# Hahn Banach Theorem implying existence of a nonzero linear functional taking 0 in a linear subspace

I am reading this paper. In the proof of theorem 1, it is stated

By the Hahn-Banach theorem, there is a bounded linear functional on $$C(I_n)$$, call it $$L$$, with the property that $$L\ne 0$$ but $$L(R) = L(S) = 0$$.

$$C(I_n)$$ is space of continuous functions on $$[0,1]^n$$. $$S$$ is a linear subspace in it. $$R$$ is the closure of $$S$$.

Can you explain to me why this statement is true?

• This is true as long as $R$ is not the entire space. What version of the Hahn-Banach Theorem do you know? There's one that's almost exactly this statement. Apr 2, 2019 at 1:59
• I was looking at Rudin 1991. Also Friedman's lemma and theorem. I see on Wikipedia they list something very similar as an important consequence.
– ztyh
Apr 2, 2019 at 2:11
• They do assume $R$ is not all of $C(I_n)$.
– ztyh
Apr 2, 2019 at 2:33

Let $$M=\{f+af_0:a\in \mathbb R\}$$ where $$f_0$$ is any fixed element not in $$R$$. Define $$T(f+af_0)=a$$. If we show that this is continuous on the space spanned by $$R \cup \{f_0\}$$ we can use Hahn Banach Theorem to get a continuous linear functional which is $$0$$ on $$R$$ and has the value $$1$$ at $$f_0$$. I will let you verify that $$T$$ is well defined. Suppose $$f_n+a_nf_0 \to g$$. If $$(a_n)$$ is unbounded it has a subsequence $${a_{n'}}$$ converging to $$\pm \infty$$. Dividing by this we get $$\frac {f_n'} {a_{n'}} +f_0=0$$ which shows that $$-f_0$$ is the limit of sequence from $$R$$ which is a contardiction. Hence $$(a_n)$$ is bounded and it has subsequence converging to some $$a$$. we then get $$f_n+a_nf_0 \to g=f+af_0$$ for some $$f \in R$$ and $$a=T(g)=\lim a_{n'} =\lim T(f_{n'}+a_{n'}f_0)$$. By arguing with subsequences we see that $$T$$ is continuous.

• What exactly do I have to say about $T$ being well defined?
– ztyh
Apr 2, 2019 at 14:47
• Also can I just say $T(\lim [f_n+a_nf_0])=\lim a_n=\lim T(f_n+a_nf_0)$ to show that it is continuous and then say $T$ is linear?
– ztyh
Apr 2, 2019 at 15:31
• @ztyh No you cannot show continuity this way because we don't know that $\lim a_n$ exists. Apr 2, 2019 at 23:06
• Can I just use bounded iff continuous? When you prove $(a_n)$ cannot tend to $\infty$, you can say $T$ is bounded and that is the end of the proof?
– ztyh
Apr 3, 2019 at 22:32
• @ztyh What Kavi has proved is that, if $f_n+a_nf_0 \to g$ for some $g\in M$, then $(a_n)$ is bounded, which doesn't imply the boundedness of $T$. And for the argument of continuity and subsequence, you may find this property useful. May 10, 2021 at 3:05