Let $X_i=(C[0, 1], d_i), i = 1, 2, 3$, be the metric spaces where
$$\begin{aligned}d_1(f, g) &= \sup_{x∈[0,1]} |f(x) − g(x)|\\
d_2(f, g) &=\int_{0}^{1}|f(x) − g(x)| \, \mathrm dx\\
d_3(f, g) &= \left(\int_{0}^{1}|f(x) − g(x)|^2 \, \mathrm dx\right)^{1/2}\end{aligned}$$
Let $\operatorname{id}$ be the identity map of $C[0, 1]$ onto itself. Pick out the true statements.
- a) $\operatorname{id} : X_1 \to X_2$ is continuous.
- b) $\operatorname{id} : X_2 \to X_1$ is continuous.
- c) $\operatorname{id} : X_3 \to X_2$ is continuous.
My attempt: I know that $d_1$ is complete in $C[0,1]$ and $d_2$ and $d_3$ are not complete in $C[0,1].$ So option a) it will not be continuous, because $d_1$ is complete and $d_2$ is not complete, they don't match.
Option b) and option c) are both correct because both are incomplete, there will be homeomorphism between $d_1$ and $d_2.$
Is my answer correct or not? I would be more thankful if my mistakes are corrected.
Best Answer
Notice that $$d_2(f,g)=\int_{0}^1|f(x)-g(x)|\mathrm{d}x\leq \sup_{x\in [0,1]}|f(x)-g(x)|=d_1(f,g).$$ Suppose $(f_n)_n$ is a convergent sequence in $(X_1,d_1)$ with limit $f$, then by the above inequality we also have that $(f_n)_n$ converges to $f$ w.r.t. the metric $d_2$. Thus $Id:X_1\rightarrow X_2$ maps converging sequences to converging sequences. Since $d_1$ is complete this is equivalent to $Id:X_1\rightarrow X_2$ maps Cauchy sequences to Cauchy sequences. Hence this map is continuous.
Consider the functions $f_n$ defined by $$f_n(x)=\begin{cases} 0 & \mbox{ if } x\in [0,\frac{1}{2}-\frac{1}{n}]\cup [\frac{1}{2}+\frac{1}{n},1]\\ n(x-\frac{1}{2}+\frac{1}{n}) & \mbox{ if } x\in [\frac{1}{2}-\frac{1}{n}, \frac{1}{2}]\\ 1-n(x-\frac{1}{2}) & \mbox{ if } x\in [\frac{1}{2},\frac{1}{2}+\frac{1}{n}] \end{cases}.$$
It is straightforward to check that $\lim_{n\rightarrow \infty}\int_{0}^1|f_n(x)|\mathrm{d}x=0$. Thus $f_n\rightarrow 0$ in w.r.t. the $d_2$-metric. But $f_n$ does not converge to $0$ in the $d_1$-metric.
Try to think in terms of sequences. Can you find the rest?