Given that $V_1$ and $V_2$ are subspaces of $V$, how do I show that $V = V_1 \oplus V_2$?

Question

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?

Answer 1

$V_2$ 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 $V_1 \cap V_2 = \{0\}$.

Answer 2

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?