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Let us consider $f:\, \mathbb{R}^n \rightarrow \mathbb{R}$ be a convex function and differentiable at a point $x_0$.

If $\partial f(x_0)$ denotes the subdifferential of $f$ I would like to prove that the only element in it is given by the gradient of f in $x_0$, i.e.

$$ \partial f(x_0)=\left\lbrace\nabla f(x_0)\right\rbrace $$

I know that it should be trivial but i can't prove the inclusion: $\subset$ ...any suggestion?

William Tomblin
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3 Answers3

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I have just found an alternative argument:

Let us consider $v\in \partial f(x_0)$. By definition of subgradient, $$ f(x)\geq f(x_0)+(v,x-x_0) \qquad \forall x \in \mathbb{R}^n$$ Then by choosing $x=x_0+\lambda z$ for any $z$ and $\lambda >0$ one gets $$(v,z) \leq \dfrac{f(x_0+\lambda z)-f(x_0)}{\lambda} \rightarrow \nabla f(x_0) \cdot z$$ that is $$(\nabla f(x_0)-v)\cdot z \geq 0 \qquad \forall z $$ and by choosing $z=-(\nabla f(x_0)-v)$ one has to conclude that $$\nabla f(x_0)=v\, .$$

William Tomblin
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(1) Recall the definition of gradient of $f$ :

$g\in T_p$ is gradient if $$ d_pf(w)\leq (g,w) $$ for all $w\in T_p$ and $$ d_pf(g)=(g,g) $$

(2) So if we have two gradients $g,\ g'\in T_p$, then $$ (g,g)=d_pf(g)\leq (g,g'),\ (g',g')=d_pf (g')\leq (g,g') $$ so that we have $$ |g-g'|^2 =(g,g)-2(g,g')+(g',g')\leq 0 $$

HK Lee
  • 19,964
  • Thank you! even if your notation has confused me a little bit... i have found another argument (that probably is the same) but with my notation – William Tomblin Oct 28 '16 at 14:07
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$\text{First part of the proof is simple. I prove the second part. Suppose d is a vector of norm 1 and u} \\ \text{is a subgradient of f at x which is not} \ c \nabla f(x) \ \text{for some constant c (in that case} \\ \text{the proof is similar and simple)} \ \\ \text{By Taylor theorem we have:} \\ f(x + \epsilon d) = f(x) + \epsilon \langle\nabla f(x) , d\rangle + o(\epsilon) =\\ f(x) + \epsilon \langle\nabla f(x) , d\rangle + \epsilon \langle u , d\rangle - \epsilon \langle u , d\rangle + o(\epsilon)= \\ f(x) + \langle u,\epsilon d\rangle + \epsilon(\langle\nabla f(x) , d\rangle - \langle u,d\rangle + \frac{o(\epsilon)}{\epsilon}) \\ \text{We can find a vector d such that}\langle\nabla f(x) , d\rangle \,< 0 \, \text{and} \, \langle u,d\rangle \, > 0. \, \text{Now by choosing a very small positive} \, \epsilon \, \text{we have} \\ f(x + \epsilon d) < f(x) + \langle u , \epsilon d\rangle \, \text{which is a contradiction.} $

Arash Jamshidi
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