Existence of Subspace so direct sum gives the orignal vector space.

Question

Let $V=\{f:X \to \mathbb{R}\}$ be the vector space of real valued functions.

Fix $x_0 \in X$

Define W=$\{f\in V|f(x_0)=0\}$

Is there a vector space $Y$-a subspace of V such that $V=W \oplus Y$

Attempt..

I think No.

Consider the set X to be $\mathbb{R}$.Then we are looking into the real valued functions from real numbers. choose $x_0=1$

Then we know that W is the subspace of $V$ which consists of functions $f:\mathbb{R} \to \mathbb{R}$ which vanish at $x=1$.

Now for the sake of contradiction I will assume the existence of such a subspace Y such that $V=W \oplus Y$.

Then $Y$ must contain all the functions which do not vanish at $x=1$. choose $f_1(x)=1+2x$ and $f_2(x)=-3x$

however $f_1(x)+f_2(x)=1-x$ which does not belong to $Y$ contradicting the additive closure property.(note-This is because $f_1(x)+f_2(x)=1-x\in W$)

Am I Correct?

If not please help me solving this one.

Thanks

Answer 1

Think of direct sums as partitioning all the information in the vector space into two completely unrelated pieces of data.

So, your original vector space is all functions. Now, you're given that one part of your description for "any function" is about its behavior at absolutely every point other than $x_0$.

What information is left to be specified about such a function?

Answer 2

No, it is possible to have such a subspace Y. I have constructed such one.

Let $$ \begin{align*} Y&=\{f\in V\mid f(x_0)=0\Rightarrow f(x)=0\text{ for all }x\in X\}\\ &=\{f\in V\mid f(x_0)\neq0\}\cup\{f\equiv 0\}. \end{align*} $$ Let $f\in V$. Consider $f_1\in W$ and $f_2\in Y$ defined by $$ \begin{align*} f_1(x)&=\begin{cases}0 & x=x_0\\ f(x) & x\neq x_0\end{cases}\\ f_2(x)&=\begin{cases}f(x) & x=x_0\\ 0 & x\neq x_0\end{cases} \end{align*} $$ Then $f(x)=f_1(x)+f_2(x)$ for all $x\in X$.

Therefore $V=W+Y$.

Now, it is easy to check that $W\cap Y=\{0\}$. So, $V=W\oplus Y$.