Is $f(\mathbf{X}) = \sup_{\|\mathbf{y}\|\leq 1 } Tr(\mathbf{X \mathbf{y}\mathbf{y}}^T)$ an affine function? Suppose $f(\mathbf{X}) = \sup_{\|\mathbf{y}\|\leq 1 } Tr(\mathbf{X \mathbf{y}\mathbf{y}}^T)$ is the trace of a resulting matrix where $\mathbf{X} \in \mathbb{R^{n\times n}}$. It is easily can be shown that $Tr(\cdot)$ is a linear function for each $\mathbf{y}$. Can we deduce the function $f(\mathbf{X})$ is an affine function?
 A: No, the supremum of a set of linear functions is very rarely affine.  It's easy to find counterexamples in the case $n=2$.
A: I would like to answer this question using a general mathy (!) proof.  
Suppose:$f(\mathbf{X}) = \sup_{\|\mathbf{y}\|\leq 1 } Tr(\mathbf{X \mathbf{y}\mathbf{y}}^T)$. For $f(\mathbf{X})$ to be an affine/a linear function, it should satisfy $f(a\mathbf{X}_{1}+b\mathbf{X}_{2}) = af(\mathbf{X}_{1})+bf(\mathbf{X}_{2})$ where $a,b \in \mathbb{R}$. Then, we have
\begin{align}
f(a\mathbf{X}_{1}+b\mathbf{X}_{2}) &= \sup_{\|\mathbf{y}\|\leq 1 } Tr((a\mathbf{X}_{1}+b\mathbf{X}_{2})\mathbf{\mathbf{y}\mathbf{y}}^T)\\
&= \sup_{\|\mathbf{y}\|\leq 1} Tr(a\mathbf{X}_{1}\mathbf{\mathbf{y}\mathbf{y}}^T+b\mathbf{X}_{2}\mathbf{\mathbf{y}\mathbf{y}}^T) \\
&=\sup_{\|\mathbf{y}\|\leq 1 } \{a\; Tr(\mathbf{X}_{1}\mathbf{\mathbf{y}\mathbf{y}}^T)+b \; Tr(\mathbf{X}_{2}\mathbf{\mathbf{y}\mathbf{y}}^T)\} \\
&\leq\sup_{\|\mathbf{y}\|\leq 1} a\; Tr(\mathbf{X}_{1}\mathbf{\mathbf{y}\mathbf{y}}^T)+\sup_{\|\mathbf{y}\|\leq 1} b \; Tr(\mathbf{X}_{2}\mathbf{\mathbf{y}\mathbf{y}}^T) \\
&= 
|a|\sup_{\|\mathbf{y}\|\leq 1} Tr(\mathbf{X}_{1}\mathbf{\mathbf{y}\mathbf{y}}^T)+|b|\sup_{\|\mathbf{y}\|\leq 1} Tr(\mathbf{X}_{2}\mathbf{\mathbf{y}\mathbf{y}}^T),
\end{align}
Hence:
$$f(a\mathbf{X}_{1}+b\mathbf{X}_{2}) =  a\sup_{\|\mathbf{y}\|\leq 1} Tr(\mathbf{X}_{1}\mathbf{\mathbf{y}\mathbf{y}}^T)+b\sup_{\|\mathbf{y}\|\leq 1} Tr(\mathbf{X}_{2}\mathbf{\mathbf{y}\mathbf{y}}^T)$$ 
is a special case of above inequality and hence it is not generally an affine function.

Notice:
$$\sup_{x} \{f(x)+g(x)\} \leq \sup_{x} f(x) + \sup_{x} g(x)$$
and
$$\sup_{x} af(x) = a \sup_x {f(x)} \quad \forall a\in \mathbb{R}_{+}.$$
