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 $span \{x,y\}$.

Thanks in advance

  • $\begingroup$ Yes, definitely @Theobendit $\endgroup$
    – pmun
    Mar 30 at 6:43
  • $\begingroup$ Well, there goes my suggestion! $\endgroup$ Mar 30 at 6:53
  • $\begingroup$ Are you saying about the current edit I made? @TheoBendit $\endgroup$
    – pmun
    Mar 30 at 6:56
  • 1
    $\begingroup$ Actually, we need to modify $span\{x,y\}$ or think of any other ways! Thank you @TheoBendit for your kind reply and time. $\endgroup$
    – pmun
    Mar 30 at 6:59
  • 1
    $\begingroup$ 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. $\endgroup$ 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.

  • $\begingroup$ Thank you so much @Theobendit. I will let you know, once I verify it. $\endgroup$
    – pmun
    Mar 30 at 7:39
  • 1
    $\begingroup$ 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. $\endgroup$ Mar 30 at 7:43
  • $\begingroup$ @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 $\|\, \cdot + x\|$. $\endgroup$ Mar 30 at 7:46
  • $\begingroup$ You are right, I missed that. The positive homogenity is easy to show, but the subaddivitivity I cannot prove on the spot. $\endgroup$ Mar 30 at 8:42
  • $\begingroup$ It follows from the convexity of the function (positive homogeneity holds for general functions). Note that$$f(h_1 + h_2)=2f\left(\frac{h_1+h_2}{2}\right)=2\lim_{t\to0^+}\frac{\phi\left(\tfrac{h_1+h_2}{2}\right)-\phi(0)}{t}\le2\lim_{t\to0^+}\frac{\frac{\phi(th_1)+\phi(th_2)}{2}-\phi(0)}{t}=f(h_1)\\+f(h_2).$$ $\endgroup$ Mar 30 at 9:51

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