Is $\|x\| = \| \overline{x} \|$ in an inner product space? 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.)
 A: 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.
A: 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
$$
\|\,\overline f\|=2\|\,f\|.
$$
