Fréchet derivative of square root as a map from L1 to L2

Question

If we pick the subset $ U = \{ f\in L^1(\Omega) | f\geq 0\ \text{in $\Omega$}\} $ of $ L^1(\Omega) $, the square root is well-defined on functions in $ U $. Furthermore, it is clear that for any $ f \in U $, we have $$ \int_\Omega |\sqrt{f}|^2 = \int_\Omega |f| < \infty $$ thus $ \sqrt{f} \in L^2(\Omega) $ for all $ f \in U $.

Making the inequality in $ U $ strict instead, let's call this set $ U' $, intuitively we should be able to consider derivatives of $ \sqrt{\cdot} : U' \to L^2(\Omega) $. For the usual $ \sqrt{\cdot} : \mathbb{R}_+ \to \mathbb{R}_+ $, it is not hard to see that the Fréchet derivative at $ x $ in direction $ h $ is given by $ h(2\sqrt{x})^{-1} $, so I'm guessing that this should carry over.

Considering $$ \left\|\sqrt{x+h} - \sqrt{x} - \frac{h}{2\sqrt{x}}\right\|_{L^2(\Omega)} $$ we can rewrite it by division and multiplication by $ \sqrt{x+h}+\sqrt{x} $ in the first two terms, which leaves us with $$ \left\|\frac{h}{\sqrt{x+h} + \sqrt{x}} - \frac{h}{2\sqrt{x}}\right\|_{L^2(\Omega)}. \tag{$\ast$} $$ Now comes the hard part, where I'm unsure. We would like to show that ($\ast$) is $ \mathcal{O}\left(\|h\|_{L^1(\Omega)}^2\right) $; or at least $ \mathcal{O}\left(\|h\|_{L^1(\Omega)}^p\right) $, where $ p > 1 $.

Answer 1

To begin with, $U$ has empty interior because one can make any function negative somewhere by subtracting a "tall narrow peak" function from it, without changing much in terms of $L^1$ norm.

So, if the definition of Fréchet differentiability requires an open set, the matter is closed. The rest of the post shows that things go wrong even without this requirement (one can define differentiability as the existence of locla linear approximation, which does not require an open domain).

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The square root map $L^1\to L^2$ is nowhere Fréchet differentiable, because of the following unpleasant property: if $f\in U\setminus \{0\}$, then $$\limsup_{g\to f}\frac{\|\sqrt{g}-\sqrt{f}\|_{2}}{\|g-f\|_1} = \infty$$ Indeed, since $f$ is not the zero function, there exists a positive number $\delta>0$ and a set of positive measure $A$ such that $\delta\le f\le 4\delta$ on $A$.

For $\epsilon\in (0,\delta)$ let $E\subset A$ be some set of measure $\epsilon$, and define $g=f+8\delta \chi_E$. Clearly, $\|g-f\|_1 = 8\epsilon\delta$. On the other hand, on the set $E$ we have $$\sqrt{g}\ge \sqrt{9\delta} = 3\sqrt{\delta} $$ which implies $$\sqrt{g}-\sqrt{f}\ge 3\sqrt{\delta}-2\sqrt{\delta} = \sqrt{\delta}$$ It follows that $$ \|\sqrt{g}-\sqrt{f}\|_2 \ge \sqrt{\int_E \delta } = \sqrt{\epsilon\delta} $$ In conclusion, $$ \frac{\|\sqrt{g}-\sqrt{f}\|_{2}}{\|g-f\|_1} \ge \frac{1}{8\sqrt{\epsilon\delta} } $$ Since $\epsilon>0$ could be arbitrarily small, the claim follows.

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Informal explanation of the above: if the derivative at $f$ existed, it would be the linear map $D_f\phi = \phi/(2\sqrt{f})$. Even if the denominator is bounded away from zero, we have a problem with the numerator, as $\phi$ is not necessarily in $L^2$.