$\displaystyle M =\{x\in\mathbb{R}^{(0,1]} : x \text{ is continuous on } (0,1] \text{ and } x(t) = o(\ln{}t)\ (t \rightarrow +0)\} $. We will consider metric $\displaystyle\rho(x, y) = \int\limits_{0}^{1}|x(t)-y(t)|dt$.

  1. Is $(M,\rho)$ complete space?
  2. Is $(M,\rho)$ separable space?
  3. If $(M,\rho)$ is not complete find its completion.

I've considered $\displaystyle x_k(t) = t^{\frac{1}{k}}\ln{}t$, $x_k \in M$. Obviously, $x_k \rightarrow \ln{}t\notin M$, as $\rho(x_k,\ln{}t) = 1- \displaystyle\frac{1}{(\frac{1}{k} + 1)^2} \rightarrow 0.$ In the same way we can show that it is Cauchy sequence. So, $(M, \rho)$ is not complete.

But I've stuck with two other questions. I think that completion of $M$ is $\mathbb{L}_1(0,1]$. How can one show it? If it is so, we will also have an answer for the second question. $\mathbb{L}_1(0,1]$ is separable, so $M$ is separable.


1 Answer 1


Integrability of $\ln t$ shows that $M$ is a subset of $L^{1}(0,1]$.

If $x \in C(0,1]$ then $x_i(t)=x(t)$ for $t \geq \frac 1 i, x_i(t)=itx_(\frac 1i)$ for $0<t<\frac 1 i$ defines a sequence in $M$ converging to $x$ in $L^{1}(0,1]$. Combined with the fact that continuous functions are dense in $L^{1}(0,1]$ it follows that $M$ is dense in $L^{1}(0,1]$. This implies that $L^{1}(0,1]$ is the completion of $M$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .