$$
\begin{array}{l} \hat {\mathcal {R}} _ {\mathcal {S}} \left(\mathcal {G} _ {T} ^ {U}\right) = \frac {1}{n} \underset {\boldsymbol {\sigma}} {\mathbb {E}} \left[ \sup  _ {g \in \mathcal {G} _ {T} ^ {U}} \sum_ {i = 1} ^ {n} \sigma_ {i} g \left(\boldsymbol {z} _ {i}\right) \right] \\ = \frac {1}{n} \mathbb {E} _ {\boldsymbol {\sigma}} \left[ \sup  _ {\mathsf {K} _ {\mathcal {S} ^ {\prime}} \in \mathcal {K} _ {T}} \sup  _ {\| \boldsymbol {\beta} \| \leq B} \sum_ {i = 1} ^ {n} \sigma_ {i} \left(\left\langle \boldsymbol {\beta}, \Phi_ {\mathcal {S} ^ {\prime}} (\boldsymbol {z} _ {i}) \right\rangle + \ell (\boldsymbol {w} _ {0}, \boldsymbol {z} _ {i})\right) \right] \\ = \frac {1}{n} \mathbb {E} _ {\boldsymbol {\sigma}} \left[ \sup  _ {\mathcal {K} _ {\mathcal {S} ^ {\prime}} \in \mathcal {K} _ {T}} \sup  _ {\| \boldsymbol {\beta} \| \leq B} \left\langle \boldsymbol {\beta}, \sum_ {i = 1} ^ {n} \sigma_ {i} \Phi_ {\mathcal {S} ^ {\prime}} (\boldsymbol {z} _ {i}) \right\rangle \right] + \mathbb {E} _ {\boldsymbol {\sigma}} \left[ \sum_ {i = 1} ^ {n} \sigma_ {i} \ell (\boldsymbol {w} _ {0}, \boldsymbol {z} _ {i}) \right] \\ = \frac {1}{n} \mathbb {E} _ {\boldsymbol {\sigma}} \left[ \sup  _ {\mathcal {K} _ {\mathcal {S} ^ {\prime}} \in \mathcal {K} _ {T}} \sup  _ {\| \boldsymbol {\beta} \| \leq B} \left\langle \boldsymbol {\beta}, \sum_ {i = 1} ^ {n} \sigma_ {i} \Phi_ {\mathcal {S} ^ {\prime}} (\boldsymbol {z} _ {i}) \right\rangle \right] \\ = \frac {B}{n} \underset {\sigma} {\mathbb {E}} \left[ \sup  _ {\mathcal {K} _ {S ^ {\prime}} \in \mathcal {K} _ {T}} \left\| \sum_ {i = 1} ^ {n} \sigma_ {i} \Phi_ {\mathcal {S} ^ {\prime}} \left(\boldsymbol {z} _ {i}\right) \right\| \right] \quad (\text {d u a l n o r m p r o p e r t y}) \\ \geq \frac {B}{n} \sup  _ {\mathcal {K} _ {\mathcal {S} ^ {\prime}} \in \mathcal {K} _ {T}} \mathbb {E} _ {\boldsymbol {\sigma}} \left[ \left\| \sum_ {i = 1} ^ {n} \sigma_ {i} \Phi_ {\mathcal {S} ^ {\prime}} (\boldsymbol {z} _ {i}) \right\| \right] \\ \geq \frac {B}{n} \sup  _ {\mathcal {K} _ {\mathcal {S} ^ {\prime}} \in \mathcal {K} _ {T}} \left\| \mathbb {E} _ {\boldsymbol {\sigma}} \left[ \left| \sum_ {i = 1} ^ {n} \sigma_ {i} \Phi_ {\mathcal {S} ^ {\prime}} (\boldsymbol {z} _ {i}) \right| \right] \right\| \quad \text {(n o r m s u b - a d d i t i v i t y)} \\ = \frac {B}{n} \sup  _ {\mathcal {K} _ {\mathcal {S} ^ {\prime}} \in \mathcal {K} _ {T}} \left(\sum_ {j \in \mathbb {N} _ {+}} \left(\underset {\sigma} {\mathbb {E}} \left[ \left| \sum_ {i = 1} ^ {n} \sigma_ {i} \left[ \Phi_ {\mathcal {S} ^ {\prime}} \left(z _ {i}\right) \right] _ {j} \right| \right]\right) ^ {2}\right) ^ {\frac {1}{2}} \quad (\text {b y}) \\ \geq \frac {B}{n} \sup  _ {\mathsf {K} _ {\mathcal {S} ^ {\prime}} \in \mathcal {K} _ {T}} \left(\sum_ {j \in \mathbb {N} _ {+}} \left(\frac {1}{\sqrt {2}} \left| \sum_ {i = 1} ^ {n} [ \Phi_ {\mathcal {S} ^ {\prime}} (\boldsymbol {z} _ {i}) ] _ {j} ^ {2} \right| ^ {\frac {1}{2}}\right) ^ {2}\right) ^ {\frac {1}{2}} \quad (\text {K h i n t c h i n e - K a h a n e i n e q u a l i t y}) \\ = \frac {B}{\sqrt {2} n} \sup  _ {\mathcal {K} _ {S ^ {\prime}} \in \mathcal {K} _ {T}} \left(\sum_ {j \in \mathbb {N} _ {+}} \left| \sum_ {i = 1} ^ {n} \left[ \Phi_ {\mathcal {S} ^ {\prime}} \left(z _ {i}\right) \right] _ {j} ^ {2} \right|\right) ^ {\frac {1}{2}} \\ = \frac {B}{\sqrt {2} n} \sup  _ {\mathcal {K} _ {\mathcal {S} ^ {\prime}} \in \mathcal {K} _ {T}} \left(\sum_ {i = 1} ^ {n} \sum_ {j \in \mathbb {N} _ {+}} [ \Phi_ {\mathcal {S} ^ {\prime}} (z _ {i}) ] _ {j} ^ {2}\right) ^ {\frac {1}{2}} \quad (\text {r e a r r a n g e t h e s u m m a t i o n s}) \\ = \frac {B}{\sqrt {2} n} \sup  _ {\mathsf {K} _ {\mathcal {S} ^ {\prime}} \in \mathcal {K} _ {T}} \left(\sum_ {i = 1} ^ {n} \| \Phi_ {\mathcal {S} ^ {\prime}} (\boldsymbol {z} _ {i}) \| ^ {2}\right) ^ {\frac {1}{2}} \\ = \frac {B}{\sqrt {2} n} \sup  _ {\mathsf {K} _ {\mathcal {S} ^ {\prime}} \in \mathcal {K} _ {T}} \left(\sum_ {i = 1} ^ {n} \mathsf {K} _ {\mathcal {S} ^ {\prime}} \left(\boldsymbol {z} _ {i}, \boldsymbol {z} _ {i}\right)\right) ^ {\frac {1}{2}} \\ = \frac {B}{\sqrt {2} n} \sup  _ {\mathsf {K} _ {\mathcal {S} ^ {\prime}} \in \mathcal {K} _ {T}} \sqrt {\operatorname {T r} (\mathsf {K} _ {\mathcal {S} ^ {\prime}} (\mathbf {Z} , \mathbf {Z}))}. \\ \end{array}
$$