I would like to show that differentiation given by $\frac{d}{dx}:C^{k+1}[a,b]\to C^k[a,b]$ is a bounded/continuous linear transformation. I have a theorem stating that boundedness, continuous at one point, and uniformly continuous are equivalent on a normed linear space, so I only need to prove one of these.
First, I think I would like to show that $\frac{d}{dx}$ is a linear transformation, and then after show that it is bounded/continuous? That$\frac{d}{dx}$ is a linear transformation seems to me immediately clear by properties of derivatives and continuous functions, so all I need to show is that $\frac{d}{dx}$ is bounded/continuous.
I would appreciate a hint as to which of these I should try to prove, and if it is boundedness, what a reasonably bound might be, and if it is continuity, then what definition of continuity I should consider.
Thank you!
Edit: To clarify, $C^k[a,b]$ is the collection of $k$-differentiable continuous functions on the interval $[a,b]\subset\mathbb{R}$.
Edit 2: Here is the definition of boundedness that I am working with. A linear transformation $T:V\to W$ where $V$ and $W$ are normed vector spaces is bounded if there exists a $C\geq 0$ such that $||T_v||_W\leq C||v||_V$ for all $v\in V$.
I was thinking maybe $C$ would relate somehow to an exponent of $k+1$ power, because when we take the derivative this becomes a scalar multiple, but there is no guarantee that this is the largest exponent, since everything in $C^{k+1}$ has to be at Least $k+1$-differentiable?
The norm on $C^k$ is usually given as $$ \|f\|_{C^k} = \|f\|_{\infty}+\|f'\|_{\infty}+\cdots+\|f^{(k)}\|_{\infty}. $$ At least it is equivalent to this norm. Therefore, $$ \|f'\|_{C^k} = \|f'\|_{\infty}+\|f''\|_{\infty}+\cdots\|f^{(k+1)}\|_{\infty} \le \|f\|_{C^{k+1}}. $$ So $Lf=f'$ is continuous from $C^{k+1}$ to $C^{k}$.