[proofplan]
We rewrite the robust constraint as a bound on the largest possible value of the affine expression $(\bar a + Pu)^\top x$ over the Euclidean ball $\{u \in \mathbb R^m : |u| \le \rho\}$. The only nonconstant part is the linear functional $u \mapsto u^\top P^\top x$, whose supremum over that ball is exactly $\rho |P^\top x|$. This gives the norm inequality form, and the auxiliary variable $t$ is then the epigraph representation of the Euclidean norm, meaning the displayed constraint $|P^\top x| \le t$ forces $t$ to lie above that norm value. In this proof, the relevant second-order cone is the set $Q_{m+1} := \{(y,t) \in \mathbb R^m \times \mathbb R : |y| \le t\}$, and a second-order cone constraint means membership in such a cone together with affine linear inequalities.
[/proofplan]
[step:Rewrite the robust constraint as a supremum over the uncertainty parameter]
Fix $x \in \mathbb R^n$ and $b \in \mathbb R$. Define the vector $q \in \mathbb R^m$ by
\begin{align*}
q = P^\top x.
\end{align*}
For each $u \in \mathbb R^m$ with $|u| \le \rho$, the corresponding uncertain vector is $a = \bar a + Pu \in \mathcal U$, and by associativity of matrix multiplication,
\begin{align*}
(\bar a + Pu)^\top x = \bar a^\top x + u^\top P^\top x = \bar a^\top x + u^\top q.
\end{align*}
Since every element of $\mathcal U$ has this form, the condition $a^\top x \le b$ for every $a \in \mathcal U$ is equivalent to
\begin{align*}
\bar a^\top x + u^\top q \le b \quad \text{for every } u \in \mathbb R^m \text{ with } |u| \le \rho.
\end{align*}
Since the set $B_\rho := \{u \in \mathbb R^m : |u| \le \rho\}$ is nonempty and, for every $u \in B_\rho$, the [Cauchy-Schwarz inequality](/theorems/432) in the Euclidean [inner product](/page/Inner%20Product) on $\mathbb R^m$ gives $u^\top q \le |u|\,|q| \le \rho |q|$, the supremum below is finite. Equivalently,
\begin{align*}
\bar a^\top x + \sup_{u \in B_\rho} u^\top q \le b.
\end{align*}
[guided]
Fix $x \in \mathbb R^n$ and $b \in \mathbb R$. The uncertainty set is parametrized by vectors $u \in \mathbb R^m$ satisfying $|u| \le \rho$, so it is useful to move the quantifier over $a \in \mathcal U$ back to a quantifier over $u$. Define $q \in \mathbb R^m$ by
\begin{align*}
q = P^\top x.
\end{align*}
If $u \in \mathbb R^m$ satisfies $|u| \le \rho$, then $a = \bar a + Pu$ belongs to $\mathcal U$. For this $a$, the left-hand side of the robust linear constraint is
\begin{align*}
a^\top x = (\bar a + Pu)^\top x = \bar a^\top x + (Pu)^\top x.
\end{align*}
Using $(Pu)^\top x = u^\top P^\top x$ and the definition $q = P^\top x$, this becomes
\begin{align*}
a^\top x = \bar a^\top x + u^\top q.
\end{align*}
Thus the robust condition is exactly
\begin{align*}
\bar a^\top x + u^\top q \le b \quad \text{for every } u \in \mathbb R^m \text{ with } |u| \le \rho.
\end{align*}
Let $B_\rho := \{u \in \mathbb R^m : |u| \le \rho\}$. This set is nonempty because $0 \in B_\rho$. For every $u \in B_\rho$, the [Cauchy-Schwarz inequality](/theorems/432) in the Euclidean inner product on $\mathbb R^m$, applied to the vectors $u$ and $q$, gives $u^\top q \le |u|\,|q| \le \rho |q|$. Therefore the supremum of $u^\top q$ over $B_\rho$ is finite. A universal upper bound over a set is the same as an upper bound on the supremum over that set, so this is equivalent to
\begin{align*}
\bar a^\top x + \sup_{u \in B_\rho} u^\top q \le b.
\end{align*}
This step isolates the only part affected by uncertainty: the support value of the Euclidean ball in the direction $q = P^\top x$.
