Extension of contractions on a Banach subspace

Question

This question is related to Contractions on a subspace but on a Banach space instead a vector space.

Let $T$ be a linear map defined on a subspace $S$ of some Banach space $V$, such that $$ d(T(v_1),T(v_2))≤d(v_1,v_2)\quad \text{for } v_1,v_2\in S $$ where $d(v_1,v_2)=\|v_1-v_2\|$ is the norm induced distance on $V$. Can we say somethig about the existence or not of a contractive linear extension $T'$, defined on $V$ such that $$ d(T'(v_1),T'(v_2))≤d(v_1,v_2)\quad \text{for all } v_1,v_2\in V $$ and that $T'(v)=T(v)$ for all $v\in S$?

Answer 1

In general, linear operators do not admit norm-preserving extension (Hahn-Banach theorem applies only to functionals). I'll mention two special cases when they do:

1. $V$ is a space in which every subspace $S$ is $1$-complemented, i.e., there exists a norm-$1$ idempotent $E:V\to V$ with range $S$. Then $T\circ E$ is the desired extension. Such spaces include: (a) Hilbert spaces; (b) two-dimensional normed spaces.
2. $V$ is an injective Banach space: one way to describe such a space is that every collection of pairwise-intersecting closed balls has nonempty intersection. Examples include $\ell_\infty$ and $L_\infty$. See Equivalent definitions of Injective Banach Spaces

---

Here is an explicit example of an operator without norm-preserving extension. Let $V=\ell_1^4$, the $4$-dimensional space with $\ell_1$ norm. Let $S=\{x\in V : \sum x_k = 0\}$, a codimension-1 subspace. Consider the operator $T:V\to V$ with the matrix (in standard basis) $$ A = \begin{pmatrix} 1 &1 & -1 & -1 \\ 1 & -1 & 1 & -1 \\ 1 & -1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} \tag1$$ I claim that:

• the norm of the restriction $T_{|S}$ is strictly less than $3$;
• any extension of $T_{|S}$ to a map on $V$ has norm at least $3$.

First claim. The norm of $T_{|S}$ is the maximum of norms $\left\|\sum_{k=1}^4 c_k v_k\right\|_1$ where $v_1,\dots, v_4$ are the columns of matrix $A$, and the coefficients satisfy $\sum c_k=0$ and $\sum |c_k| = 1$. It is clear that at least two of the coefficients $c_k$ must be nonzero. Whether they have same sign or opposite sign, forming the sum $\sum_{k=1}^4 c_k v_k$ entails cancellation, numbers of different signs added together. (This is because for any pair of columns of $A$, some row has elements of different sign and some row has elements of the same sign). Hence, $$ \left\|\sum_{k=1}^4 c_k v_k\right\|_1 < \sum_{k=1}^4 |c_k| \|v_k\|_1 = 3 $$

Second claim. Let $T'$ be any extension of $T_{|S}$. Then $T'-T$ vanishes on $S$, which means it is a rank-one operator with matrix of the form
$$ B = \begin{pmatrix} a&a&a&a \\ b&b&b&b \\ c&c&c&c \\ d&d&d&d \end{pmatrix} \tag2$$ Since every row of $A$ has median zero, adding any constant to that row does not decrease its $\ell_1$ norm. Therefore, the sum of absolute values of all entries of $A+B$ is at least $12$ (the value of this sum for $A$). Hence, at least one column of $A+B$ has $\ell_1$ norm $\ge 3$, and the claim is proved.