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Let $f:\mathbb{R}^n\to(-\infty,+\infty]$ be convex and $x \in \mathrm{dom}f = \{x^* \mid f(x^*) < +\infty\}$. I am trying to prove the next relation:

\begin{equation} \mathrm{cl}\bigcup_{\lambda > 0}\frac{\mathrm{epi}f - (x,f(x))}{\lambda} \subset \bigcup_{\lambda > 0}\frac{\mathrm{epi}f - (x,f(x))}{\lambda} - (0,\varepsilon) \qquad (\varepsilon > 0) \tag{1}\label{1} \end{equation}

where $\mathrm{epi}f = \{(x,\alpha) \in \mathbb{R}^n\times\mathbb{R}\mid f(x)\le\alpha\}$ and $\mathrm{cl}A$ is the closure of a set $A$. Since the left-hand-side is a tangent cone to $\mathrm{epi}f$ at $(x, f(x))$, the relation seems obvious but I want a rigorous proof. I am struggling to handle the closure, but couldn't get a result so far. Would you give me any hint or a reference?

Edit:

Since \eqref{1} turns out to be false by the counter example of copper.hat, I redirect the question to the original problem that bears the question of \eqref{1}. In the book that I am studying by myself, the next relation is given without a proof:

Let $f:\mathbb{R}^n\to(-\infty,+\infty]$ be convex and $x\in\mathrm{dom}f$. Then, \begin{equation} \mathrm{epi}Df(x)(\cdot) = T((x,f(x)),\mathrm{epi}f) \tag{2}\label{2} \end{equation} where \begin{align} \mathrm{epi}Df(x)(\cdot) &= \left\{(v,\alpha)\in\mathbb{R}^n\times\mathbb{R} \,\middle\vert\, Df(x)(v)=\lim_{\lambda\downarrow0}\frac{f(x+\lambda v)-f(x)}{\lambda}\le\alpha\right\} \\ T((x,f(x)),\mathrm{epi}f) &= \mathrm{cl}\bigcup_{\lambda>0}\frac{\mathrm{epi}f - (x,f(x))}{\lambda}. \end{align}

In my attempt to prove \eqref{2}, I reached to the point that $T((x,f(x)),\mathrm{epi}f) \subset \mathrm{epi}Df(x)(\cdot)$ if \eqref{1} is true.

In fact, the copper.hat's example also shows that \eqref{2} is false because \begin{align} \mathrm{epi}Df(0)(\cdot) &= \{(v,\alpha)\mid v\le0, \alpha\in\mathbb{R}\}\setminus(\{0\}\times(-\infty,0)) \\ T((0,f(0)),\mathrm{epi}f) &= \{(v,\alpha)\mid v\le0, \alpha\in\mathbb{R}\}. \end{align}

It might be necessary to assume that $x$ is an interior point of $\mathrm{dom}f$ to make \eqref{2} true.

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  • $\begingroup$ Out of curiosity, what led you to this question? Is this for a class you're taking? $\endgroup$ – littleO Sep 15 '16 at 23:04
  • $\begingroup$ Also, it is not true, unless you add some assumptions such as $f$ being closed & proper. $\endgroup$ – copper.hat Sep 15 '16 at 23:37
  • $\begingroup$ @littleO I added the original problem that made the question. $\endgroup$ – flyingwith Sep 16 '16 at 7:43
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Convex functions can have peculiar behaviour at the boundary of $\mathbb{dom} f$.

Let $n=1$ and define $f(x) = \begin{cases} 0,& x < 0 \\ 1, & x=0 \\ +\infty, & x>0 \end{cases}$.

Fix $x=0$ and let $A=\bigcup_{\lambda > 0}\frac{\mathrm{epi}f - (0,f(0))}{\lambda}$, and note that $A= \{ (x,\phi) | x \le 0\} \setminus (\{0\} \times (-\infty,0) )$.

We see that $\overline{A} = \{ (x,\phi) | x \le 0\}$, in particular, we have $(0, t) \in \overline{A}$ for all $t <0$.

If we choose $t=-2 \epsilon$, we see that $(0, t) \notin A - \{(0,\epsilon) \}$.

As an aside, note that if $f$ has a subgradient at $x$ then the result is true.

Very loosely speaking, our intuition for convex functions holds for $\mathbb{ri} \ \mathbb{dom} f $, we should walk slowly on $\mathbb{rb} \ \mathbb{dom} f $.

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  • $\begingroup$ @cooper.hat Thanks a lot. Your example also solves my original problem that made the question above. $\endgroup$ – flyingwith Sep 16 '16 at 7:40

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