0
$\begingroup$

Let $X$ be a normed space and $Y$ be a Banach space. The bounded linear extension theorem states that a bounded linear operator $T : M \to Y$, where $M \subseteq X$ is dense, can be extended to a bounded linear operator $T_0 : \overline{M} \to Y$.

To show that the completeness of $Y$ is essential, I considered the spaces $X = c_0 (\mathbb{N})$ and $Y = c_{00} (\mathbb{N})$, where $c_{00} (\mathbb{N})$ denotes the space of sequences with finitely many non-zero entries.

Clearly, the identity $I : Y \to Y$ is bounded. However, I have difficulties to show that $I$ cannot be extended to a bounded operator $I_0 : X \to Y$.

Any help is appreciated. Thanks in advance.

$\endgroup$
1
$\begingroup$

There's nothing to see, actually. You can find a sequence $(x_n) \subset Y$ such that $x_n \to x \in X \setminus Y$. Then $(I_0(x_n))$ must converge to $I_0(x) \in Y$ (because we want $I_0$ continuous). But $I_0(x_n) = I(x_n) = x_n \to x \not \in Y$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy