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Your set is indeed a convex cone.

Since it is a cone, it suffices to show that the $m=1$ section is convex.

But this is equivalent to show that the (symmetric matrix valued) "function" $u\mapsto P(u)=S(u)^*S(u)$ is "convex", i.e. $P((u+v)/2)\prec (P(u)+P(v))/2$, because the set is basically the "epigraph" $(u,L)$ : $L\succ P(u)$.

Now it is easily checked that the quadratic function $P$ satisties the parallelogram identity $P((u+v)/2)+P((u-v)/2)=(P(u)+P(v))/2$, which does the job since $P$ is positive.

EDIT : in fact any set of $(x,y)$ defined by an inequality $S(y)-A(x)^*A(x) \succ 0$, with $S(y)$ $n\times n$ symmetric and linear in $y$, and $A(x)$ $n\times d$ and linear in $x$, is convex and moreover defined by a Linear Matrix Inequality.

Indeed this is equivalent to

$$\left(\begin{matrix} S(y) & A(x)^* \\ A(x) & I \end{matrix}\right) \succ 0$$

as easily seen by row and column operations. In fact substituting $I$ by $\lambda I$, you can "re-homogenize" the problem.

1

Your set is indeed a convex cone.

Since it is a cone, it suffices to show that the $m=1$ section is convex.

But this is equivalent to show that the (symmetric matrix valued) "function" $u\mapsto P(u)=S(u)^*S(u)$ is "convex", i.e. $P((u+v)/2)\prec (P(u)+P(v))/2$, because the set is basically the "epigraph" $(u,L)$ : $L\succ P(u)$.

Now it is easily checked that the quadratic function $P$ satisties the parallelogram identity $P((u+v)/2)+P((u-v)/2)=(P(u)+P(v))/2$, which does the job since $P$ is positive.