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When trying to prove that a linear functional is bounded iff it is Lipschitz continuous, is it true that if we assume that a linear functional is Lipschitz continuous on a normed linear space $X$, then it is also continuous on $X$?

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I don't understand this question. Any Lipschitz continuous function on any metric space is (even uniformly) continuous by the triangle inequality. Also, what kind of linear spaces do you have in mind? I guess normed, but you should specify this in your question. – t.b. Apr 22 '11 at 17:27
No problem, but my first point still remains and trivially answers the question you ask. What do you really want to know? – t.b. Apr 22 '11 at 17:35
@Theo You basically answered my question. What I really wanted to know was quite trivial actually, i.e. does Lipschitz continuity imply continuity in general? – Libertron Apr 22 '11 at 17:36
Yes, and you don't even need the triangle inequality. Simply use the fact that a function $f: X \to Y$ is continuous iff it is sequentially continuous. Then for an $L$-Lipschitz continuous function and $x_n \to x$ we have $d_Y (f(x_n), f(x)) \leq L \cdot d_X (x_n, x) \to 0$. Or, more precisely, let $\varepsilon \gt 0$. If $\delta = \varepsilon/L$ then $d_X(x,x') \lt \delta$ implies $d_Y(f(x),f(x')) \leq L \cdot d_X(x,x') \lt \varepsilon$. – t.b. Apr 22 '11 at 17:37
@Theo Thank you for guiding me! – Libertron Apr 22 '11 at 17:40
up vote 1 down vote accepted

In view of t.b.'s comment, the answer is concluded to be yes. In essence, a sequential continuity argument seems sufficient.

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