suppose $u, w \in H^1 (R^2)$. I'd like to know where does the following inequality come from (it appears in a proof I've been reading and I can't figure it out) $$ | \nabla (u |u|^2) - \nabla(w|w|^2)| \leq C | \nabla(u-w)| \cdot(|u|^2 + |w|^2) + C|u-w|\cdot(|\nabla u| + |\nabla w|)(|u|+|w|)$$ $u$ and $w$ are complex valued. Can anyone provide some suggestions? C is some constant, its value isn't importnat.

Perhaps I should mention that I'm going to use it with integrals ($L^2$), but I'm pretty sure it should be a pointwise inequality


1 Answer 1


Let $f:\mathbb{C}\to\mathbb{C}$ be the function $f(x,y) = (x+iy) (x^2 + y^2)$. Its derivative is

$$ \mathrm{d}f = (x^2 + y^2) \mathrm{d}(x + i y) + 2(x + i y) (x \mathrm{d}x + y \mathrm{d}y) $$

which satisfies

$$ |\mathrm{d}f| \leq C(x^2 + y^2) $$

Integrating along the straight line joining $v,w\in \mathbb{C}$ gives

$$ |f(w) - f(v) | \leq C \max(|v|^2,|w|^2) |v-w| \leq C(|v|^2 + |w|^2) |v-w|$$

Now let $w,v$ be differentiable $\mathbb{C}$ valued functions.

Since you are applying to functions in $L^2$ this is okay by density.

Since there seems to be some confusion, let me do this in a bit more detail.

Let $u_s = (1-s) v + s w$ so that $u_0 = v$ and $u_1 = w$ be a one parameter family of functions. What we should do is to compute

$$ \nabla f(v) - \nabla f(w) = \nabla f(u_0) - \nabla f(u_1) = \int_1^0 \partial_s \nabla f(u_s) \mathrm{d}s $$

The term (writing $u = u_s$ and $\dot{u} = \partial_s u_s$)

$$ \partial_s \nabla f(u) = \nabla \left[ 2\bar{u} u \dot{u} + u^2 \dot{\bar{u}}\right] $$

Putting in the $\nabla$ you see that by the product rule you have

$$ \partial_s \nabla f(u) = 2 \nabla \bar{u} u \dot{u} + 2 \bar{u} \nabla u \dot{u} + 2 \bar{u} u \nabla \dot{u} + 2 u \nabla u \dot{\bar{u}} + u^2 \nabla \dot{\bar{u}} $$

Now, since $u$ is linear in $s$, for each component of the $\nabla$ the term with $\dot{u}$ is signed. So putting in absolute value signs and integrating both sides give you the desired inequality.

  • $\begingroup$ nice, thank you for your answer. however, I'm not quite sure I fully understood. I think I can follow up to the point $|df| \leq C(x^2 + y^2)$ and then later when you apply it to $f(w) - f(v)$ for $w, v \in \mathbb{C}$ but I'm a bit confused as to whether I can just say that it will also work if I make $w$ and $v$ differentiable functions. Also, your answer doesn't seem to include $\nabla w$ or $\nabla v$, how do I get to the RHS of the inequality in my question? I think that your estimate should do fine for my uses anyway, but I'm just not sure if I understand it properly $\endgroup$
    – mm-aops
    Commented Nov 21, 2013 at 14:22

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