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Suppose we have an invertible, bounded linear operator $K$ on a Hilbert space $H$. Is there a constant $c \in \mathbb{R}_+$ such that $$ ||Ku|| \geq c||u|| $$ for all $u \in H$ ?

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  • $\begingroup$ Let $H=\mathbb R$, $K = x\mapsto x^3$. I'm guessing your actual question is about linear invertible operators? $\endgroup$ – AlexR May 25 '15 at 20:01
  • $\begingroup$ @AlexR I am sorry, I am considering linear bounded invertible operators, I just edited the question $\endgroup$ – the8thone May 25 '15 at 20:21
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    $\begingroup$ If by "invertible" you mean "bijective" then this is a well known theorem. $\endgroup$ – Nate Eldredge May 25 '15 at 20:53
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Let $H=\ell^2(\mathbb{N})$ be the space of square-summable series, i.e. $$ \big\langle (a_i)_{i\in\mathbb{N}},\;(b_i)_{i\in\mathbb{N}}\big\rangle = \sum_{i=1}^\infty a_ib_i $$ and therefore $$ \|(a_i)_{i\in\mathbb{N}}\|^2 = \big\langle(a_i)_{i\in\mathbb{N}},\;(a_i)_{i\in\mathbb{N}}\big\rangle =\sum_{i=1}^\infty a_i^2. $$

Now look at the operator $$ K \;:\; H\to H\;:\; (a)_{n\in\mathbb{N}} \to (\tfrac{1}{n}a)_{n\in\mathbb{N}}. $$ and in particular at the images of the vectors $$ \mathbf{u}_i = (\underbrace{0,\ldots,0}_{i-1\text{ zeros}},1,0,\ldots). $$

Edit: Note, however, that $K$ isn't surjective, only injective. So $K$ isn't bijective, although it does have a left-inverse (but no right-inverse).

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  • $\begingroup$ I am sorry, I just edited the question. I am asking this question for bounded linear operators $\endgroup$ – the8thone May 25 '15 at 20:23
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    $\begingroup$ @the8thone: $K$ is a bounded linear operator. $\endgroup$ – Nate Eldredge May 25 '15 at 20:47
  • $\begingroup$ Nitpick: you don't really mean $H = \mathbb{R}^{\mathbb{N}}$ which is the set of all sequences, not just those which are square-summable. The usual notation for the space of square-summable sequences is $\ell^2(\mathbb{N})$. $\endgroup$ – Nate Eldredge May 25 '15 at 20:48
  • $\begingroup$ And the second display should have $\|(a_i)_{i \in \mathbb{N}}\|^2$. $\endgroup$ – Nate Eldredge May 25 '15 at 20:48
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    $\begingroup$ It depends on what exactly is meant by "invertible". It has an inverse which is not defined on all of $H$. If "invertible" means "bijective" then indeed your operator is not bijective, and in fact every bijective bounded linear operator on a Hilbert space does have a bounded inverse. $\endgroup$ – Nate Eldredge May 25 '15 at 21:04

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