Let $H$ be a Hilbert space, $T_+(H)$ the set of positive self-adjoint trace-class operators on $H$, and $f : T_+(H) \to [0,m]$ a non-negative, bounded, convex functional. I don't necessarily know that it's continuous or semicontinuous.

Assume that a minimizer $x_0 \in \operatorname{argmin}_{x \in T_+(H)} f(x)$ exists. Given some $y\in T_+(H)$, the one-sided directional derivative of $f$ at $x_0$ in the direction $y-x_0$ is defined over the set $S:=T_+(H)- x_0$ as $$D_{x_0}(y-x_0) := \lim_{t\to 0^+} \frac{1}{t}\left[ f(x +t(y-x_0) ) - f(x)\right].$$ I am wondering if there are simple conditions on $f$ that guarantee that $D_{x_0} : S\to \mathbb{R}$ is a linear functional.


1 Answer 1


A sufficient condition and a necessary and sufficient condition for the linearity of a functional derivative were given by Mikhail Mordukhovich Vaĭnberg in his well known monograph [1], chapter 1 §3.2, pp. 37-40. Before briefly describing his results it is necessary to say that Vaĭnberg

  • proves the results for general operators, considering functionals as particular cases, and
  • he works on Banach spaces, thus his results are applicable to Hilbert spaces operators.

Said this, the statements are the following ones, adapted to the question context

Theorem 3.1 ([1], pp. 37-39) Suppose the two following conditions are satisfied:

  1. The functional derivative $Vf_{x}(y-x)$ of an operator $f$ on a given Banach space exists and is continuous respect to $x\in U_{x_0}$ where $U_{x_0}$ is some neighborhood of $x_0$, and
  2. $Vf_{x_0}(y-x_0)$ is continuous respect to $y-x_0$ at the point $y-x_0=0\iff y=x_0$

Then $Vf_{x_0}(y-x_0)=Df_{x_0}(y-x_0)$ is a linear operator respect to $y-x_0$ and thus a is Gâteaux derivative.

For the statement of the necessary and sufficient condition, we need a definition: an operator $f$ for which there exists a $\delta(y-x_0)>0$ such that if $|t|<\delta(y-x_0$), then $$ \|f\big(x_0+t(y-x_0)\big)-f(x_0)\|\le C\|t(y-x_0)\| $$ where the constant $C$ does not depend on $y-x_0$ is said to be weakly Lipschitz at the point $x_0$.

Theorem 3.2 ([1], pp. 39-40). In order for the functional derivative $Df_{x_0}(y-x_0)$ of the operator $f$ to be a linear operator respect to $y-x_0\triangleq h$ it is necessary and sufficient that

  1. $f$ is weakly Lipschitz at the point $x_0$, and
  2. $\Delta^2_{th_1,th_2} f(x_0)= o(t)$, where $$ \Delta^2_{h_1,h_2} f(x_0)=f(x_0+h_1+h_2) - f(x_0+h_1) - f(x_0+h_2)+ f(x_0) $$


  • From what is known on the functional $f: T_+(H)\to [0,m]$ it seems to me that the most fruitful way to proceed in order to prove the linearity of $D_{x_0}$ is to use theorem 3.1 above: the convexity of $f$ implies the inequality $$ f(y) \geq f(x_0) + D_{x_0}(y-x_0) $$ which, jointly with its boundedness, seems suitable to prove the two required continuity conditions given above.
  • These two results are not due to Vaĭnberg: the sufficient condition is due to George Marinescu, while the necessary and sufficient one is due to N. A. Ivanov (for functionals) and to Vaĭnberg himself (for the more general case of operators). See [1] for the references.

[1] Vaĭnberg, Mikhail Mordukhovich, Variational methods for the study of nonlinear operators. With a chapter on Newton’s method by L.V. Kantorovich and G.P. Akilov, translated and supplemented by Amiel Feinstein, Holden-Day Series in Mathematical Physics. San Francisco-London- Amsterdam: Holden-Day, Inc. pp. x+323 (1964), MR0176364, ZBL0122.35501.

  • 1
    $\begingroup$ Thank you for your really useful and insightful comment! I have to digest some, but from a quick glance I'm not sure Thm 3.1 can be applied in my case. For me, the natural space to consider is the Banach space of self-adjoint trace-class operators with the trace norm topology, $(T(H),\Vert \cdot \Vert_1)$. In this space, $\mathrm{dom} f = T_+(H)$ is not an open set, so for any $x \in T_+(H)$ it is not guaranteed that there is a neighborhood of $x$ that belongs to $T_+(H)$, so $f$ may not even be defined (much less continuous) on that neighborhood... $\endgroup$
    – Artemy
    Oct 22, 2020 at 7:02
  • $\begingroup$ So, it's more accurate to say that $D_{x_0}$ in my question should be defined over a certain domain. I will update the question. $\endgroup$
    – Artemy
    Oct 22, 2020 at 7:03
  • $\begingroup$ Regarding Thm 3.2, should there be a $t$ in the definition of $\Delta^2_{th_1,th_2}$? $\endgroup$
    – Artemy
    Oct 22, 2020 at 7:07
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    $\begingroup$ @Artemy, no it is a typo and should be removed (as is in the original reference): $\Delta^2_{th_1,th_2}$ is obtained from $\Delta^2_{h_1,h_2}$ by applying the transformation $(h_1, h_2)\mapsto (th_1,th_2)$ for a given $t$. $\endgroup$ Oct 22, 2020 at 7:12
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    $\begingroup$ @Artemy a further advice is to remember that, while $Df$ can be linear or not, it is always $1$-homogeneous, i.e. $$ Df_{x_0}(\lambda h)=\lambda D f_{x_0} (h)\quad \forall \lambda.$$ This could be a useful property in many non regular situations. $\endgroup$ Oct 22, 2020 at 11:52

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