Are $L$ and $M$ linear transformations? I want to determine whether the following transformations are linear:
$$L:\mathcal C^1(\Bbb R) \rightarrow \Bbb R, g \mapsto g'(0)-g(0)$$
$$M:\mathcal C^1(\Bbb R) \rightarrow \mathcal C^0(\Bbb R), g \mapsto g'-g$$
In order for a transformation to be linear it needs to satisfy the following two criteria:

  
*
  
*$T(\lambda v)=\lambda T(v)$
  
*$T(v+w)=T(v)+T(w)$
  

This is what I have done so far:
I tried to check the criteria for the transformation $L$:
$$L(\lambda g)=\lambda g'(0)-\lambda g(0)=\lambda(g'(0)-g(0))=\lambda L(g) \checkmark$$
$$L(a+b)=?$$
How do I check the second criterion?
Is the information that the function is differentiable and the derivative is continuous ($\mathcal C^1)$ important here? 
Thanks in advance.

$$\begin{align} L(f+g) &= (f+g)'(0) - (f+g)(0) \\ &= (f'+g')(0) - (f+g)(0) \\ &= f'(0)+g'(0) - f(0)-g(0) \\ &= f'(0) - f(0) + g'(0) - g(0) \\ &= L(f) + L(g) \quad \checkmark\end{align}$$

Is the information that the function is differentiable and the derivative is continuous (C1) important here?

Not relevant to the computations. This is just to make sure that $f'$ really does exist.

Follow your nose and use the definition: $$f \in \ker L \iff L(f) = 0 \iff f'(0) - f(0) = 0 \iff f'(0) = f(0),$$ so: $$\ker L = \{ f \in {\cal C}^1(\Bbb R) \mid f'(0) = f(0) \}.$$ On the other hand, $L$ is clearly surjective. Take $b \in \Bbb R$ and put $f(x) = b + 2bx$. So $L(f) = f'(0)-f(0) = 2b - b = b$ and $f \in {\cal C}^1(\Bbb R)$. Think alike for $M$.

I claim that $M$ is surjective. That is, given a continuous function $f \in {\cal C}^0(\Bbb R)$, I want $g \in {\cal C}^1(\Bbb R)$ such that: $$M(g) = g'-g = f.$$ We use an integrating factor: $$g'(x)-g(x) = f(x) \implies e^{-x}g'(x)-e^{-x}f(x) = e^{-x}f(x) \implies \frac{\rm d}{{\rm d}x}(e^{-x}g(x)) = e^{-x}f(x),$$ so $$g(x) = e^x \int e^{-x}f(x)\,{\rm d}x$$ does the job (I took the constant of integration being $0$, we actually have infinite $g$ satisfying $M(g) = f$).

Consider $\pi: \Bbb R^3 \to \Bbb R^3$ given by $\pi(x,y,z) = (x,y,0)$. Recall the rank-nullity theorem: $$3 = \dim \ker \pi + \dim {\rm Im}\,\pi.$$ Look at $\Bbb R^2 \hookrightarrow \Bbb R^3$ in the obvious way, and let $\Bbb R{\bf e}_3$ be the $z$ axis. We have that $\Bbb R{\bf e}_3 \subset \ker \pi$, and also $\Bbb R^2 \subset {\rm Im}\,\pi$. From this $1 \leq \dim \ker \pi$ and $2 \leq \dim {\rm Im}\,\pi$. From the rank-nullity theorem we get: $$\dim \ker \pi = 1 \quad \dim{\rm Im}\,\pi,$$ so by finite-dimensionality we get: $$\ker \pi = \Bbb R{\bf e}_3 \quad {\rm Im}\,\pi = \Bbb R^2.$$
