# The norm of the extension of an operator

If $T_0$ is the identity operator from $E$ to $E$, where $E$ is a subspace of $F$ and both of them are Banach spaces (maybe not needed). If the bounded linear operator $T$ is an extension of $T_0$ from $F$ to $E$, is it true that the norm of $T$ can be bounded by the norm of $T_0$?

Given the Banach spaces $\mathcal{c}_0$ and $\ell^\infty$.

Consider the identity: $$T_0:\mathcal{c}_0\to\mathcal{c}_0:\quad T_0:=\mathbb{1}$$

It has No continuous extension: $$T:\ell^\infty\to\mathcal{c}_0:\quad T\restriction_{\mathcal{c}_0}=T_0$$

For the details see: Werner

• Thanks, but in my question, the existence of continuous extensions is assumption. Aug 22, 2015 at 8:50
• @ShijieDong: The answer to that question is no for any Banach space due to Hahn-Banach. Aug 23, 2015 at 4:38

No. Consider $F=\mathbb R^2$. And $T(e_1)=e_1$, $T(e_2)=2 e_2$. The subspace generated by $e_1$ could be your $E$. The norm of $T_0$ then is 1 and the norm of $T$ is 2.

Even better: Let $F$ be some Banach space and $T:F\to F$ a bounded non-zero operator. Take $E=\{0\}$. Then we have norm 0 for $T_0$ and a positive norm for $T$ because $T$ is non-zero.

Edit: Since we want $T(F)\subseteq E$ let $e_1,e_2$ be a basis of $F$ and $e_1$ be a basis of $E$. We define $T(e_1)=e_1$ and $T(e_2)=2e_1$. Then we have $|T|=2$ and $|T_0|=1$.

• Thanks, but your examples are not consistent with my question. In my question, the image of the extension operator $T$ is still in the subspace $E$. Aug 22, 2015 at 8:54
• @ShijieDong You are right. Consider my edit! Aug 22, 2015 at 10:44
• That's cool, thanks. Aug 22, 2015 at 11:14