= Solution
For $\mathbf a,\mathbf b\in\mathbb R^{n+1}$, the <triangle inequality> and the <Cauchy-Schwarz inequality> give
$$
\begin{aligned}
|S(\mathbf a)-S(\mathbf b)|
&\leq \lVert T(\mathbf a-\mathbf b)\rVert_\infty\\
&\leq \sum_{r=0}^n|a_r-b_r|
\leq \sqrt{n+1}\,\lVert\mathbf a-\mathbf b\rVert_2.
\end{aligned}
$$
Thus $S$ is a <continuous function>. The <unit sphere> in the finite-dimensional <Euclidean normed vector space> $\mathbb R^{n+1}$ is <compact>. Moreover, $S(\mathbf a)=0$ implies that the <polynomial> $T\mathbf a$ vanishes identically, so every coefficient $a_r$ is zero. Hence $S$ is strictly positive on the unit sphere. By the <extreme value theorem>, it has a positive minimum
$$
\delta=\min_{\lVert\mathbf a\rVert_2=1}S(\mathbf a)>0.
$$
The <supremum norm> is homogeneous. For $\mathbf a\ne0$,
$$
\lVert T\mathbf a\rVert_\infty
=\lVert\mathbf a\rVert_2
S\left(\frac{\mathbf a}{\lVert\mathbf a\rVert_2}\right)
\geq\delta\lVert\mathbf a\rVert_2.
$$
It follows directly that $\lVert T\mathbf a\rVert_\infty\to\infty$ whenever $\lVert\mathbf a\rVert_2\to\infty$.
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