数学 Mathematics B GENG0002W1-01

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这是一份southampton南安普敦大学GENG0002W1-01作业代写的成功案例

数学 Mathematics B GENG0002W1-01

from which we see that $f+(-f)=0$. The additive inverse law follows. For the distributive laws note that for real numbers $a, b$ and continuous functions $f, g \in V$, we have that for all $0 \leq x \leq 1$,
$$
a(f+g)(x)=a(f(x)+g(x))=a f(x)+a g(x)=(a f+a g)(x),
$$
which proves the first distributive law. For the second distributive law, note that for all $0 \leq x \leq 1$,
$$
((a+b) g)(x)=(a+b) g(x)=a g(x)+b g(x)=(a g+b g)(x),
$$
and the second distributive law follows. For the scalar associative law, observe that for all $0 \leq x \leq 1$,
$$
((a b) f)(x)=(a b) f(x)=a(b f(x))=(a(b f))(x),
$$
so that $(a b) f=a(b f)$, as required. Finally, we see that
$$
(1 f)(x)=1 f(x)=f(x),
$$

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GENG0002W1-01 COURSE NOTES :

First let $f(x), g(x) \in V$ and let $c$ be a scalar. By definition of the set $V$ we have that $f(1 / 2)=0$ and $g(1 / 2)=0$. Add these equations together and we obtain
$$
(f+g)(1 / 2)=f(1 / 2)+g(1 / 2)=0+0=0 .
$$
It follows that $V$ is closed under addition with these operations. Furthermore, if we multiply the identity $f(1 / 2)=0$ by the real number $c$ we obtain that
$$
(c f)(1 / 2)=c \cdot f(1 / 2)=c \cdot 0=0 .
$$
It follows that $V$ is closed under scalar multiplication. Now certainly the zero function belongs to $V$, since this function has value 0 at any argument. Therefore, $V$ contains an additive identity element. Finally, we observe that the negative of a function $f(x) \in V$ is also an element of $V$, since
$$
(-f)(1 / 2)=-1 \cdot f(1 / 2)=-1 \cdot 0=0 .
$$