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It's false. Take $X = B = \mathbb{C}^2 \cong l^\infty(\{0,1\})$, and define $f: X \to B$ by $f(a,b) = (a,.5(a-b))$. Then $f(1,1) = (1,0) \geq (0,0)$ and $\|f\| = \|f(1,1)\|_\infty = 1$, but $f(0,1) = (0, -.5) \not\geq 0$.