Continuity of an inner-product mapping I want to show that $Q_{T}(v):S \rightarrow \mathbb{R}$ defined by $Q_{T}(v) := \langle T(v),v \rangle$ is uniformly continuous, where $V$ is a finite-dimensional inner-product space, $T\epsilon Hom(V,V)$ is self-adjoint and $S:=\{v \epsilon V: \| v \| = 1 \}$. I'm given a hint to show that there exists a $c_{T}\epsilon \mathbb{R}$ such that $\|T(v)\| \leq c_{T} \|v\|$ for all $v \epsilon V$. Since norms map into $\mathbb{R_{\geq 0}}$, I'm presuming $c_{T} \geq 0$. 
The hint reminds me of Lipshitz continuity, which does implies uniform continuity, however, I'm confused as to how this hint actually satisfies Lipshitz continuity for $Q_{T}$. Wouldn't I rather need to show that $$d_{\mathbb{R}}(\langle T(v),v\rangle , \langle T(v^{'}),v^{'} \rangle) \leq c_{T} d_{S}(v,v^{'})$$ or equivalently, $$|\langle T(v),v\rangle - \langle T(v^{'}),v^{'} \rangle | \leq c_{T} \|v-v^{'}\|$$ for all $v,v{'} \epsilon S$ (rather than the vectors being in $V$, as the original hint said)? So my question concisely is how does the hint help me (if at all)? Is there a better way to go about this?
Furthermore, I've already proven that $S$ is compact in $V$ and $\|v\|$ is continuous.
My miscellaneous thoughts that I don't think go anywhere productive or at least I have failed to see it yet:
1) There's a result I might be able to invoke which says the following: 
Let $f$ be a continuous mapping of a compact metric space $X$ into a metric space $Y$. Then $f$ is uniformly continuous on $X$.
Should $X = S$ (with the norm acting as the metric) and $Y = \mathbb{R}$, then it is sufficient to show $Q_{T}$ is continuous (Unless I've misread something). I have, however, had equal difficulty doing that with any sort of $\delta - \varepsilon$ argument.
2) While I'm still stuck on how the hint helps, I've noticed if it is sufficient to show the hint for all $v$ merely in $S$ (where the hint calls for $V$), then $\|v\| = 1$, so it is equivalent to show there exists some positive, real $c_{T}$ such that for all $v \epsilon S, \|T(v)\| \leq c_{T}$, namely that $T(S)$ is bounded in $\mathbb{R}$?
Any help forward is greatly appreciated!
 A: I shall assume that the condition in the definition of $S$ is $\lVert v \rVert = 1$, rather than $\lVert v \rVert < 1$ as shown.
(The same argument also works, with only one minor change at the end, if the condition is $\lVert v \rVert \leqslant 1$. It even works if the condition is $\|v\| < 1$, as stated in the question; however, the question also states that $S$ is compact, so it is hard to know what is intended.)
For all $v, w \in V$,
\begin{align*}
Q_T(v + w) & = \langle T(v) + T(w), v + w \rangle \\
& =
\langle T(v), v \rangle +
\langle T(v), w \rangle +
\langle T(w), v \rangle +
\langle T(w), w \rangle.
\end{align*}
Therefore, using the Cauchy-Schwarz inequality,
\begin{align*}
|Q_T(v + w) - Q_T(v)| & \leqslant
|\langle T(v), w \rangle| +
|\langle T(w), v \rangle| +
|\langle T(w), w \rangle| \\
& \leqslant
\lVert T(v) \rVert \lVert w \rVert +
\lVert T(w) \rVert \lVert v \rVert +
\lVert T(w) \rVert \lVert w \rVert \\
& \leqslant
c_T(2\lVert v \rVert + \lVert w \rVert)\lVert w \rVert.
\end{align*}
Therefore, for all $u, v \in S$,
\begin{align*}
|Q_T(u) - Q_T(v)| & \leqslant
c_T(2\lVert v \rVert + \lVert u - v \rVert)\lVert u - v \rVert  \\
& \leqslant
c_T(3\lVert v \rVert + \lVert u \rVert)\lVert u - v \rVert \\
& = 4c_T\lVert u - v \rVert.
\end{align*}
I haven't used the hypothesis that $T$ is self-adjoint, so perhaps something is wrong.

I have been asked (in the comments) to provide a proof of the existence of $c_T$, to make this answer more self-contained. In this special case (a finite-dimensional real inner product space, $V$), there is a quick proof, avoiding considerations of continuity and compactness, as follows:
Choose an orthonormal basis $(e_1, \ldots, e_n)$. Then, for each $v \in V$, there exist $a_1, \ldots, a_n \in \mathbb{R}$ such that $v = \sum_{i=1}^n a_ie_i$. Therefore:
$$
\|T(v)\| = \left\lVert\sum_{i=1}^n a_iT(e_i)\right\rVert
\leqslant
\sum_{i=1}^n |a_i|\|T(e_i)\|.
$$
But $\|v\|^2 = \sum_{i=1}^n |a_i|^2$, therefore $|a_i| \leqslant \|v\|$ ($i = 1, \ldots, n$), therefore
$$
\|T(v)\| \leqslant \left(\sum_{i=1}^n \|T(e_i)\|\right)\|v\|.
$$
