I'm struggling to follow a certain proof in Luenberger's book "Optimization by Vector Space Methods". If anyone could help shed some light on what I'm missing, I would certainly appreciate it. Let me start with a couple of definitions.

Definition 1 Let $S$ be a subset of a normed linear space $X$. The orthogonal complement of $S$, denoted $S^\perp$, consists of all elements $x^* \in X^*$ ($X^*$ denotes the continuous dual space) orthogonal to every vector in $S$.

Definition 2 Given a subset $U$ of the dual space $X^*$, we define the orthogonal complement of $U$ in $X$ as the set $^\perp U \subset X$ consisting of all elements in $X$ orthogonal to every vector in $U$.

Definition 3 The vectors $x \in X$ and $x^* \in X^*$ are said to be orthogonal if $\langle x, x^*\rangle =0$.
$\langle \cdot,\cdot \rangle$ denotes the dual pairing.

Theorem Let $M$ be a closed subspace of a normed space $X$. Then $^\perp[M^\perp] = M$.

Proof It is clear that $M \subset{}^\perp[M^\perp]$. To prove the converse, let $x \notin M$. On the subspace $[x + M]$ generated by $x$ and $M$, define the linear functional $f(\alpha x + m) = \alpha $ for $m \in M$. Then \begin{equation} \|\,f\|=\sup_{m\in M}\frac{f(x+m)}{\|x+m\|}=\frac{1}{\inf_{m}\|x+m\|} \end{equation} and since $M$ is closed, $\|\,f\| < \infty$. Thus by the Hahn-Banach theorem, we can extend $f$ to an $x^* \in X^*$. Since $f$ vanishes on M, we have $x^* \in M^\perp$ But also $\langle x, x^*\rangle = 1$ and thus $x \notin{}^\perp[M^\perp]$.

Comments I'm struggling to understand the necessity of the subspace being closed. Would bounded not suffice?

My understanding of the proof is as follows. Using the definition of the norm of an $f$ in the dual space and positivity we get the first equality in the displayed equation. Noticing that $\alpha=1$ and interchanging $\sup$ and $\inf$ by taking the $\inf$ into the denominator gives the second equality. I would have thought from here we would be able to guarantee finiteness of the norm. We then use the Hahn-Banach Theorem to extend $f$ to $x^* \in X^*$. Then using $\alpha=0$ implies $f=0$ on $M$. Thus $f(x_1)=\langle x^*,x_1\rangle=0$ where $x_1 \in M$ so they are orthogonal. However, $\alpha=1$ implies $f(x)=1=\langle x^*,x\rangle$ hence $x \notin{}^\perp[M^\perp]$.

  • $\begingroup$ Which subspace of $X$ is bounded? $\endgroup$ – Paul Frost Aug 19 '18 at 14:39
  • $\begingroup$ Check out my edit to see how to prevent a left superscript from being attached to the symbol to its left. $\endgroup$ – joriki Aug 19 '18 at 14:53

If $M$ is not closed, then for each $x \in \overline{M} \backslash M$ there is a sequence $(m_n)$ in $M$ such that $\lVert x - m_n \rVert \to 0$ as $n \to \infty$. But then $\inf_m \lVert x + m \rVert = 0$ since $-m_n \in M$.

Note that no linear subspace except the trivial one (having $0$ as its only element) is bounded.

  • $\begingroup$ I see. Sorry. I realise now that that was a somewhat silly thing not to notice. Thanks. $\endgroup$ – mark Aug 19 '18 at 21:28

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.