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Let $X$ and $Y$ be two Banach Spaces and $X_1$ be a subspace of $X$. If $T$ is a bounded linear operator from $X_1\to Y$, then,

  1. this is an extension of $T$ from $X\to Y$ such that $\|T\|_{X}\le \|T\|_{X_1}$.

  2. The extension is unique if $X_1$ is dense in $X$.

Could some please help with the proof? Or provide a reference where I can find an answer. I have been flipping through all my text and urgently need something quick when studying the chapter of Sobolev space.(There are quite a few extension theorem I've seen in books. However every version is slightly different. Hence I hope someone can guide me through as an example.)

math101
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  • I don't quite understand the first part of the question. You have $T$ defined on the $X_1 \subset X$. Do you want to extend $T$ now on the whole space? In your sentence, do you mean: '1. this is a restriction of $T$ from $X\to Y$'? –  May 16 '15 at 07:06
  • First, yes, want T on whole space. Second, an extension of T (no typos there) – math101 May 16 '15 at 08:15
  • I read it as $T: X_1 \to Y$ is an extension of $T: X\to Y$. For me, this makes no sense. Please clarify this. –  May 16 '15 at 08:16
  • take a look at page 5 lemma 1.3 in http://www.math.pitt.edu/~xfc/PDEII/Notes01Sp.pdf – math101 May 16 '15 at 08:20
  • I think it is not correct. 1. How can it be an extension if it is defined on a smaller set than the operator you want to extend. As the name suggest, the domain of the extension should be larger than the previous domain. 2. Assume there is a typo, i.e. 'this' meant to be 'there', then the statement is still false as my answer suggests. –  May 16 '15 at 08:51
  • Anything still unclear? –  May 17 '15 at 12:34

1 Answers1

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The following claim is not true in general.

Let $X$ and $Y$ be Banach spaces and $X_1 \subset X$ a subspace. If $T: X_1 \to Y$ is a continuous bounded operator, then there exists a continuous linear extension to the whole space $X$, i.e. there exists a bounded linear operator $\hat T: X \to Y$ such that $ \hat{T}\restriction_{X_1} = T$.

Take for example $X=\ell^\infty$ $X_1 = c_0$ and $Y=c_0$. Now, the identity $\text{id}: c_0\to c_0$ is clearly a bounded linear operator but has no bounded extension to $\ell^\infty$. If we had one, this would be a continuous decomposition of the form $\ell^\infty = c_0 \oplus V$. This cannot happen due to Phillip's lemma, see this MSE thread.

However, if $X_1\subset X$ is a dense subspace then there is a unique continuous extension. This is essentially the continuous linear extension theorem. In general, there exist also unbounded extensions, however, only one bounded extension.

Community
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  • Without a dense subspace $X_1$, I think the first part of the conclusion still hold. Right? – math101 May 16 '15 at 04:40
  • What I really asked for help is to prove just the first statement, i.e. existence of such an extension, not necessarily uniqueness yet. – math101 May 16 '15 at 04:43
  • The existence statement still holds even if the subspace is not dense, and it's basically the Hahn-Banach Theorem. – Curran Apr 06 '21 at 16:44
  • @Curran As this answer shows, this is not true. Hahn-Banach only holds if the codomain is the scalar field. For a given space $Y$, this property holds for all $X_1\subset X$ and $T\colon X_1\to Y$ if and only $Y$ is a so-called $1$-injective Banach space, which in turn is equivalent to $Y$ being isometrically isomorphic to $C(K)$ for some extremally disconnected compact space $K$. – MaoWao Aug 11 '21 at 11:22