Show that the map $B(f) = f^{\prime}(0)$ defines a linear functional on the normed vector space $(C^{(1)}[0,1], \| \cdot \|)$ that is not bounded.

Let $C^{(1)}[0,1]$ be the space of all continuous functions on $[0,1]$ that have continuous derivative. (This space is called the space of continuously differentiable functions.) Define a norm $\|\cdot \|$ on $C^{(1)}[0,1]$ by $\| f \|=\max _{t \in T}|f(t)|.$ Show that the map $$B(f) = f^{\prime}(0)$$ defines a linear functional on the normed vector space $(C^{(1)}[0,1], \| \cdot \|)$ that is not bounded.

My attempt:

For any $f,g \in C^{(1)}[0,1],$ we have $B(f+g) = (f + g)^{\prime}(0) = f^{\prime}(0) + g^{\prime}(0) = B(f) + B(g).$ Therefore, $B$ is a linear functional.

To show that $B$ is not bounded, we wish to show that for any $M > 0$, there exists $f \in C^{(1)}[0,1]$ such that $\| f \| \leq 1$ but $|B(f)| > M.$

Let $M > 0$ be given. I have trouble constructing a function with maximum $\leq1$ and has derivative at $x = 0$ greater than $M.$

I plan to construct $f$ which looks similar to $\sin(x)$. For any given $M > 0$, I can draw $\sin(x)$ such that it is above the line $y = Mx$. However, I do not know how to formulate this phenomena using mathematical symbols.

Define $f_n(x) = \sin (nx)$ then $\sup|f_n(x)| =1$ however $f_n'(x) = n\cos nx$ hence $$|f'_n(0)| =n.$$