Suppose $X$ is a reflexive, smooth Banach space and $Y$ is a proper non zero closed subspace of $X$ such that $Y$ is the range of a norm 1 projection map $P: X \to Y.$ Then, the Banach adjoint $i^\times$ of the inclusion map $i: Y \hookrightarrow X$ is given by $i^\times : X^* \to Y^*$ by $i^\times (f)= f|_Y,\quad f\in X^*.$
Since, $Y$ is a closed subspace of a reflexive space $X,$ $i^\times (f)$ attains its norm at some point of $Y.$
My doubt: Let $x\in X\setminus Y.$ Then, can we say $i^\times (f_x)= f_{_{P(x)}}\quad$ where $f_x$ means the support functional at $x?$ If not, can we say $i^\times(f)=f_z$ where $z$ is a scalar multiple of $P(x)?$
Support functional definition: A functional $f: X\to \mathbb{C}$ is said to be support functional at $x\in X$ if $f(x)=\|x\|^2$ and $\|f\|=\|x\|.$
Smooth Banach space: A Banach space is said to be smooth if for every point $x\in X,$ there is unique support functional at $x.$
I have found that the same is true when $X$ is a Hilbert space.
Because, $i^\times(f_x)(a)=\langle a, x\rangle= \langle a, P(x)+(I-P)(x)\rangle = \langle a, P(x)\rangle = f_{_{P(x)}}\;\;(a),\quad a\in Y.$
Any counter example or suggestion is appreciated. Thanks in advance.