Why is $(f_n)_{n \ge 1 }$ in $X^*$ a bounded sequence?

Question

Let $X$ be a real Banach space, $(\epsilon_n)_{n \ge 1 }$ a sequence of real positive numbers converging to $0$, and $(f_n)_{n \ge 1 }$ a sequence in $X^*$ for which there exists $r > 0$ so that for any $x \in B_r(0)$

$f_n(x) \leq \epsilon_n \Vert f_n \Vert + C(x)$, $n \ge 1$ for some $C(x) \gt 0$.

Then the sequence $(f_n)_{n \ge 1 }$ is bounded.

But why is this sequence bounded? Can someone explain it to me?

Answer 1

For each $n\geq 1,x\in B_r(0)$, $$ \left(1-\frac{\varepsilon_n}{r}\right)f_n(x)= f_n(x) - \varepsilon_n f_n(x/r) \leq f_n(x)-\varepsilon_n \|f_n\|.$$

Either $f_n(x)\geq 0$ or $f_n(x)<0$. Consider the case where $f_n(x)\geq 0$. Then by the assumption (eventually, when $\varepsilon_n/r$ is below 1):

$$0\leq (1-\varepsilon_n/r)f_n(x)\leq f_n(x)-\varepsilon_n\|f_n\|\leq C(x)$$ Namely,

$$0\leq f_n(x)\leq (1-\varepsilon_n/r)^{-1}C(x)$$

Since $\varepsilon_n\to 0$, then $(1-\varepsilon_n/r)^{-1}C(x)\leq 2C(x)$ for all but finitely many $n$, so there is some $M_x$ where $0\leq f_n(x)\leq M_x$ for all $n.$ By linearity $f_n(-x)\geq -M_x$ so $|f_n(x)|\leq M_x$ for all $n$.

Then by the uniform boundedness principle on $(f_n)$, you get what you want.