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I've seem similar questions asked about this topic but I've got some additional conditions to prove that I haven't seen and am a bit confused about.

For a linear/antilinear functional $l$ defined on a normed vector space $V$, I'm trying to prove that the following conditions are equivalent:

(i) $l$ is continuous on $V$

(ii) $l$ is sequentially continuous on $V$

(iii) $l$ is continuous at $0$ (zero vector)

(iv) $l$ is sequentially continuous at $0$

(v) $l$ is $bounded$, i.e. there exists $C > 0$ s.t. $$|l(v)| \leq C||v||_V$$ where $||v||_V$ is the norm in V.

I would imagine that the argument for the first two would be the same as for a function; continuity implies sequential continuity because as a sequence $x_k$ converges to a point $x_0$, $F(x_k)$ converges to $F(x_0)$. And by contradiction the converse can be proven.

Then, because V is a metric space, if it's sequentially continuous, it's sequentially compact and therefore compact - leading to boundedness.

But I'm wondering, how is continuity at $0$ different than (i), and how does it imply the other conditions?

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    I'm not sure what you are asking exactly, but since $l(x)-l(y) = l(x-y) - l(0)$, you can see that continuity at $y$ and continuity at $0$ are the same thing, as such. – 2017-02-04
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    Yep I just saw that somewhere else and of course it makes sense, thank you. I'm sorry, I'm new to functional analysis and trying to work through these proofs as a newbie. Thank you! – 2017-02-04
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    The problem with learning something new is that there is so much accumulated material that took decades to evolve and now you need to work through it to get the the next round of evolving material. – 2017-02-04
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    Yeah, it's tough, and although the class I'm in didn't list prerequisites, it uses a lot of functional analysis so I'm having to learn it in bits and pieces. It helps that there are some good introductory books out there at least that I can try to read through. – 2017-02-04
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    Good luck! ${}{}$ – 2017-02-04

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