# 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.

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Consider the space $\mathcal{F}$ of all complex valued functions on the real line. That is - easily seen - an $\mathbb{R}$ vector space as well as a $\mathbb{C}$ vector space. $V \subset \mathcal{F}$ is an $\mathbb{R}$-subspace. What condition must an $\mathbb{R}$-subspace of a $\mathbb{C}$ vector space fulfill to be a $\mathbb{C}$-subspace? – Daniel Fischer Jul 22 '13 at 12:12
i) If $c \not \in \mathbb{R}$ then $c\not=\bar c$ and $c\overline{f(t)}\not=\overline{cf(t)}$. – Angela Richardson Jul 22 '13 at 12:14
For (i) you forgot to show that $V\ne \emptyset$, but that is clear anyway from the example in (ii). – Hagen von Eitzen Jul 22 '13 at 13:03
Thank you all... – Jackson George Jul 22 '13 at 21:20

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:

$$\forall\,t\in\Bbb R\;,\;\;f(t):=it\implies f(-t)=-it=\overline{it}=\overline{f(t)}$$

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$f(x)=i$ does not belong to $V$. – Christian Bueno Jul 22 '13 at 14:42
Why not, @ChristianBueno ? – DonAntonio Jul 22 '13 at 14:50
@DonAntonio : Hi DonAntonio, Please edit the above answer to: $(if)(-x)=i(f(-x))=i\overline{f(x)}=i(-i)=(-1)(-1)=1$ $\overline{(if)x}=\overline{i(f(x))}=\overline i\overline{f(x)}=-i(-i)=(1)(-1)=-1$ $\\$ Thanks. – Jackson George Jul 22 '13 at 17:24
@DonAntonio Because $f(-x)=i\neq -i = \overline{i}=\overline{f(x)}$. Since it fails this criterion it fails to be in $V$. On the other hand, a constant real function would work. – Christian Bueno Jul 22 '13 at 19:50
Thanks @ChristianBueno , you're right. – DonAntonio Jul 22 '13 at 20:02

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.

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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 $f(-t)=-it=\overline{it}=\overline{f(t)}$.

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