prove that the derivate operator is unbounded

Question

Consider the vector space \begin{equation*} C^1[0,\frac{1}{2}]=\{f:[0,\frac{1}{2}] \rightarrow \mathbb{C} \, : \, f \text{ is differentiable and $f'$ is continuous }\} \end{equation*} Then $C^1[0,\frac{1}{2}]$ is a subspace of $C[0,\frac{1}{2}]$. We now equip both spaces with the supremums-norm, and consider the mapping \begin{equation*} \mathcal{D}: C^1[0,\frac{1}{2}] \rightarrow C[0,\frac{1}{2}], \quad (\mathcal{D}f)(x):=f'(x), \, x\in[0,\frac{1}{2}]. \end{equation*} I want to prove $\mathcal{D}$ is a linear unbounded operator. It is obviously that $\mathcal{D}$ is linear and to prove that it's unbounded, the book (where I take this problem) says I have to take $f_n(x)=x^n$. My idea now is prove: \begin{equation*} \|\mathcal{D}f_n\|_{\infty} > n \|f_n\|_{\infty} \text{ for all } n\in \mathbb{N}. \end{equation*} I know that $\frac{1}{\|f_n\|}\geq 2^n$, but I don't know what to do with $\|\mathcal{D}f_n\|_{\infty}$

Answer 1

Just calculate it:

$$\|\mathcal Df_n\|_\infty \ =\ \max_{x\in [0,\frac12]} nx^{n-1}\ =\ n\frac1{2^{n-1}}\,. $$

So, $\|\mathcal Df_n\|_\infty=2n\|f_n\|_\infty$.