-2\tau (a+b^Tx). $

Here is what I have tried:

$C1.$ is equivalent to the fact that $M \succcurlyeq 0$.
If $C3.$ holds then, under $a+b^Tx \leq 0$, $l(x) \geq 1-2\tau (a+b^Tx) \geq 1.$ Hence, $C2.$ holds.

The following exact statement from paper concludes that $C2.$ gives $C3.$

With $C1.$ in force, an application of the classical strong duality result for convex programs under the Slater assumption shows that the above condition is sufficient, provided there exists an $x_0$ such that $a+b^Tx_0 <0.$

I am having trouble here. I tried writing a convex problem $$ \min_{x} l(x)-1 \\ \text{s.t.} a+b^Tx \leq 0. $$ So we get its Lagrange function as $L(x,\tau) = l(x)-1 + \tau (a+b^Tx) $. A KKT point will satisfy $C3.$, but I am unable to show it for every $x$. Moreover I am not using $C1. $ but the statement takes it into account.

This question is asked here.

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1 Answer 1

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Let $$ \begin{array}{lrl} p^\star = &\min& l(x)-1 \\ &\text{s.t.}& a+b^Tx \leq 0. \end{array} $$ Assumption C1 tells us that this is a convex optimization problem, and existence of a Slater point, $x_0$, satisfying $a + b^T x_0 < 0$, tells that this convex optimization problem satisfies strong duality. That is, $$ \max_{\tau \geq 0} \min_x L(x,\tau) = p^\star. $$

where $L(x,\tau) = l(x)-1 + \tau (a+b^Tx)$ is the Lagrange function. By assumption C2, telling us that $p^\star \geq 0$, we thus have that the optimal dual multiplier, $\tau^\star \geq 0$, satisfies $$ \min_x L(x,\tau^\star) \geq 0 $$ and thus we conclude $L(x,\tau^\star) \geq 0$ for all $x \in \mathbb{R}^n$, which is equivalent to C3.

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