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Suppose $X$ is a complex inner product space of complex valued functions that is closed under conjugation.

Is it true that $\|x\| = \| \overline{x} \|$ for all $x$?

If not, is there a simple counterexample?

(This question was triggered by Making a complex inner product symmetric.)

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    $\begingroup$ What does $\bar{x}$ mean when $x$ is a vector in some generic inner product space? If $\overline{\left\langle x,x\right\rangle}=\left\langle\bar{x},\bar{x}\right\rangle$, then the result is true from i.p.s. axioms, but aside from vector spaces with coordinates, I'm not sure what $\bar{x}$ means. $\endgroup$ – alex.jordan Oct 19 '14 at 6:56
  • $\begingroup$ @alex.jordan: Good point, I have elaborated a bit. $\endgroup$ – copper.hat Oct 19 '14 at 7:03
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I believe I have a counterexample to your statement about the equality $\|x\|=\|\bar{x}\|$.

Let $S=\{a, b\}$ be a set with two entries. Define a complex-valued function $f$ on $S$ by $$f(a)=1+i, \quad f(b)=1$$ and consider the two-dimensional space $V=\operatorname{Span}\{f, \bar{f}\}$ over $\mathbb{C}$ equipped with the inner product $$(\phi, \psi)=[\phi]_{\mathcal{B}}^*P[\psi]_{\mathcal{B}}$$ where $\mathcal{B}=\{f, \bar{f}\}$ and $*$ denotes the conjugate transpose and $P=\begin{bmatrix}1&1\\1&2\end{bmatrix}$.

Now consider $g=3if+\bar{f}\in V$. Then $$\|g\|^2= \begin{bmatrix}-3i&1\end{bmatrix} \begin{bmatrix}1&1\\1&2\end{bmatrix} \begin{bmatrix}3i\\1\end{bmatrix}=11 $$ whereas, $$\|\bar{g}\|^2= \begin{bmatrix}1&3i\end{bmatrix} \begin{bmatrix}1&1\\1&2\end{bmatrix} \begin{bmatrix}1&-3i\end{bmatrix}=19. $$ The same example shows that the function defined by $\theta(x, y)=(\bar{x}, y)$ is not symmetric in general which is a counterexample to the statement of the problem linked to this question.

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Not true. Take $f(x,y)=x+iy$ and $X$ the space spanned by $f(x)$ and $\overline{f}(x,y)=x-iy$ over $\mathbb C$. Now, every element o $X$ is of the form $u=af+b\overline{f}$.

Define the inner product $$ \langle u_1,u_2\rangle=a_1\overline a_2+4 b_1\overline b_2. $$ Then $$ \|\,\overline f\|=2\|\,f\|. $$

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