Show the differential operator is bounded I need to show the differential operator is bounded. I believe it is bounded on $C[0,1]$ with the sup norm by 1, but I am struggling to show this. What I have so far:
NTS: $\|Tf\|\leq\|T\|\|f\|$ Where:

\begin{equation} \|Tf\|=\sup_{\|f\|=1}\|Tf\|=\sup_{\|f\|=1} \sup_{x\in [0,1]} |f'(x)|=\sup_{\|f\|=1} \sup_{x\in [0,1]} \left|\frac{\lim_{h\rightarrow 0} f(a+h)-f(a)}{h}\right|
\end{equation}

Not entirely sure if I should have written the derivative like that in the last equality above. I need a hint to see where to go next.

Thanks in advance!
Hint: Denote by
$$D:(C^{1}[0,1],\Vert\cdot\Vert_{\infty})\rightarrow(C[0,1],\Vert\cdot\Vert_{\infty})$$
the differential operator. Let $f_{n}(x)=\sin(nx)$. Note that $\Vert f_{n}\Vert_{\infty}=1$ (for $n\geq 2$). However, $(Df_{n})(x)=n\cos(nx)$, and hence $\Vert Df_{n}\Vert_\infty=n$. What does this mean?
What about the effect of differentiating a sequence f sub n = sin(2pi* n* x) where n is integer &gt; 0?
You get 2pi *n *cosine, whose supremum is infinite