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Here $\Omega$ is a bounded domain.

Define $f:H^1 \times H^1 \to \mathbb{R}$ by $$f(u,v) = \int_\Omega (v-u)^+.$$ (Note the $H^1$, not $L^2$).

I want to find an expression for the (directional or Frechet) derivative $f'(u,v)(h_1, h_2)$.

Define $I(w) = \int_\Omega w$, then $I'(w)h = \int_\Omega h$. Now I think we need to differentiate $g(x) : = (x)^+$ and somehow combine, but I don't know the details. Could someone help please?

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    I would expect $f'(u,v)(h_1, h_2) = \int_{\{v-u > 0\}} (h_2 - h_1) + \int_{\{v - u = 0 \}} (h_2-h_1)^+$. Note that this is not linear in $(h_1,h_2)$. – 2017-01-12
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    @PostName: Working with $w^+$ is cumbersome, maybe it is easier to work with $\frac 1 2 (w + |w|)$. Therefore, the problem becomes finding the derivative of $w \mapsto \int _\Omega |w|$. Not sure that it helps, though. – 2017-01-12

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We have the functions $f(u,v) = ∫_Ω (v-u)^+, I(w) := ∫_Ω w, g(w):= (w)^+, h(u,v) := v-u$ with the relationship $$ f(u,v) = I\circ g\circ h(u,v)$$ We use the notation $dF(x_0)[ε_0]$ to mean the derivative of $F$ at $x_0$ evaluated with increment $ε_0$. By chain rule the derivative at $u,v$ is $$ df(u,v)[(ε,\delta)] = dI(g\circ h(u,v))[dg(h(u,v))[dh(u,v)[(ε,\delta)]] ] $$ We have the individual derivatives $$dI(w)[h] = I(h), \quad dh(u,v)[(ε,\delta)] = h(ε,\delta)$$ by linearity, and $dg(w)[ε] = (\mathbb 1_{w\geq 0}) ε $. Combining,

$$ df(u,v)[(\epsilon,\delta)] = ∫_Ω \mathbb 1_{v(x) - u(x) ≥ 0} (ε(x)- δ(x) ) dx $$

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    thanks but I think you made a mistake. Isn't the derivative of $g$ at $w$ in direction $\epsilon$ actually $1_{h \geq 0}h$???? – 2017-01-13
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    @PostName Its almost everywhere classically differentiable with derivative $g'(x_0) = \mathbb 1_{x_0≥0}$. For such functions the Frechet derivative at $x_0$ is the linear map $h↦ g'(x_0)h$. – 2017-01-13
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    But let's work at the point $0$. Then $\frac{g(0+th)-g(0)}{t} =\frac{g(0+th)}{t} = h1_{h \geq 0}$. Isn't it right? – 2017-01-13
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    0 is a problem point, I would recommend not working there. Try graphing the function; the classical derivative should be piecewise constant. – 2017-01-13
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    Ok, but in the last equation you have taken out the directions $\epsilon $ and $\delta$. Aren't they $H^1$ functions? – 2017-01-13
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    @PostName oops yes I saw it again later and forgot that they were functions. Thanks for the correction – 2017-01-13