Understanding relationships between $L^1$ and $L^2$ spaces

Question

In both my Fourier and Applied differential equations course, we have discussed the notions of treating certain function spaces as vector spaces. In the DE course, we defined the $L^2(I)$ space as the set $\{f: I \rightarrow \mathbb{R} : \int_I f^2 < \infty\}$. This definition is in line with the one we specified during our Fourier analysis course too (in this context we discussed it in terms of "$L^2$ convergence"). I am also aware that this is not technically the correct definition, as we need to define the space in terms of an equivalence relationship on almost everywhere functions. But this was mostly left out as we're not covering any measure theory.

One of the things we proved in the DE course, was that $L^2(I) \subset L^1(I)$. I don't know very much linear algebra, and so most of this has been self-taught from online materials or discussed with a tutor, but it is my understanding that for any $L^p$ space, the only one on which an inner product is defined is for $p =2$. I.e., that $L^1$ does not have any inner product.

In our DE course, we worked almost exclusively with real numbers, and so within $L^2$ we defined the inner product as $\langle .,.\rangle: L^2 \times L^2 \rightarrow \mathbb{R}$, with $\langle f,g\rangle = \int_I fg$. I also know that any inner product space is a normed space, and the norm is defined $\|f\| = \sqrt{\langle f,f\rangle}$.

Given all of this, the part I am really not following is coming from my Fourier analysis course. We have defined the set $\mathcal{R} = \{f: S^1 \rightarrow \mathbb{C}: f \text{ is Riemann integrable}\}$. On this set we have said that there exists an inner product given by $$\langle f,g\rangle = \frac{1}{2\pi}\int_{-\pi} ^{\pi} f(\theta)\overline{g(\theta)} d\theta$$

This is really confusing me because apart from the function now being complex-valued (and having the constant term in front), it looks exactly like the inner product in the $L^2$ case. So now it seems like there is an inner product defined on $L^1$, and it is in fact the same as that for the $L^2$ case. But this seems like it is contradicting the fact that the $L^1$ space has no inner product? What exactly is going on here?

I double-checked my notes and on video, and it really does seem like we are just asserting in the Fourier class that the functions are merely Riemann integrable. I was wondering if maybe it's because $L^1 \subset L^2$, since at least in real analysis, if $f$ is Riemann integrable, then so is $f^2$. But then because we proved $L^2 \subset L^1$, we would have $L^2(I) = L^1(I)$. And so again we have an inner product defined on $L^1$ because now we can just use the $L^2$ inner product! I am really not sure what is happening, but I understand that perhaps my lack of linear algebra knowledge is making me miss something extremely obvious. I would greatly appreciate any pointers as to where I am going wrong.

Answer 1

The spaces $L^1$ and $L^2$ are usually defined with respect to the Lebesgue integral. But your set $\mathcal{R}$ is defined to consist only of functions which are Riemann integrable. In particular, it can be shown that every Riemann integrable function is bounded, and so the space $\mathcal{R}$, although it is contained in $L^1$, is only a subset of it (and also of $L^2$). So you only have an inner product on a subset of $L^1$, and you will not be able to extend it to all of $L^1$.

It's probably more useful to view $\mathcal{R}$ as a subset of $L^2$, endowed with the (restriction of the) $L^2$ inner product.

Answer 2

What you are missing here is the Hölder-inequality. This states the following. Let $p,q> 1$, s.t. $p^{-1} + q^{-1} = 1$ and let $f\in L^p$ and $g\in L^q$. Then $fg \in L^1$, furthermore $$ ||fg||_{L^1} \leq || f ||_{L^p} ||g||_{L^q}. $$ If $p=q=1/2$, then this is just the Cauchy-Schwarz inequality. Now, if $I$ is a measurable set of finite measure and $f\in L^2$, then, due to the H-inequality $$ ||f||_{L^1(I)} = || f \cdot 1|| _{L^1(I)} \leq ||f||_{L^2(I)} ||1||_{L^2(I)} = ||f||_{L^2} \sqrt{\mu(I)}, $$ where $\mu(I)$ is the measure of the set. If $\mu=\lambda$, i.e. the measure is the Lebesgue-measure and $I$ is a finite interval, then $\lambda(I)$ is just the length of $I$. Hence $f\in L^1(I)$.