0

I know how to show that when $f$ is proper and $\mu$-strongly convex, its subgradient $\partial f$ is $\mu$-strongly monotone. Is the converse true?

Let $H$ be a real Hilbert space and $f : H \longrightarrow \overline{\mathbb R}$ proper. I claim that the following are equivalent:

  1. $f$ is $\mu$-strongly convex, that is $$ f[(1-t)x+ty] \leq (1-t)f(x) + tf(y) - \frac{t(1-t)}{2}\mu |x-y|^2 $$

  2. $ f(y) \geq f(x) + (x^* , y-x) + \frac{\mu}{2} |x-y|^2,\quad \forall x,y \in H,\quad \forall x^* \in \partial f(x)$

  3. $\partial f$ is $\mu$-strongly monotone, that is $$ (x^*-y^* , x-y) \geq \mu |x-y|^2,\quad \forall x,y \in H,\quad \forall x^* \in \partial f(x),\quad \forall y^* \in \partial f(y). $$

Proof of $1 \Longrightarrow 2$ : Suppose that $f$ is $\mu$ strongly convex, let $x,y \in H$, $x^* \in \partial f(x)$ and $0 < t < 1$. Apply the inequality to $x + t(y-x)$ in place of $y$: $$ f(x) + (x^* , t(y-x)) \leq f(x + t(y-x)) \leq (1-t)f(x) + tf(y) - \frac{t(1-t)}{2}\mu |x-y|^2. $$ Simplifying by $f(x)$, dividing by $t>0$ and letting $t$ going to $0$ we get 2.$\square$

Proof of $2 \Longrightarrow 3$ : Suppose that $f$ satisfy the estimate 2, let $x,y \in H$, $x^* \in \partial f(x)$ and $y^* \in \partial f(y)$. We have $$ f(y) \geq f(x) + (x^* , y-x) + \frac{\mu}{2}|x-y|^2,\quad f(x) \geq f(y) + (y^* , x-y) + \frac{\mu}{2}|x-y|^2 $$ so summing up we get $\partial f$ to be $\mu$ strongly monotone.$\square$

Now I experience some troubles to show that 3 implies 1 even in the $C^1$ case. Any help would be appreciated.

blamethelag
  • 2,154
  • It's not true in general, the subdifferential might be empty if you don't at least assume it's convex. – Jürgen Sukumaran Nov 22 '22 at 13:10
  • Then how about the convex and $C^1$ case? – blamethelag Nov 22 '22 at 13:23
  • 1
    By definition of subdifferential $u\in\partial f(x)$,

    $$f(x) \geq f(y) + \langle u, x-y\rangle$$

    Now add and subtract a subgradient of $f$ at $y$ in the inner product and use the strong monotonicity: $$f(x)\geq f(y) + \langle u, x-y\rangle\ =f(y) + \langle u-v+v, x-y\rangle\ =f(y) + \langle u-v, x-y\rangle + \langle v, x-y\rangle\ \geq f(y) + \mu|x-y|^2 + \langle v, x-y\rangle $$

    Which gives

    $$f(y)\leq f(x) + \langle v,x-y\rangle +\mu|x-y|^2$$

    i.e., 3 implies 2

     – Jürgen Sukumaran Nov 22 '22 at 13:35
  • Adapting this proof https://math.stackexchange.com/questions/3996183/equivalent-definitions-of-convexity-for-f-in-mathcal-c1-mathbb-rn to the strongly monotone $C^1$ case is very straightforward, you just need to use strong monotonicity instead of monotonicity when looking at $g'(t_2)-g'(t_1)$. – Jürgen Sukumaran Nov 22 '22 at 13:43
  • @mordecaiiwazuki in your second comment, the first inequality is wrong. You should have taken $u \in \partial f(y)$. – blamethelag Dec 27 '22 at 17:17

1 Answers1

-1

The answer to the original question is negative, even for $f\in \Gamma_0(\mathbb R)$. To see it consider the following facts. There $f \in \Gamma_0(H)$.

  1. If $f$ is $\mu$ strongly convex, then $\partial f$ is $\mu$ strongly monotone.

  2. $\partial f$ is $\mu$ strongly monotone if and only if $\partial f^*$ is $\mu$ co coersive.

  3. If $f$ is $C^1(H;\mathbb R)$ and $\nabla f$ is $L$ Lipschitz, then $\partial f$ is $\frac 1 L$ co coersive (this is the Baillon Haddad theorem).

  4. $f$ is $\mu$ strongly convex if and only if $f^*$ is $C^1(H ; \mathbb R)$ and $\nabla f^*$ is $\frac 1 \mu$ Lipschitz.

Then consider $f \in \Gamma_0(\mathbb R)$ defined by $$ f(x) = \left\lbrace \begin{array}{ccc} x^2 & \text{if} & -1 \leq x \leq 1 \\ + \infty & \text{else} \end{array} \right. $$ We can compute its subgradient $$ \partial f(x) = \left\lbrace \begin{array}{ccc} \{2x\} & \text{if} & -1 \leq x \leq 1 \\ \emptyset & \text{else} \end{array} \right. $$ which is cocoersive $$ (2x-2y,x-y) = 2|x-y|^2 \geq \frac1 2|2x-2y|^2. $$ However $f$ is not $C^1$, so the reciprocal of Baillon Haddad is false. Moreover this is a counter example to the original question because if the original question had a positive answer, $g := f^*$ would have a strongly monotone subgradient hence be strictly convex and its conjugate $g^* = f^{**}=f$ would be $C^1$.

blamethelag
  • 2,154
  • 2
    This counterexample is incorrect. The subdifferential of f at 1 is 2x plus the normal cone to the constraint, that is R+. By taking x=0, x=0, y=1 and y=1+t for positive t, the cocoercivity inequality becomes impossible because the lower term grows faster with t than the upper bound. Overall, it is a general fact that cocoercive operators are single-valued and Lipschitz continuous, which is not true here. – Guillaume Mar 18 '24 at 08:24