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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!

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  • $\begingroup$ Is $V$ finite-dimensional? $\endgroup$ – paf Aug 23 '16 at 20:47
  • $\begingroup$ Yes, sorry! I'll edit that in $\endgroup$ – Xodus Aug 23 '16 at 21:09
  • $\begingroup$ Hint for the hint: consider the action of $T$ on the unit sphere. $\endgroup$ – Neal Aug 24 '16 at 13:48
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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 $$ \|T(v)\| \leqslant \left(\sum_{i=1}^n \|T(e_i)\|\right)\|v\|. $$

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  • $\begingroup$ In order to complete your answer, you should explain why there is a $c_T$ s.t. $\forall v\in V,\;\| T(v)\|\leq c_T\|v\|$. $\endgroup$ – paf Aug 24 '16 at 10:26
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    $\begingroup$ @paf (Correcting typo in first version of this comment:) I thought it was OK to assume this, because: (a) the OP would be doing it as part of the exercise; and (b) it is well-known, in the form $\|T(v)\| \leqslant \|T\| \|v\|$, where $\|T\| = \sup\{ \|T(x)\| : \|x\| \leqslant 1\}$. I wasn't sure whether it was necessary to spell out the fact that all linear operators on finite-dimensional normed spaces are continuous (i.e. bounded), but I didn't worry about it. If the answer really should give all the details, rather than just those explicitly called for in the question, I will fill them in. $\endgroup$ – Calum Gilhooley Aug 24 '16 at 11:01
  • $\begingroup$ @CalumGilhooley I think for this answer, you should show that all operators on finite-dimensional normed spaces are bounded, as showing this in the hint is where OP got stuck. $\endgroup$ – Neal Aug 24 '16 at 13:48
  • $\begingroup$ @Neal In view of the quick proof for this special case which I've just appended to my answer (simultaneously with your comment), do you still think that this is necessary? In the general case, there is a quick proof of boundedness with respect to the $\|\cdot\|_1$ norm, but one still has to show that all norms on $V$ are equivalent. The way I've gone, that necessity is avoided (at the price of getting only a crude upper bound for $\|T\|$, but that is enough for the purposes of this question.) $\endgroup$ – Calum Gilhooley Aug 24 '16 at 13:56
  • $\begingroup$ Thank you! This clarifies the question considerably. I did mean for S to have $v$ such that $\|v\| = 1$, and have edited in appropriately. $\endgroup$ – Xodus Aug 24 '16 at 17:29

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