Norm in $l^1$ is not differentiable

Question

How to prove that norm in $l_1$ is not differentiable? The norm is $\|x\| = \sum\limits_{n=1}^{\infty} |x_n|$.

I know the definition of derivative: $\lim\limits_{h \to 0} \frac{\|x+h\| - \|x\| - A(x)(h)}{\|h\|} =0 $, but I don't know how to use it here. For example, after plugging $x=e_1$ and $h=te_2$ I get $\lim\limits_{h \to 0} \frac{\|e_1 +te_2\| - \|e_1\| - A(e_1)(te_2)}{\|te_2\|}$ but what to do with $A(e_1)(te_2)$?

Answer 1

Some remarks:

• Every norm is non-differentiable at the origin, but presumably this isn't what you are asking
• There are multiple notions of differentiability (Frechét and Gâteaux being best known).

Anyway, one can show that there is no derivative of the norm at $e_1$ in the direction of $e_2$. This is because $\|e_1+te_2\|=1+|t|$ which is a non-differentiable function at $0$.

More formally: suppose you had a derivative at $e_1$, i.e., a linear operator $A$ such that $$\lim\limits_{t \to 0} \frac{\|e_1+te_2\| - \|e_1\| - A(te_2)}{\|te_2\|} =0$$ (This means using $h=te_2$ where $t\in\mathbb{R}$). Simplify the above to $$\lim\limits_{t \to 0} \frac{1+|t|-1 - tA(e_2)}{|t|} =0$$ and you will see a contradiction: $$\lim\limits_{t \to 0} \frac{-tA(e_2)}{|t|} =-1$$ which cannot hold for any value of $A(e_2)$.