Prove that T is a linear transfomation I facing some difficulties to prove that T is a linear transformation according to the statements below:
V = C[−a, a] and a > 0 ,a ∈ R

Let  T : C[−a, a] → C[−a, a],
     T(g(x)) = 1/2 (g(x) + g(−x))

We say that function :
f ∈ C[-a, a] , f != 0 is eigenfunction with eigenvalue λ , if T(f) = λf.

Your help is appriciated.

 A: By definition, $T$ is a linear transformation if
\begin{align}
T(f+g) &= T(f) + T(g)\\
T(\alpha f) &= \alpha T(f)
\end{align}
for all $f,g \in C[-a,a]$ and $\alpha \in \mathbb{R}$.
I don't want to do the entire problem for you, but to show the second property:
Let $f \in C[-a,a]$ and $\alpha \in \mathbb{R}$. Then
\begin{align}
T(\alpha f)(x) &= 1/2(\alpha f(x)+\alpha f(-x))\\
&= \alpha \cdot 1/2(f(x) + f(-x))\\
&= \alpha T(f)(x)
\end{align}
(where the jump from the first to the second line uses the distributive property for real numbers)
so $T(\alpha f) = T(f)$.
A: For every $\lambda,\mu \in\mathbb{R}$ and for every $f,g\in C[-a,a]$, $$T[\lambda f(x)+\mu g(x)]=\frac{1}{2}[\lambda f(x)+\mu g(x)+\lambda f(-x)+\mu g(-x)]$$ $$=\lambda\left[\frac{1}{2}\left(f(x)+f(-x)\right)\right]+\mu\left[\frac{1}{2}\left(g(x)+g(-x)\right)\right]=\lambda T[f(x)]+\mu T[g(x)].$$ So, $T$ is a linear map. Besides, for $f$ even function, $$T[f(x)]=\frac{1}{2}[f(x)+f(-x)]=\frac{1}{2}[2f(x)]=1f(x),$$ that is, $\lambda=1$ is an eigenvalue of $T.$ Idem $T[f(x)]=0f(x)$ for $f$ odd function.