[/guided]
[/step]
[step:Compute the supremum of the linear functional on the Euclidean ball]
Let $B_\rho := \{u \in \mathbb R^m : |u| \le \rho\}$. We claim that
\begin{align*}
\sup_{u \in B_\rho} u^\top q = \rho |q|.
\end{align*}
For every $u \in \mathbb R^m$ with $|u| \le \rho$, the Euclidean inner product satisfies
\begin{align*}
u^\top q \le |u|\,|q| \le \rho |q|.
\end{align*}
Here the [first inequality](/theorems/2897) is the Euclidean [Cauchy-Schwarz inequality](/theorems/432) applied to the vectors $u$ and $q$. Hence the supremum is at most $\rho |q|$.
If $q = 0$, then $u^\top q = 0$ for every admissible $u$, so the supremum is $0 = \rho |q|$. If $q \ne 0$, define $u_0 \in \mathbb R^m$ by
\begin{align*}
u_0 = \rho \frac{q}{|q|}.
\end{align*}
Then $|u_0| = \rho$, so $u_0$ is admissible, and
\begin{align*}
u_0^\top q = \rho \frac{q^\top q}{|q|} = \rho |q|.
\end{align*}
Therefore the upper bound is attained when $q \ne 0$, and in all cases the supremum equals $\rho |q|$.
[guided]
We now compute the support value of the closed Euclidean ball $B_\rho := \{u \in \mathbb R^m : |u| \le \rho\}$ in the direction $q \in \mathbb R^m$. The expected answer is $\rho |q|$: the Cauchy-Schwarz inequality gives the upper bound, and choosing $u$ in the direction of $q$ gives equality.
For every $u \in B_\rho$, the [Cauchy-Schwarz inequality](/theorems/432) in the Euclidean inner product on $\mathbb R^m$, applied to the two vectors $u$ and $q$, gives
\begin{align*}
u^\top q \le |u|\,|q|.
\end{align*}
Because $u \in B_\rho$, we have $|u| \le \rho$, and therefore
\begin{align*}
u^\top q \le \rho |q|.
\end{align*}
This proves
\begin{align*}
\sup_{u \in B_\rho} u^\top q \le \rho |q|.
\end{align*}
It remains to show that this upper bound is actually attained, or at least approached. If $q = 0$, then $u^\top q = 0$ for every $u \in B_\rho$, so
\begin{align*}
\sup_{u \in B_\rho} u^\top q = 0 = \rho |q|.
\end{align*}
If $q \ne 0$, define the vector $u_0 \in \mathbb R^m$ by
\begin{align*}
u_0 = \rho \frac{q}{|q|}.
\end{align*}
This definition is valid because $|q| > 0$. Its Euclidean norm is
\begin{align*}
|u_0| = \rho \frac{|q|}{|q|} = \rho,
\end{align*}
so $u_0 \in B_\rho$. Evaluating the linear functional at this admissible point gives
\begin{align*}
u_0^\top q = \rho \frac{q^\top q}{|q|} = \rho |q|.
\end{align*}
Thus the upper bound from Cauchy-Schwarz is attained when $q \ne 0$, and the case $q = 0$ was already handled. Hence
\begin{align*}
\sup_{u \in B_\rho} u^\top q = \rho |q|.
\end{align*}
[/guided]
[/step]
[step:Obtain the norm form of the robust constraint]
Let $B_\rho := \{u \in \mathbb R^m : |u| \le \rho\}$. Substituting the computed supremum with $q = P^\top x$ gives
\begin{align*}
\bar a^\top x + \sup_{u \in B_\rho} u^\top q \le b
\end{align*}
if and only if
\begin{align*}
\bar a^\top x + \rho |q| \le b.
\end{align*}
Since $q = P^\top x$, this is precisely
\begin{align*}
\bar a^\top x + \rho |P^\top x| \le b.
\end{align*}
Combining this with the first step proves the equivalence between the robust linear constraint and the single second-order cone inequality.
[guided]
The first step reduced the robust constraint to the scalar inequality
\begin{align*}
\bar a^\top x + \sup_{u \in B_\rho} u^\top q \le b,
\end{align*}
where $B_\rho := \{u \in \mathbb R^m : |u| \le \rho\}$ and $q = P^\top x$. The previous step computed this supremum exactly:
\begin{align*}
\sup_{u \in B_\rho} u^\top q = \rho |q|.
