I'm currently on a theorem showing a general representation of linear functionals on the space of continuous functions on the interval $[0,1]$.

My problem is on the beginning of the proof.

First we define a linear functional $f(x)$ on the space $C[0,1]$.

After that using the Cantor theorem we show that every continuous function on a compact interval is actually bounded and thus $C[0,1]$ is a subspace of $M[0,1]$ - bounded functions on the interval $[0,1]$.

Using the Hahn-Banach theorem we show that our functional $f$ can be extended to a functional $F$ on the whole space $M[0, 1]$ with the same norm.

Maybe I've missed something foundamental, but how do we know that the functional $f$ is bounded, so that it has a norm and it's norm is the same with the extension $F$?

Thanks in advance!


No, there exist linear functionals on $C[0,1]$ that are not bounded.

In fact, for every infinite dimensional space there exists a discontinuous (and therefore unbounded) linear functional, see here.

I would suspect that there is an additional assumption somewhere that the linear functional is bounded.

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  • $\begingroup$ I guess it's just presumed. I think this theorem is a partial case of the Riesz representation theorem, in the end it is just shown that every linear functional can be expressed as a Stiltes integral with a diferential a function with bounded variation. $\endgroup$ – Nikola Jan 4 '19 at 11:12
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    $\begingroup$ @Nikola No, the Riesz theorem does not show what you said! It's not true that every linear functional is given by a Stieltjes integral. Every bounded linear functional is given by a Stieltjes integral. $\endgroup$ – David C. Ullrich Jan 4 '19 at 13:32
  • $\begingroup$ Then boundedness of the functional is just presumed in the book I"m using. $\endgroup$ – Nikola Jan 4 '19 at 14:03

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