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Royden leaves the following as an exercise:

Let $X$ be a linear subspace of $C[0,1]$ that is closed as a subset of $L^2[0,1]$. $X$ is closed, and there is a constant $M$ such that $\|f\|_\infty\le M\|f\|_2,\,\|f\|_2\le\|f\|_\infty$ for all $f\in X$.

Show that for all $y\in[0,1]$, there is a function $k_y\in L^2[0,1]$ with $f(y)=\int_0^1k_y(x)f(x)\,\mathrm{d}x$ for all $f\in X$.

This comes after a chapter on basic linear operator theory, with theorems like the open mapping and closed graph theorems covered, and some lemmas on when we can know if an operator is continuous/open, and also some theorems on the isomorphy of finite dimensional linear spaces with $\Bbb R^n$.

I have studied more measure theory than is covered thus far in Royden's book, and I have seen the proof of the Riesz representation theorem and I can tell you that since $T_y\in L^2[0,1]^*$, $T_y:f\mapsto f(y)$ is a continuous linear functional on the subset $X$ (by extreme value theorem), it must have the representation $k_y\in L^2[0,1]$ as the $L^p$-conjugate of $2$ is again $2$ - I hope I'm using this theorem right.

So there is an immediate and rather too powerful solution. Royden does not cover this theorem until much later in the book I believe, yet he expects students to find, or show the existence of, such a delta-esque function $k_y$ using basic linear operator theory.

What was the solution he had in mind? I'd be happy with any linear theoretic solution really, since knowing exactly what Royden intended is hard! I just expect one can get away with arguments weaker than the Riesz Representation theorem.

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  • $\begingroup$ Is there a previous section on Hilbert spaces??? $\endgroup$ Commented Feb 5, 2022 at 11:44
  • $\begingroup$ @DavidC.Ullrich Maybe later on, but not at this stage in the text! I know the Hilbert space version of Riesz representation also $\endgroup$ Commented Feb 5, 2022 at 11:45
  • $\begingroup$ That being said, how would you do it using the Hilbert space version? @DavidC.Ullrich $\endgroup$ Commented Feb 5, 2022 at 11:52
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    $\begingroup$ $X$ (with the $L^2$ norm) is a Hilbert space. The inequality $||f||_\infty\le c||f||_2$ shows that $\lambda$ is a bounded linear functional on $X$, if $\lambda f=f(0)$. So there exists $K\in X$ with $\lambda f=\langle f, K\rangle$. $\endgroup$ Commented Feb 5, 2022 at 11:56
  • $\begingroup$ Royden (in at least the third ed.) proves RRT for $L_p$-spaces in section 6.5 (the section on Classical Banach Spaces). Your problem comes later in this edition. $\endgroup$ Commented Feb 5, 2022 at 12:30

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The key is that $\dim X<\infty$. Suppose that $\dim X=\infty $. Fix $m\in\mathbb N$. Then there exist $g_1',\ldots,g_m'$, linearly independent. Using Gram-Schmidt, we get an orthonormal set $\{g_1,\ldots,g_m\}\subset X$. Given $c_1,\ldots,c_m\in\mathbb C$ and $x\in[0,1]$ $$ \Big|\sum_{n=1}^mc_ng_n(x)\Big|^2\leq\Big\|\sum_{n=1}^mc_ng_n\Big\|_\infty^2\leq M^2\Big\|\sum_{n=1}^mc_ng_n\Big\|_2^2=M^2\,\sum_{n=1}^m|c_n|^2. $$ In particular, with $c_n=\overline{f_n(x)}$, $$ \Big|\sum_{n=1}^m|g_n(x)|^2\Big|^2\leq M^2\,\sum_{n=1}^m|g_n(x)|^2, $$ so $$ \sum_{n=1}^m|g_n(x)|^2\leq M^2. $$ Integrating, $$ m=\sum_{n=1}^m\|g_n\|^2\leq M^2, $$ which is a contradiction as $m$ is arbitrary. Thus $X$ is finite-dimensional, with orthonormal basis $\{g_1,\ldots,g_m\}$. Now let $$ k_y=\sum_{n=1}^m g_n(y)\,\overline{g_n}. $$ Then $$\tag1 \int_0^1k_y(x)\,g_n(x)\,dx=g_n(y). $$ By linearity, $(1)$ holds for all $f\in X$.

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  • $\begingroup$ How do you assume a countable orthonormal basis exists? $\endgroup$ Commented Feb 5, 2022 at 12:55
  • $\begingroup$ By definition of dimension, and the fact that $L^2[0,1]$ is separable. In any case, the argument only needs that the set is orthonormal, so the fact that $X$ is separable (which it is) is irrelevant. $\endgroup$ Commented Feb 5, 2022 at 12:58
  • $\begingroup$ The space may have uncountable dimension for all I know. I’ve not studied infinite dimensional linear algebra! As it happens, the whole point of Royden’s exercise is to show that because $k_y$ exists, the space $X$ is finite dimensional... I’ll upvote your answer but presently don’t understand it fully $\endgroup$ Commented Feb 5, 2022 at 13:00
  • $\begingroup$ You are focusing on the wrong point. As I said, all that matters is that if $X$ is not finite-dimensional then it has orthonormal subset of any finite cardinality. I have edited the answer to phrase it that way. $\endgroup$ Commented Feb 5, 2022 at 13:05

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