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I am asked to show the following:

Proposition. There is a unique injection $j: L^2([0, 1]) \hookrightarrow L^1([0,1])$ which continuously extends $Id: C([0, 1]) \to C([0, 1])$.

Here $L^1([0, 1])$ and $L^2([0, 1])$ are defined as completions of $C([0, 1])$ in the usual $L^1$ and $L^2$ norms, respectively (but in this case we can simply use Riemann integral in place of Lebesgue integral). So the first step of proving the above proposition (without worrying about injectivity for now) is essentially demonstrating (for continuity)

Proposition. Every Cauchy sequence in $C([0, 1])$ in the $L^2$ norm is also Cauchy in the $L^1$ norm. Moreover, two Cauchy sequences that converge to the same function in $L^2([0,1])$ also converge to a same function in $L^1([0,1])$.

If this proposition is proved, then the extension $j$ is uniquely determined, and we are left with injectivity.

However, I have trouble proving the above proposition. Since the above is false for $L^1(\mathbb{R}^d)$ and $L^2(\mathbb{R}^d)$, there must be something peculiar about the compactness of $[0, 1]$, but I could not figure out how. (I tried uniform continuity to no avail.) Could anyone give me a hint on this? Thanks in advance.

4ae1e1
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    The interval $[0,1]$ has finite measure. That is all you need for the continuous inclusion $L^2 \hookrightarrow L^1$. And a hint: Cauchy-Schwarz inequality. – Daniel Fischer Feb 26 '14 at 20:03
  • @DanielFischer So it's a plain Cauchy-Schwarz! I can't believe I'm so dumb as to not see it. Thank you very much. – 4ae1e1 Feb 26 '14 at 20:07

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I think the essential result you need is:

$L^p$ and $L^q$ space inclusion

So you have $L^{2}$'s inclusion inside of $L^{1}$ as a special case. This result appeared in analysis very often.

For your question, you can show that $[\int^{1}_{0}f^{p}]^{1/p}$ is an increasing function for $p$ on $[1,+\infty)$ using Holder inequality. So I think if you can show a sequence $f_{i}$ is Cauchy in $L^{2}$, then it should follow that $|f_{m}-f_{n}|<\epsilon,\max(m,n)>N$ in $L^{1}$. Then you can show $f_{i}$ is Cauchy in $L^{1}$ as well.

But I think the question requires you to approach a function in $L^{1}$ by a continuous function in $C[0,1]$, and you need to fill in some details on $C[0,1]$ dense in $L^{p}$. I do not really know what does it mean "continuously extends" at here except using this interpretation.

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Bombyx mori
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  • Thanks a lot. In fact I'm not that advanced in measure theory yet. In particular I only learned Lebesgue measure and haven't touched general measure theory, so the linked question is a bit hard for me to follow here and there. Anyway, for this simple special case, a Cauchy-Schwarz on $L^2$ will do: $\int_0^1 |f_m - f_n| \cdot 1, dx \le \sqrt{\int_0^1 (f_m - f_n)^2 dx} \sqrt{\int_0^1 1^2 dx}$. I somehow didn't realize this before it is pointed out in the comments that the only thing necessary is $m([0, 1]) < \infty$. – 4ae1e1 Feb 26 '14 at 20:19