Let $K \subset R^n$ be a closed, convex, nonempty set. Define $$ I[x] := \begin{cases}0&\mbox{If }x\in K\\ \infty&\mbox{If }x\notin K\end{cases} $$ Explicitly determine $A=\partial I$,$J_{\lambda}=(I+{\lambda}A)^{-1}$,$A_{\lambda}=\frac{I-J_{\lambda}}{\lambda}({\lambda >0})$ in terms of the geometry of $K$.
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I have no idear on the queation! Wish someone help me solve this question,thanks a lot – fx0123 Jul 04 '13 at 13:23
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The most important thing is determining $A=\partial I$. Once you have done this the rest is easier. Try to work out the case in which $K$ is the unit circle in the plane. Clearly, if $x$ is interior to the circle, then $A[x]={0}$, and if $x$ is outside the circle then $A[x]=\varnothing$. What happens when $x$ is on the boundary? Once you understand this, redo the reasoning with a square instead of the circle. – Giuseppe Negro Jul 04 '13 at 14:01
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@GiuseppeNegro, do you know what is the meaning of the notation $(I+\lambda A)^{-1}$? – Tomás Jul 04 '13 at 14:25
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Is $I$ in $(I+\lambda A)^{-1}$ the indicator function of $K$, or is it $Iu=u$? – Tomás Jul 04 '13 at 14:31
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@Tomás: $I$ is the identity. Indeed $(I+\lambda A)^{-1}$ is called nonlinear resolvent, if I recall correctly. (I'm taking my copy of Evans's book to check.) – Giuseppe Negro Jul 04 '13 at 21:51
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Yes it is the identity. The point is that, even if $A=\partial I$ (there is a notational conflict here, this $I$ is the indicator function) is a multi-valued function, it happens that the equation (with $\lambda >0$ a fixed parameter) $$u+\lambda A[u]\ni w $$ has a unique solution $u$ for each $w$, and so the operator $(I+\lambda A)^{-1}$ is well-defined and single-valued. Of course here $I$ is the identity. – Giuseppe Negro Jul 04 '13 at 21:56