Let $X$ be a normed linear space, $M$ a linear subspace of $X$. The following proposition

If $x_0 \in X$ and not in the closure of $M$, then there exists $T \in X^{\ast}$ such that $Tx_0 \neq 0$ and $T_{|M} = 0$

is mentioned as a corollary of this version Hahn-Banach theorem:

If $S$ is a bounded linear functional on $M$, then there exists $T \in X^{\ast}$ such that $Tx = Sx$ for all $x \in M$, and $||T|| = ||S||$.

Why is this? I don't see why I can't just argue this directly. If $x \not\in \overline{M}$, then I can find a complement $W$ to $\overline{M}$ such that $X = W \oplus \overline{M}$ such that $x \in W$, a bounded linear functional $S$ on $W$ for which $Sx \neq 0$, and then extend $S$ to all of $X$ by defining $S(0,m) = 0$ for $m \in M$.


One does not always have a complement $W$ (If it does, even the Hahn Banach Theorem is trivial). See a counterexample here.

To prove this, consider $M' = M \oplus \langle x\rangle$. This is a closed subspace of $X$. Define $T(m + ax) =a$ on $M'$ and use Hahn Banach Theorem to extend this $T$ to $X$.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.