# Invertible , bounded linear operator on a Hilbert space

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 $u \in H$ ?

• Let $H=\mathbb R$, $K = x\mapsto x^3$. I'm guessing your actual question is about linear invertible operators? – AlexR May 25 '15 at 20:01
• @AlexR I am sorry, I am considering linear bounded invertible operators, I just edited the question – the8thone May 25 '15 at 20:21
• If by "invertible" you mean "bijective" then this is a well known theorem. – Nate Eldredge May 25 '15 at 20:53

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).
• @the8thone: $K$ is a bounded linear operator. – Nate Eldredge May 25 '15 at 20:47
• 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 $\ell^2(\mathbb{N})$. – Nate Eldredge May 25 '15 at 20:48
• And the second display should have $\|(a_i)_{i \in \mathbb{N}}\|^2$. – Nate Eldredge May 25 '15 at 20:48
• 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. – Nate Eldredge May 25 '15 at 21:04