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Given $u, v\in \mathbb R^d$, I am trying to figure out the factor $M(u,v)$ in the bound

$$\|u\otimes u - v\otimes v\| \leq M(u,v) \cdot \| u - v\|_2,$$

if that's possible. The first norm is the standard operator norm. I think it must be something like $M(u,v) = \|u\| + \|v\|$ but I'm having trouble not getting bogged down in the battle with the indices. I suppose that this is a fairly standard result but I couldn't find it so far.

I'm also interested in the generalization the Hilbert space case with

$(u \otimes v)(w) := u \cdot \langle v, w\rangle.$

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    What is $||u-v||_2$?? – 2017-02-02
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    The standard euclidean norm in $\mathbb R^d$ or the norm induced by the Hilbert space in the general setting. – 2017-02-02
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    Hint: $u\otimes u - v\otimes v = u\otimes (u-v) + (u-v)\otimes v$. – 2017-02-02
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    Which norm do you use on matrices? Operatornorm or Frobenius? – 2017-02-02

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The way I see it $$\|u\otimes u - v\otimes v\|=\sup_{||x||=1}\|\langle x,u\rangle u-\langle x,v \rangle v \|=\sup_{||x||=1}\|\langle x,u\rangle u-\langle x,u\rangle v+\langle x,u\rangle v-\langle x,v \rangle v \| \le \sup_{||x||=1}\|\langle x,u\rangle u-\langle x,u\rangle v\|+\sup_{||x||=1}\|\langle x,u\rangle v-\langle x,v \rangle v \| \le \|u-v\|\|u\|+\|u-v\|\|v\|$$ $$=(\|u\|+\|v\|)\|u-v\|$$

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    In the second line of the equation I think there is a typo. You should replace the $v$ by an $u$ – 2017-02-02
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    I don't see it. Where exactly?? – 2017-02-02
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    In $$ \sup_{\|x\| = } – 2017-02-02
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    In $$ \sup_{\|x\| = 1} \| \langle x, u \rangle u - \langle x, u \rangle \underbrace{v}_{\text{Here,= u? }} \|$$ – 2017-02-02
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    No. I have inserted it knowingly and then took it out in the next thing. Look carefully. – 2017-02-02
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    My mistake! Sorry. – 2017-02-02