Define the functional $I:L^2(\Omega) \to \mathbb{R}$ by $$I(u) = \int_\Omega |u|$$ I want to know if $I$ is strictly differentiable. I.e, does the limit $$\lim_{u \to \bar u,\\ v \to \bar u}\frac{I(u)-I(v)-I'(\bar u)(u-v)}{\lVert u-v\rVert }=0$$ for some continuous linear $I'(\bar u):L^2(\Omega) \to \mathbb{R}$. I guess it is not true since the derivative without the integrals is not linear.
Strict differentiability of a functional
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functional-analysis
derivatives
1 Answers
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No. It is not even differentiable: Consider $\bar u = 0$ and $v = 1$. Then, $$\lim_{t \searrow 0} \frac{I(\bar u + t v)-I(\bar u)}{t} = |\Omega|,$$ and $$\lim_{t \searrow 0} \frac{I(\bar u - t v)-I(\bar u)}{t} = |\Omega|.$$ Hence, the directional derivative in direction $v$ is equal the directional derivative in direction $-v$.