# How to determine the $\delta$ in the open mapping theorem?

Let $X$ and $Y$ be Banach spaces and $T\in\mathcal{L}(X,Y)$ be a bounded linear operator from $X$ to $Y$.

If $T$ is surjective, then the open mapping theorem says that there is a positive $\delta$ such that $TB_1\supset\delta B_2$, where $B_1$ and $B_2$ are open unit balls in $X$ and $Y$ respectively.

My question is how is the $\delta$ related to the norm of $T$, which gives a (sharp) bound for the norm of the inverse of $T$ if $T$ is also injective.

Thanks!

And another related question: If $\mu$ is at a positive distance to $\sigma(T)$, the spectrum of $T$, how is $\|(\mu-T)^{-1}\|$ related to the distance from $\mu$ to $\sigma(T)$? Obviously we have an lower bound, but what I need is an upper bound.

Thanks!

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Short answer: $0< \delta\le \|T\|$ is as much as we can say, if $\|T\|$ is all we know.
Slightly longer answer. To an operator $T$ we can associate a lot of numbers, such as: the norm $\|T\|=\sup_{\|x\|=1}\|Tx\|$, the lower bound $m_T=\inf_{\|x\|=1}\|Tx\|$, and the covering number $\delta_T=\sup\{r>0\colon TB_1\supset r B_2\}$. The first measures boundedness, the second injectivity, the third surjectivity. For the adjoint operator the norm stays the same: $\|T^*\|=\|T\|$ but the other two trade places: $m_{T^*}=\delta_T$ and $\delta_T=m_{T^*}$. So, $\delta$ is directly related to the lower bound for $T^*$.
Thanks!Actually I am thinking about another problem. I know that $\mu$ is at a positive distance to $\sigma(T)$, and I am wondering whether we can bound the norm of $(\mu-T)^{-1}$ using the norm of $T$ and the distance between $\mu$ and $\sigma(T)$. Do you have any suggestion on this? –  Hui Yu Jun 13 '12 at 18:08
@HuiYu There is an easy estimate in the wrong directions: the norm of $(\mu-T)^{-1}$ is at least $1/\mathrm{dist}(\tau,\sigma(T))$. But I don't know how to estimate the norm of the inverse from above. –  user31373 Jun 13 '12 at 20:02