\end{align*}
Substituting this value into the reduced inequality gives
\begin{align*}
\bar a^\top x + \rho |q| \le b.
\end{align*}
Finally, using the definition $q = P^\top x$, this becomes
\begin{align*}
\bar a^\top x + \rho |P^\top x| \le b.
\end{align*}
Therefore the original robust condition $a^\top x \le b$ for every $a \in \mathcal U$ is equivalent to the displayed norm inequality, which is the single second-order cone form of the constraint.
[/guided]
[/step]
[step:Introduce the epigraph variable for the Euclidean norm]
We prove the auxiliary-variable formulation. The epigraph interpretation here means that the scalar variable $t \in \mathbb R$ is constrained to satisfy $t \ge |P^\top x|$, so the pair $(P^\top x,t)$ lies above the graph of the Euclidean norm. Suppose first that
\begin{align*}
\bar a^\top x + \rho |P^\top x| \le b.
\end{align*}
Define $t \in \mathbb R$ by
\begin{align*}
t = |P^\top x|.
\end{align*}
Then $|P^\top x| \le t$, and the inequality above gives
\begin{align*}
\bar a^\top x + \rho t \le b.
\end{align*}
Conversely, suppose there exists $t \in \mathbb R$ such that
\begin{align*}
\bar a^\top x + \rho t \le b
\end{align*}
and
\begin{align*}
|P^\top x| \le t.
\end{align*}
Since $\rho \ge 0$, multiplying the [second inequality](/theorems/2136) by $\rho$ preserves the inequality:
\begin{align*}
\rho |P^\top x| \le \rho t.
\end{align*}
Adding $\bar a^\top x$ to both sides and using $\bar a^\top x + \rho t \le b$ gives
\begin{align*}
\bar a^\top x + \rho |P^\top x| \le b.
\end{align*}
Thus the norm inequality is equivalent to the existence of $t \in \mathbb R$ satisfying the displayed system consisting of one affine inequality and the cone membership condition $(P^\top x,t) \in Q_{m+1}$, equivalently $|P^\top x| \le t$. This is the asserted second-order cone formulation. This completes the proof.
[guided]
We now show that the norm inequality and the auxiliary-variable system are equivalent. The epigraph idea is to replace the nonlinear term $|P^\top x|$ by a scalar upper bound $t$. The relevant second-order cone is
\begin{align*}
Q_{m+1} := \{(y,t) \in \mathbb R^m \times \mathbb R : |y| \le t\}.
\end{align*}
Thus the constraint $|P^\top x| \le t$ is exactly the cone membership condition $(P^\top x,t) \in Q_{m+1}$.
First suppose that
\begin{align*}
\bar a^\top x + \rho |P^\top x| \le b.
\end{align*}
Define $t \in \mathbb R$ by
\begin{align*}
t = |P^\top x|.
\end{align*}
Then $|P^\top x| \le t$ holds with equality, and substituting this choice of $t$ into the norm inequality gives
\begin{align*}
\bar a^\top x + \rho t \le b.
\end{align*}
So the auxiliary-variable system is feasible.
Conversely, suppose there exists $t \in \mathbb R$ such that
\begin{align*}
\bar a^\top x + \rho t \le b
\end{align*}
and
\begin{align*}
|P^\top x| \le t.
\end{align*}
The hypothesis $\rho \ge 0$ is used here: multiplying an inequality by $\rho$ preserves its direction. Hence
\begin{align*}
\rho |P^\top x| \le \rho t.
\end{align*}
Adding $\bar a^\top x$ to both sides gives
\begin{align*}
\bar a^\top x + \rho |P^\top x| \le \bar a^\top x + \rho t.
\end{align*}
Using the assumed affine inequality $\bar a^\top x + \rho t \le b$, we obtain
\begin{align*}
\bar a^\top x + \rho |P^\top x| \le b.
\end{align*}
Therefore the single norm inequality is equivalent to the existence of $t \in \mathbb R$ satisfying the affine inequality and the cone membership condition $(P^\top x,t) \in Q_{m+1}$. This proves the auxiliary-variable second-order cone formulation and completes the proof.
[/guided]
[/step]
