# Equivalent condition for indicator function

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.

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.