What is the Gateaux derivative of the norm $||\cdot||_p$ on $\mathcal{L}^{\mathbb{R}}_{p}([0,1])$?

Question

To get the Gateaux derivative of the norm $||\cdot||_p$ on $\mathcal{L}^{\mathbb{R}}_{p}([0,1])$, I have to do a manual computation of the derivative and, then, use the definition of Gateaux differentiablity. Now, kindly check out what I have done so far. Define $f:U\longrightarrow\mathbb{R}:\,x\mapsto$ $||x||_p,\,p\in(1,\infty),$ where $U$ is a closed and convex subset of $\mathcal{L}^{\mathbb{R}}_{p}([0,1]).$ Let $x\in U\setminus\{0\}.$ Then, by direct computation, \begin{align} Df(x)&=D\left(\int_{0}^{1}|x(t)|^pdt \right)^{\frac{1}{p}}\\&= \frac{1}{p}\left(\int_{0}^{1}|x(t)|^pdt \right)^{\frac{1}{p}-1} D\left(\int_{0}^{1}|x(t)|^pdt \right)\\&= \frac{1}{p}\left(\int_{0}^{1}|x(t)|^pdt \right)^{\frac{1-p}{p}} \int_{0}^{1}p\,|x(t)|^{p-1}D|x(t)|dt \\&= ||x||_p^{1-p} \int_{0}^{1}|x(t)|^{p-1}\frac{x(t)}{|x(t)|}dt\\&= ||x||_p^{1-p} \int_{0}^{1}|x(t)|^{p-1}\text{sign}(x(t)) dt. \end{align} Applying $(Df)(x)$ to any $h\in U$ close to $x$, we get \begin{align} \langle Df(x),h\rangle&=||x||_p^{1-p} \int_{0}^{1}|x(t)|^{p-1}\text{sign}(x(t))h(t)dt. \end{align} Let me start my question here: I was thinking I should get \begin{align} \langle Df(x),h\rangle&=\dfrac{1}{||x||_p } \int_{0}^{1}x(t)h(t)dt. \end{align} Please, what is wrong with my thinking. Any way out?

Answer 1

The direct computation is correct, your expectations were wrong. To see this, calculate the derivative of $$ x\mapsto \sqrt[p]{|x|^p+1}. $$ The chain rule will produce this factor $sign(x) |x|^{p-1}$. Also, no cancellation can happen between $|x|^{p-1}$ and powers of $|x|^p+1$.