# Relation between subdifferential and functionals

Let $$X$$ be a complex Banach space and let $$x,y\in X$$ be non-zero. Suppose that

$$\|x+ty\|\geq \|x\|$$ for all $$t\in \mathbb{R}$$. Let $$\phi: \mathbb{R}\to \mathbb{R}$$ be defined by $$\phi(t) := \|x+ty\|$$. It is easy to see that $$\phi$$ attains minimum $$t=0$$. Consequently, the subdifferenetial of $$\phi$$ at $$0$$ (written as $$\partial \phi(0)$$) contains $$0$$.

From this can we conclude the following:

$$\exists$$ $$x^*\in X^*$$ such that $$\|x^*\|=1$$ with $$x^*(x)=\|x\|$$ and $$Re~x^*(y)=0$$.

I am a beginner in functional analysis and a detailed answer will be of great help.

Edit: Actually, in real case, the result can be obtained by a straighforward application of Hahn-Banach Theorem, by defining a functional $$f: span\{x,y\}\to \mathbb{R}$$ by $$f(\alpha x+\beta y)= \alpha\|x\|$$. However, in case of complex Banach spaces, we cannot use $$span \{x,y\}$$.

• Yes, definitely @Theobendit
– pmun
Mar 30 at 6:43
• Well, there goes my suggestion! Mar 30 at 6:53
– pmun
Mar 30 at 6:56
• Actually, we need to modify $span\{x,y\}$ or think of any other ways! Thank you @TheoBendit for your kind reply and time.
– pmun
Mar 30 at 6:59
• Essentially, it's much the same. Define $\phi$ over the span of $y$; it has a minimum at $0$, so the $0$ functional on $\operatorname{span}\{y\}$ is a subgradient on this space. Define the sublinear function $f(h)$ by the directional derivative of the norm function at $x$ in direction $h$. Extend this sublinear function to all of $X$, and the properties should fall out. Mar 30 at 7:11

The following is a proof in the real case, as requested in the comments.

Let $$Y = \operatorname{span}\{y\}$$ and define $$\phi : Y \to \Bbb{R} : z \mapsto \|x + z\|.$$ Then $$\phi$$ is convex. Note that $$\phi$$ attains a minimum at $$0$$, hence the $$\mathbf{0}$$ functional in $$Y^*$$ minorises $$\phi$$. That is, $$\mathbf{0}(z) = 0 \le \phi(z),$$ for all $$z \in Y$$. Let $$f : X \to \Bbb{R} : h \mapsto \lim_{t \to 0^+} \frac{\|x + th\| - \|x\|}{t}.$$ That is, $$f(h)$$ is the directional derivative of $$\|\cdot\|$$ at $$x$$, in direction $$h$$. In particular, this means that $$f$$ is sublinear. Further, clearly $$f(h) \ge 0$$ for $$h \in \operatorname{span} Y$$.

Using Hahn-Banach theorem, we can extend $$\mathbf{0}$$ to a linear functional $$x^* \in X$$ such that $$x^*(h) \le f(h)$$ for all $$h \in X$$. I claim that $$x^*$$ satisfies the conditions we want.

Note that $$x^*(y) = 0$$ for $$z \in Y$$, since $$x^*(z) = \mathbf{0}(z) = z$$ for all $$z \in Y$$. We also have, $$x^*(x) \le f(x) = \lim_{t \to 0^+} \frac{\|x + tx\| - \|x\|}{t} = \lim_{t \to 0 +}\frac{t\|x\|}{t} = \|x\|$$ and $$x^*(-x) \le f(-x) = \lim_{t \to 0^+} \frac{\|x - tx\| - \|x\|}{t} = \lim_{t \to 0 +}\frac{(1 - t - 1)\|x\|}{t} = -\|x\|,$$ where we restrict our attention to $$t \in (0, 1]$$. Together, we get $$x^*(x) = \|x\|,$$ as required.

Note that this implies $$\|x^*\| \ge 1$$. We also have, for $$z \in X$$, $$x^*(z) \le f(z) = \lim_{t \to 0^+} \frac{\|x + tz\| - \|x\|}{t} \le \lim_{t \to 0^+} \frac{\|x\| + t\|z\| - \|x\|}{t} = \|z\|,$$ thus $$\|x^*\| = 1$$ as required.

• Thank you so much @Theobendit. I will let you know, once I verify it.
– pmun
Mar 30 at 7:39
• It might be useful to recall that one-sided derivatives of convex functions do exist (see math.stackexchange.com/questions/1816383/…), so $f$ is indeed well-defined. Mar 30 at 7:43
• @SeverinSchraven Thanks! We need even more though: that $f$ is sublinear, and I was sort of taking it for granted that the reader would know this. Alternatively, one could use the Hahn-Banach separation theorem to separate the line $Y \times \{0\}$ from the interior of the epigraph of $\|\, \cdot + x\|$. Mar 30 at 7:46
• You are right, I missed that. The positive homogenity is easy to show, but the subaddivitivity I cannot prove on the spot. Mar 30 at 8:42
• It follows from the convexity of the function (positive homogeneity holds for general functions). Note that$$f(h_1 + h_2)=2f\left(\frac{h_1+h_2}{2}\right)=2\lim_{t\to0^+}\frac{\phi\left(\tfrac{h_1+h_2}{2}\right)-\phi(0)}{t}\le2\lim_{t\to0^+}\frac{\frac{\phi(th_1)+\phi(th_2)}{2}-\phi(0)}{t}=f(h_1)\\+f(h_2).$$ Mar 30 at 9:51