Let $V$ be the vector space over $\R$ of all real valued function on the interval $[0, 1]$ and let
\[W=\{ f(x)\in V \mid f(x)=f(1-x) \text{ for } x\in [0,1]\}\]
be a subset of $V$. Determine whether the subset $W$ is a subspace of the vector space $V$.

We claim that $W$ is a subspace of $V$.
To show the claim, we need to check that the following subspace criteria.

Subspace Criteria.

The zero vector in $V$ is in $W$.

For any two elements $f(x), g(x) \in W$, we have $f(x)+g(x) \in W$.

For any scalar $c$ and any element $f(x) \in W$, we have $cf(x) \in W$.

The zero vector of $V$ is the zero function $\theta(x)=0$.
Since we have
\[\theta(x)=0=\theta(1-x)\]
for any $x\in [0, 1]$, the zero vector $\theta$ is in $W$, hence condition 1 is met.

Let $f(x), g(x)$ be arbitrary elements in $W$. Then these functions satisfy
\[f(x)=f(1-x) \text{ and } g(x)=g(1-x)\]
for any $x\in [0,1]$.
We want to show that the sum $h(x):=f(x)+g(x)$ is in $W$. This follows since we have
\[h(x)=f(x)+g(x)=f(1-x)+g(1-x)=h(1-x).\]
Thus, condition 2 is satisfied.

Finally, we check condition 3. Let $c$ be a scalar and let $f(x)$ be an element in $W$.
Then we have
\[f(x)=f(1-x).\]
It follows from this that
\[cf(x)=cf(1-x),\]
and this shows that the scalar product $cf(x)$ is in $W$.

Therefore condition 3 holds, and we have proved the subspace criteria for $W$. Thus $W$ is a subspace of the vector space $V$.

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