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?

  • $\begingroup$ 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. $\endgroup$
    – Jose27
    Apr 2, 2019 at 1:59
  • $\begingroup$ 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. $\endgroup$
    – ztyh
    Apr 2, 2019 at 2:11
  • $\begingroup$ They do assume $R$ is not all of $C(I_n)$. $\endgroup$
    – ztyh
    Apr 2, 2019 at 2:33

1 Answer 1


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.

  • $\begingroup$ What exactly do I have to say about $T$ being well defined? $\endgroup$
    – ztyh
    Apr 2, 2019 at 14:47
  • $\begingroup$ 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? $\endgroup$
    – ztyh
    Apr 2, 2019 at 15:31
  • 1
    $\begingroup$ @ztyh No you cannot show continuity this way because we don't know that $\lim a_n$ exists. $\endgroup$ Apr 2, 2019 at 23:06
  • $\begingroup$ 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? $\endgroup$
    – ztyh
    Apr 3, 2019 at 22:32
  • $\begingroup$ @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. $\endgroup$
    – Sam Wong
    May 10, 2021 at 3:05

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