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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
@Sachin: This is unrelated to the question, but it is worth mentioning: If ever someone provides an answer to one of your questions, you can "Accept" it by clicking on a green checkmark underneath the upvote icon. (This checkmark only appears on question that you ask) Perhaps no answer to any question you have asked so far has been satisfactory, but if they have been I suggest going back and accepting them. It is just a way to say "thanks" to the answerer, and changes the fact that you have a "0% accept rate." (This only applies to actual answers, not comments) –  Eric Naslund Apr 22 '11 at 18:27

1 Answer 1

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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