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I'm asked to state whether or not $(X, d_u)$ and $(X, d_L)$ are complete where $d_u$ is the uniform metric and $d_L$ is the $L^1$ metric. All I need is to give the name of a supporting theorem or counterexample. $X$ is taken to be the space of continuous real-valued functions over the interval $[0, 1]$.

My thoughts are

  • if I can show compactness then this holds (using perhaps heine-borel)
  • arzela-ascoli directly
  • completeness directly

But since all I'm required is a theorem name - I'm supposing that it should be obvious. So how would one go about showing completeness or lack of for these metric spaces?

Tito
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1 Answers1

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The sup norm case is easy. If $X$ is a Hausdorff space, $C_b(X)$ - that is, continuous & bounded $\mathbb{R}$ or $\mathbb{C}$-valued functions form a banach space. Addition and scalar multiplication are obviously defined pointwise.

Fix $\epsilon>0$ and consider a Cauchy sequence $\{f_n\}$. Then there exists $N$ s.t. for all $m,n \geq N$, $||f_m - f_n || <\epsilon$. Then, by definition of sup norm, for any $x \in X$, $|f_m(x) - f_n(x)| \leq ||f_m - f_n || < \epsilon$. So, $\{f_n\}$ is pointwise Cauchy sequence in $\mathbb{R}$ or $\mathbb{C}$, and thus we can define the function $f(x) = \lim_{n \to \infty} f_n(x)$ for all $x \in X$. Now, by triangle inequality, $|f(x) - f_n(x)| \leq |f(x) - f_m(x)| + ||f_m - f_n || < |f(x) - f_m(x)| + \epsilon$. Letting $m \to \infty$, we see $|f(x) - f_n(x)| < \epsilon$ for $n \geq N$ independent of $x$ (and thus, $f_n \to f$ uniformly, showing $f$ is continuous). Taking sup over $x$ shows $||f - f_n|| \to 0$. Another application of triangle inequality shows $||f|| \leq ||f-f_N|| + ||f_N|| < \epsilon' + ||f_N||<\infty$. Thus, every cauchy sequence converges and is thus complete.

Batman
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  • The compactly supported continuous functions don't form a Banach space in general. $C_c(\mathbb{R}^n)$ is a dense subspace of $C_0(\mathbb{R}^n)$, the space of continuous functions vanishing at infinity. The space of bounded continuous functions on $X$, $C_b(X)$ or $C^\ast(X)$, is a Banach space with the sup-norm. – Daniel Fischer May 20 '14 at 19:30
  • Right, thanks.. – Batman May 20 '14 at 19:31