A function in a Real Vector Space V (where V is the set of all complex-valued functions f on the real line ), which is NOT real-valued. Question:
Let $V$ be the set of all complex-valued functions $f$ on the real line such that
(for all $t \in \Bbb R) \ \ f(-t) = \overline {f(t)} $.
The bar denotes complex conjugation.
(i) Show that $V$, with the operations
$(f + g)(t) = f(t) + g(t)$ and
$(cf)(t) = cf(t)$
is a vector space over the field of real numbers but NOT a vector space over the filed of complex numbers.
(ii) Give an example of a function in $ V$
which is  NOT  real-valued.
My Try:
(i) $ (f + g)(-t) = f(-t) + g(-t) = \overline {f(t)} + \overline {g(t)} = \overline {f(t) + g(t)} = \overline {(f + g)(t)}$ .
$ (cf)(-t)=cf(-t)=c \overline{f(t)}= \overline{c f(t)}      $ .
Hence, a subset $V$ of the real vector space of all functions from $\Bbb R $ to $\Bbb C $ is closed
under addition and multiplication by real numbers. This means that $V$ is a
subspace and satisﬁes all properties of a vector space.
Hence  $V$ is a vector space over the field of real numbers.
BUT I DON'T KNOW HOW TO PROVE that $V$ is NOT a Vector Space over the filed of complex numbers. Please help me solve this.
(ii) An example of a non-real-valued function in $V$ is $f(t) = it$
Now, how will I prove that this function is in $V$ ?
My attempt is given below:
$f(t) = it$
$\Rightarrow f(-t)= i(-t)=-it$
or $f(t) = it$
$\Rightarrow f[-(-t)]=it$
$\Rightarrow \overline {f(-t)}=it$
$f(t)+f(-t)=0 \in V$
This is University exam question and I'm preparing for the same exam. Part (i) is for 10 marks and Part (ii) is for 5 marks. 
If there is any mistake, please correct the solution given by me and help me complete the solution.
Thanks.
 A: If $\,V\,$ were a complex linear space then it'd fulfill
$$(if)(-x)=\overline{(if)(x)}\;,\;\;\forall\,f\in V$$
Yet for $\,f(x)=1\;,\;\;\forall\,x\in\Bbb R\;$ , we get:
$$\begin{align*}(if)(-x)&=i(f(-x))=i(1)=i\\
\overline{(if)x}&=\overline{i(f(x)}=\overline i\overline{f(x)}=-i(1)=-i\end{align*}$$
For part two I think it's easier and simpler:
$$\forall\,t\in\Bbb R\;,\;\;f(t):=it\implies f(-t)=-it=\overline{it}=\overline{f(t)}$$
A: It will be easier to answer your questions in reverse order.
Part (ii)
The necessary and sufficient condition for a function $f$ to belong to $V$ is that $f(-t)=\overline{f(t)}$. If $f(t)=it$ then we have that $f(-t)=-it=\overline{it}=\overline{f(t)}$ as desired.
Part (i)
Once again let $f(t)=it$. From before we know that $f\in V$, but if $V$ were truly a complex vector space, it would follow that $g(t)=(a+bi)f(t)=-bt+ati$ would also belong to $V$ (for all $a$ and $b$). However, this is not the case because $g(-t)=bt-ati\neq -bt-ati = \overline{g(t)}$. So we must conclude that $V$ is not a complex vector space.
A: Let $f\in V$, $c\in\mathbb C$. Then $(cf)(-x)=c\cdot f(-x)=c\cdot \overline{f(x)}$ and this is in general $\ne\overline{(cf)(x)}=\overline c\cdot\overline{f(x)}$. In concreto, it suffices to find $c\in\mathbb C$ with $c\ne\overline c$ (for example $c=i$) and $f\in V$, $x\in \mathbb R$ with $f(x)\ne 0$ (for example $x=1$ with  the function used as example in (ii) or simlpy $t\mapsto t^2$).

For (ii) you seem to be showing $f(t)+f(-t)=0$, but what you really need is (per definition of $V$), $f(-t)=\overline{f(t)}$. And indeed $f(-t)=-it=\overline{it}=\overline{f(t)}$.
