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I'm working on the following question in Royden:

Define $\phi: C[a,b] \rightarrow \mathbb{R}$ by $\phi(f)=f(a)$ for all $f\in C[a,b]$. Show that there is no functional $h \in L^1[a, b]$ for which: $$\phi(f) = \int_a^bhf \text{ } \forall f \in L^{\infty}[a,b]$$

The problem says to use a corollary about Hahn Banach that every bounded linear functional has an extension $\psi$ so that $\psi(x) = \|x\|$ (for each $x$ in the extension) and $\|\psi\| = 1$. I'm getting confused by this question though:

  1. Is the question asking whether evaluation of $f$ at $a$ can be represented by that integral formula for some $h$ in $L^1$? I'm having trouble seeing how to begin.

  2. In general, I'm getting confused by the relations between evaluation functionals in $X^{**}$, elements on $X^*$, and $X$ as they pertain to introductory examples to Hahn-Banach extensions. I think the confusion I have at the end of this post stems from the same issue.

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2 Answers 2

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You have written $\forall f \in L^{\infty}[a,b]$ whereas the left side $\phi(f)$ is defined only for $f \in C[0,1]$. So the corrected version should have ' for all $f \in C[0,1]$'. To prove the result let $a<c<d<b$. Let $\epsilon$ be a small positive number and define $f$ by the following properties: $f=1$ on $[c,d]$ $f=0$ on $[a,c-\epsilon]$, $f=0$ on $[d+\epsilon,b]$ and $f$ linear in $[c-\epsilon, c]$ as well as $[d,d+\epsilon]$. [ 'Linear' means the graph is a straight line. Observe that $0\leq f(x) \leq 1$ for all $x$]. Since $\phi(f)=f(a)=0$ we get $\int_c^{d} h(x)dx+\int _{[c-\epsilon, c]} f(x)h(x) dx+\int _{[d,d+\epsilon]} f(x)h(x) dx=0$. The sedond and the third terms tend to 0 as $\epsilon \to 0$ (because $0\leq f(x) \leq 1$). Hence $\int_c^{d} h(x)dx=0$ for every sub-interval $(c,d)$ of $(a,b)$. This implies $h=0$ almost everywhere. But then $\int_a^{b} f(x)h(x)dx=0$ for all $f \in C[0,1]$. You get a contradiction by taking $f \equiv 1$.

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Note that $\phi$ is bounded linear functional on $C[a,b]$ with norm 1.

So by corollary to Hahn-Banach theorem, $\phi$ can be extended linearly to $L^\infty[a,b]$ by keeping norm same (which is 1).

Now if function $h$ Satisfying above condition exists then its $L^1$-norm should be 1.

But as proved in above answer $h=0$ a.e.

Which is a contradiction.

Hence there cannot exist $h$ satisfying the condition.

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  • $\begingroup$ @yoshi Are you convinced by the answer I have given? $\endgroup$
    – Mayuresh L
    May 5, 2018 at 14:21

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