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I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help to confirm my thinking on Proposition 11.3.5 - 1 ...

Garling's statement and proof of Proposition 11.3.5 reads as follows:

View attachment 8963

In Proposition 11.3.5 - 1, given the definition of \(\displaystyle A^{ \bot }\), we are required to prove that

\(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle a, x \rangle = 0\) for all \(\displaystyle a \in A \}\)

We need to show

\(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle x, a \rangle = 0\) for all \(\displaystyle a \in A \}\)

By definition \(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle x, a \rangle = 0\) for all \(\displaystyle a \in A \}\)

\(\displaystyle \Longrightarrow A^{ \bot } = \{ x \in V \ : \ \overline{ \langle x, a \rangle } = 0\) for all \(\displaystyle a \in A \}\)

Now suppose \(\displaystyle \langle a, x \rangle = u + i v\)

Then \(\displaystyle \overline{ \langle x, a \rangle } = \langle a, x \rangle = u + i v\) by definition of an inner product ...

and so \(\displaystyle \langle x, a \rangle = u - i v\) Thus if \(\displaystyle \langle a, x \rangle = 0\) then \(\displaystyle u = v = 0\)

... which implies \(\displaystyle \langle x, a \rangle = 0\) Thus if

\(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle a, x \rangle = 0 for all a \in A \}\)

then

\(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle x, a \rangle = 0 for all a \in A \} \)

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help to confirm my thinking on Proposition 11.3.5 - 1 ...

Garling's statement and proof of Proposition 11.3.5 reads as follows:

View attachment 8963

In Proposition 11.3.5 - 1, given the definition of \(\displaystyle A^{ \bot }\), we are required to prove that

\(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle a, x \rangle = 0\) for all \(\displaystyle a \in A \}\)

We need to show

\(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle x, a \rangle = 0\) for all \(\displaystyle a \in A \}\)

**Proof:**By definition \(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle x, a \rangle = 0\) for all \(\displaystyle a \in A \}\)

\(\displaystyle \Longrightarrow A^{ \bot } = \{ x \in V \ : \ \overline{ \langle x, a \rangle } = 0\) for all \(\displaystyle a \in A \}\)

Now suppose \(\displaystyle \langle a, x \rangle = u + i v\)

Then \(\displaystyle \overline{ \langle x, a \rangle } = \langle a, x \rangle = u + i v\) by definition of an inner product ...

and so \(\displaystyle \langle x, a \rangle = u - i v\) Thus if \(\displaystyle \langle a, x \rangle = 0\) then \(\displaystyle u = v = 0\)

... which implies \(\displaystyle \langle x, a \rangle = 0\) Thus if

\(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle a, x \rangle = 0 for all a \in A \}\)

then

\(\displaystyle A^{ \bot } = \{ x \in V \ : \ \langle x, a \rangle = 0 for all a \in A \} \)

*Peter***Is the above proof correct? Is there a better way to prove the above ...?**#### Attachments

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