Properties of a linear operator given an inequality

Question

Let $X$ be a Banach** space and let $X'$ be its dual. Let $x_0 \in X$ and assume that there is $L \in X'$ such that for every $x \in X$ $$\frac{1}{2}\|x_0\|_X^2 - L(x_0) \le \frac{1}{2}\|x\|_X^2 - L(x).$$ I want to prove that $L(x_0) = \|x_0\|_X^2$ and furthermore that $\|L\|_{X'} = \|x_0\|_X$.

**I am not sure that we really need the space to be Banach (normed should be enough), but it is also true that performing only algebraic operations on the above inequality I could not prove anything.. So even if it is irrelevant for the formulation of the problem it is not impossible that we really need the space to be Banach.

Answer 1

For $\lVert\xi\rVert_X = 1$, on the line through $0$ and $\xi$, the minimal value of

$$h_\xi(t) = \frac{\lVert t\xi\rVert_X^2}{2} - L(t\xi) = \frac{1}{2}t^2 - t\cdot L(\xi)$$

is $-\frac{1}{2}L(\xi)^2$, attained at $t = L(\xi)$. Thus if

$$\frac{\lVert x_0\rVert_X^2}{2} - L(x_0) \leqslant \frac{\lVert x\rVert_X^2}{2} - L(x)\tag{1}$$

for all $x\in X$, then if $x_0 = 0$, we must have $L = 0$, and otherwise

$$\frac{\lVert x_0\rVert_X^2}{2} - L(x_0) \leqslant -\frac{1}{2} L\left(\frac{x_0}{\lVert x_0\rVert_X}\right)^2 \iff \left(L(x_0) - \lVert x_0\rVert_X^2\right)^2\leqslant 0,$$

so $L(x_0) = \lVert x_0\rVert_X^2$, whence $\lVert L\rVert_{X'} \geqslant \lVert x_0\rVert_X$. Furthermore we must then have

$$\lVert L\rVert_{X'}^2 = \sup_{\lVert\xi\rVert_X = 1} L(\xi)^2 \leqslant 2L(x_0) - \lVert x_0\rVert_X^2 = \lVert x_0\rVert^2,$$

and that means $\lVert L\rVert_{X'} \leqslant \lVert x_0\rVert_X$.

A completeness assumption for $X$ is not needed at any step. Furthermore, the above shows that $(1)$ holds for all $x\in X$ if and only if $\lVert L\rVert_{X'} = \lVert x_0\rVert_X$ and $L(x_0) = \lVert x_0\rVert_X^2$, and by the Hahn-Banach theorem(s), there is such an $L\in X'$ for every $x_0 \in X$.