Suppose $f(x)$ is continuously differentiable, $\partial f(x)$ is the subdifferential. Is the following true? See 8.8(c) on page 304 Variational Analysis Book

$\partial f(x)=\{f'(x)\} \quad (1)$

Here is my disproof. We know that $x^\ast$ is a minimizer of $f(x)$ if and only if $0 \in \partial f(x^\ast)$. Thus equation (1) says if and only if $f'(x^\ast)=0$. But the "if condition" is invalid. Thus (1) can't be true. Did I misunderstand something here?

  • 1
    $\begingroup$ If $f$ is continuously differentiable and convex, then $\partial f(x) = \{ \nabla f(x) \}$. But if $f$ is not assumed to be convex, then equation (1) might not be true, as you noted. $\endgroup$
    – littleO
    Commented Aug 25, 2019 at 20:29
  • $\begingroup$ The book I cited is a famous one. Problem 8.8 has no answer. $\endgroup$
    – jsmath
    Commented Aug 25, 2019 at 20:41
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    $\begingroup$ Exercise 8.8 (p. 304) discusses $\hat \partial f(\bar x)$, rather than the more familiar subdifferential $\partial f(x)$. The expression $v \in \hat \partial f(\bar x)$ is defined in Definition 8.3 (p. 301). With this definition, it is true that if $f$ is differentiable at $x$, then $\hat \partial f(x) = \{ \nabla f(x) \}$. $\endgroup$
    – littleO
    Commented Aug 25, 2019 at 21:17
  • $\begingroup$ $\hat\partial f(\bar x)$ is for part (a). How about part (b) which states $\partial f(x)=\{\nabla f(x)\}$. $\endgroup$
    – jsmath
    Commented Aug 25, 2019 at 22:30
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    $\begingroup$ Oh, the notation on this book is not what I'm used to. Usually $z \in \partial f(x_0)$ means that $f(x) \leq f(x_0) + \langle z, x - x_0 \rangle$ for all $x \in \mathbb R^n$. But, in this book $\partial f(x_0)$ has a different meaning, given on p. 301. I suspect that using the definition of $\partial f(x_0)$ which appears on p. 301 will resolve the problem, but I haven't worked out the details yet. $\endgroup$
    – littleO
    Commented Aug 25, 2019 at 22:45


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