If $[T f](x) = f(x) \cos(x)$, show that $T$ is a linear map. Let $C(\mathbb{R})$ be the vector space (over $\mathbb{R}$) of continous functions $f : \mathbb{R} \to \mathbb{R}$.
Define $T : C(\mathbb{R}) \to C(\mathbb{R})$ by
$$[T f](x) = f(x) \cos(x)$$
Show that $T$ is a linear map.
I know I should prove the additivity and homogeneity but I don't know how to prove it for this specific problem.
 A: Hint: You have to prove linearity in $f$, not in $x$.
A: It's just like any other problem with the same goal. In order to show that, for a vector space $V$, the mapping $T:V\to V$ is linear, you need to:


*

*Prove that for all vectors $v, w\in V$, you have $T(v+w) = T(v) + T(w)$. In your case, $v,w$ are functions, so take two continuous functions $g,f\in \mathcal C(\mathbb R)$. Therefore, you need to prove, for an arbitrary pair $f,g$ of continuous functions, that $T(f+g)$ is equal to $Tf + Tg$. Now, remember that two functions on the same domain are equal if and only if their values are equal on all elements of the domain. Now, ask yourself: for a given $x\in\mathbb R$, what is $(T(f+g))(x)$? What is $(Tf)(x) + (Tg)(x)$?

*Similarly, prove that for a scalar $\alpha\in\mathbb R$ and a function $f$ you have $T(\alpha f) = \alpha Tf$.

A: Well, we have
$$T(f(x)) = f(x) \cos(x)$$
For linearity of $T$ we have to show
\begin{align*}
T(f(x)+g(x)) &=T(f(x))+T(g(x)) \\
T(\alpha f(x)) &= \alpha T(f(x))
\end{align*}
for some  $\alpha \in \mathbb{R}$. This is quite straightforward, i.e.
$$T(f(x)+g(x)) = (f(x)+g(x)) \cos(x) = f(x)\cos(x) + g(x)\cos(x) = T(f(x))+T(g(x))$$
$$T(\alpha f(x)) = (\alpha f(x)) \cos(x) = \alpha f(x)\cos(x) = \alpha (f(x) \cos(x)) = \alpha T(f(x))$$
and you're done.
