To show that the set of all continuous functions $F[a,b]=\{f(x)\mid x\in [a,b],{\int }_a^{b}f(x)dx=0\}$ is a subspace of $C[a,b]$ (the set of all continuous functions on $[a,b]$), we need to demonstrate three things:
1. The zero function, $0(x)=0$ for all $x\in [a,b]$, is in $F[a,b]$.
2. $F[a,b]$ is closed under addition, meaning that if $f(x)$ and $g(x)$ are in $F[a,b]$, then $f(x)+g(x)$ is also in $F[a,b]$.
3. $F[a,b]$ is closed under scalar multiplication, meaning that if $f(x)$ is in $F[a,b]$ and $c$ is a scalar, then $c·f(x)$ is also in $F[a,b]$.
Let's prove each of these statements:
1. To show that the zero function is in $F[a,b]$, we need to verify that ${\int }_a^{b}0dx=0$. Since the integral of the constant function $0$ over any interval is zero, this condition is satisfied. Therefore, $0(x)=0$ is in $F[a,b]$.
2. Now let's prove closure under addition. Suppose $f(x)$ and $g(x)$ are in $F[a,b]$. We need to show that $f(x)+g(x)$ is also in $F[a,b]$.
By the linearity of integration, we have:
${\int }_a^b(f(x)+g(x))dx={\int }_a^{b}f(x)dx+{\int }_a^{b}g(x)dx.$
Since ${\int }_a^{b}f(x)dx=0$ and ${\int }_a^{b}g(x)dx=0$ (since $f(x)$ and $g(x)$ are in $F[a,b]$), we have:
${\int }_a^b(f(x)+g(x))dx=0+0=0.$
Thus, $f(x)+g(x)$ satisfies the condition for being in $F[a,b]$, and $F[a,b]$ is closed under addition.
3. Finally, let's prove closure under scalar multiplication. Suppose $f(x)$ is in $F[a,b]$ and $c$ is a scalar. We need to show that $c·f(x)$ is also in $F[a,b]$. Using the linearity of integration, we have:
${\int }_a^b(c·f(x))dx=c{\int }_a^{b}f(x)dx.$
Since ${\int }_a^{b}f(x)dx=0$ (since $f(x)$ is in $F[a,b]$), we have:
${\int }_a^b(c·f(x))dx=c·0=0.$
Thus, $c·f(x)$ satisfies the condition for being in $F[a,b]$, and $F[a,b]$ is closed under scalar multiplication.
Since we have shown that $F[a,b]$ contains the zero function, is closed under addition, and is closed under scalar multiplication, we can conclude that $F[a,b]$ is a subspace of $C[a,b]$.
Now, let's provide an example of a subset of $C[a,b]$ that is not a subspace. Consider the subset $G[a,b]=\{f(x)\mid x\in [a,b],f(x)>0\}$. This subset consists of continuous functions on $[a,b]$ that are strictly positive for all $x$ in the interval.
To show that $G[a,b]$ is not a subspace, we need to demonstrate that it violates at least one of the three subspace properties.
Let's consider closure under addition. Take two functions $f(x)=1$ and $g(x)=2$, both defined on $[a,b]$. These functions are in $G[a,b]$ since they are positive for all $x\in [a,b]$. However, their sum, $f(x)+g(x)=1+2=3$, is not in $G[a,b]$ because it takes the value $3$, which is not strictly positive for all $x\in [a,b]$.
Since $G[a,b]$ fails to satisfy closure under addition, it is not a subspace of $C[a,b]$.