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Q) Let $\mu(X)<\infty$ and $V$ be the space of all measurable functions on $X$. Let $V_0 = \{f\in V| f=0, \mu \text{ -a.e. }\}$ and

$$d(f,g)= \int_X\frac{|f-g|}{1+|f-g|}d\mu$$

Show that $V/V_0$ is a complete metric space.

I've shown that $d$ is a metric but to show that $V/V_0$ is complete, I have to show that every Cauchy sequence converges to a limit in $V/V_0$ in metric $d$ but am not sure how to show that?

1 Answers1

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It suffices to show the completeness for $V$.

Since the metric on $V$ is invariant, the completeness for $V$ is equivalent to showing that \begin{align*} \sum_{n}\|f_{n}\|<\infty~~~~\rightarrow\sum_{n}|f_{n}|\in V. \end{align*} First we look at the inequality \begin{align*} \dfrac{|f|+|g|}{1+|f|+|g|}=\dfrac{|f|}{1+|f|+|g|}+\dfrac{|g|}{1+|f|+|g|}\leq\dfrac{|f|}{1+|f|}+\dfrac{|g|}{1+|g|}, \end{align*} then it is easy to see that \begin{align*} \dfrac{\displaystyle\sum_{n}|f_{n}|}{1+\displaystyle\sum_{n}|f_{n}|}\leq\sum_{n}\dfrac{|f_{n}|}{1+|f_{n}|}, \end{align*} and hence \begin{align*} \int_{X}\dfrac{\displaystyle\sum_{n}|f_{n}|}{1+\displaystyle\sum_{n}|f_{n}|}d\mu\leq\int_{X}\sum_{n}\dfrac{|f_{n}|}{1+|f_{n}|}d\mu=\sum_{n}\|f_{n}\|<\infty, \end{align*} we are done.

user284331
  • 55,591
  • Understood that $\int_{X}\dfrac{\displaystyle\sum_{n}|f_{n}|}{1+\displaystyle\sum_{n}|f_{n}|}d\mu\leq\sum_{n}|f_{n}|$. Why is completeness of $V$ equivalent to showing that $\sum_{n}|f_{n}|<\infty$? –  Nov 24 '19 at 02:40
  • Not showing this, it is that assuming this holds true, and show that $\displaystyle\sum f_{n}$ exists a.e. and belongs to the space. – user284331 Nov 24 '19 at 02:43
  • Go and check out Robert E. Megginson, An Introduction To Banach Space Theory, page 20. – user284331 Nov 24 '19 at 03:04
  • Sorry for revisiting this, but we are saying $|f_n| = \int_X \frac{|f_n|}{1+|f_n|}$ but $|f_n|$ defined this way is not a norm because $|cf_n| \neq |c||f_n|$? –  Dec 11 '19 at 04:55
  • I knew that some time ago, as you may see that I deleted the statement. But still, for invariant metric, the above argument sounds. – user284331 Dec 11 '19 at 05:10