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Suppose that $f : X → ]−\infty, +\infty]$ is convex and proper, that $\bar{x} \in dom f$, and that $\lambda > 0$. Show that

$$\partial^{\infty}(\lambda f)(\bar{x}) = \lambda\partial^{\infty}f(\bar{x}).$$

I have this property but I am still unsure where to start with the proof.

$$v\in \partial f(\bar{x})\iff(\forall\;x\in X)f(\bar{x})\langle v,x-\bar{x}\rangle\leq f(x). $$

Jeremy
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mathjl
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  • It is really hard to follow what you are asking. Are you asking to show that the subdifferential is positive homogeneous? It does follow, almost immediately, from the definition. – copper.hat Oct 29 '15 at 18:45
  • Sorry, the formatting didn't work out for some reason. I need to prove ∂^infty (λf)(x ̄) = λ ∂^infty f(x ̄). I am unsure what the notation ∂^infty implies, this is where I am stuck – mathjl Oct 29 '15 at 18:49
  • Sorry, I missed the $\infty$ entirely, how do you define $\partial^\infty f(x)$? – copper.hat Oct 29 '15 at 18:51
  • It could be Rockafellar's 'horizontal subgradient' or the asymptotic generalised gradient, but I suspect that you would know what these are if you are dealing with them (certainly not introductory complex analysis). Are you dealing with normal cones, etc? – copper.hat Oct 29 '15 at 19:05
  • It is introductory - i believe it is defined as the singular subdifferential (according to a textbook I found) – mathjl Oct 29 '15 at 19:12
  • Surely the professor defined it? – copper.hat Oct 29 '15 at 19:19
  • For the formatting issue, you can put $ on either side of the symbolic math to get the desired formatting. Alternatively, you may want to consider learning LaTex. – Jeremy Nov 16 '15 at 01:11

1 Answers1

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For this type of problem you typically want to choose some element you know to exist on the left hand side and algebraically show it must also be contained on the right hand side. For this problem specifically we want to chose $v\in\partial^{\infty}(\lambda f)(\bar{x})$ and conclude that $v\in\lambda\partial^{\infty}f(\bar{x}).$ This can be done entirely with iff statements. Also, it will be useful to recall that singular sub differential at a point x is equal to the normal cone over the domain of f at x. That is to say $$\partial^{\infty}f(\bar{x})=N_{dom(f)}(\bar{x})$$ where the normal cone is defined as $$N_{dom(f)}(\bar{x})= \{v\in\mathbb{R}:\langle v,x-\bar{x}\rangle\leq0\;\forall x\in dom(f)\}.$$

Now that we have all the required information I will get you started with the actual proof:

Observe that \begin{align} v\in\partial^{\infty}(\lambda f)(\bar{x})&\iff v\in N_{dom(\lambda f)}(\bar{x})\\ &\iff \langle (v,0),(x,( f)(\lambda x))-(x,( f)(\lambda \bar{x})) \rangle\leq0\\ &\quad\vdots\\ &\iff v\in\lambda\partial^{\infty}f(\bar{x}). \end{align}

Jeremy
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