So $V=C([-1, 1])$, the space of all real valued and continuous functions over the interval $[-1, 1]$, and $V_1 = \left\{ f_1 \in V \mid \int\limits_{-1}^{1} f(t) dt = 0\right\}$ and $V_2 = \left\{ f_2 \in V \mid f(t)=k, k \in \mathbb{R} \right\}$.
Now I know that if a function is odd, its integral, over an interval centered in the origin, is 0, so I thought that the $V_1$ is the subspace of odd functions. How every constant function is even, I said that $V_2$ is the subspace off even functions. To show that $V_1 \cap V_2 = \{0\}$ and $V_1 + V_2 = V$ I used the idea from this questions.
Was my reasoning right?
Asked
Active
Viewed 151 times
1
-
Not quite. You're right that every odd function is in $V_1$, but other functions are as well, such as $f(t) = 3t^2-1$. You're also right that every constant function is even, but not all even functions are in $V_2$. Still, you should be able to show straight from the definitions of $V_1$ and $V_2$ that $V_1\cap V_2={0}$ and $V_1+V_2=V$. – Greg Martin Nov 30 '19 at 21:22
-
Note that $V_2$ does not contain every even function, i.e. $\cos:[-1,1]\to \Bbb R$ is even but not constant. – Nightgap Nov 30 '19 at 21:23
2 Answers
1
Suppose you are given a function $f\in C([-1,1])$. Then $C:=\int_{-1}^1f(t)\mathrm{d}t<\infty$ exists. Can you modify $f$ such that the integral is zero?
Nightgap
- 1,261
0
$V_2$ is the set of constant functions. Every function can be written as a sum of a function that integrates to $0$ and a constant function. Suppose that $\int_{-1}^1 f = C$ then $f - \frac C2$ integrates to $0$ and
$$f = \underbrace{\left(f - \frac C2\right)}_{\in V_1} + \underbrace{\frac C2}_{\in V_2}. $$
Check what I just said and also show that $V_1 \cap V_2 = \{0\}$.
Trevor Gunn
- 27,041