Comparing different norms on $C[0,1]$ 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.
 A: 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?
A: Every metric space is first countable thus sequential continuity inplies continuity.

$a)$Let $f_n \in X_1$ such that $f_n \to f$.
In other words $\sup_{x \in [0,1]}|f_n(x)-f(x)| \to 0$.
Thus $d_2(I(f_n),I(f))=d_2(f_n,f)= \int_0^1|f_n(x)-f(x)|dx \leq \sup_{x \in [0,1]}|f_n(x)-f(x)| \to 0$
So $I:X_1 \to X_2$ is continuous.

.

$c)$Let $f_n \in X_3$ such that $f_n \to f$ in $X_3$
In other words $\sqrt{\int_0^1|f_n(x)-f(x)|^2dx} \to 0$
Now $$d_2(I(f_n),I(f))=d_2(f_n,f)=\int_0^1|f_n(x)-f(x)|dx=$$ $$\int_0^11|f_n(x)-f(x)| \ dx \leq \sqrt{\int_0^1 1^2dx} \sqrt{\int_0^1|f_n(x)-f(x)|^2dx} \to 0$$
Thus $I:X_3 \to X_2$ is continuous

.

$b)$Here take $f_n=x^n$ and $f=0$
We have that $d_2(f_n,f) \to 0$ but $d_1(I(f_n),I(f))=d_1(f_n,f)=1$ which does not go to zero.
So $I:X_2 \to X_1$ not continuous at the zero function.

A: An answer to part of it.
To show that $d_3\geq d_2$: Let $h(x)=|f(x)-g(x)|.$ Then $$d_3(f,g)\geq d_2(f,g)\iff$$ $$\iff  2\int_0^1h(x)^2dx\geq 2\int_0^1h(x)dx \cdot \int_0^1h(y)dy \iff$$ $$\iff \int_{x=0}^1\int_{y=0}^1 h(x)^2dxdy+\int_{x=0}^1\int_{y=0}^1h(y)^2dxdy\geq 2\int_{x=0}^1\int_{y=0}^1 h(x)h(y)dxdy \iff$$ $$\iff \int_{x=0}^1\int_{y=0}^1(h(x)-h(y))^2dxdy\geq 0.$$
Hence id$_{[0,1]}:X_3\to X_2$ is continuous.
A: a) and c) are continuous but b) is not. The question has no relation to completeness. Just use sequential continuity.
A: Essentially everything has been said to understand the problem, but let me point out how your problem relates to measure theory and functional analysis. This makes the solution very clear in my eyes. 
Note that in this case (certainly not always) $\vert\vert f \vert\vert_i = d_i(f,0)$ ($i=1,2,3$) define norms on the vector space $C[0,1]$. These norms are just the $L^p-$norms for $p=\infty$ (in case of $d_1$), $p=1$ (in case of $d_2$) and $p=2$ (in case of $d_3$). Since $(0,1)$ is a finite measure space, the Hölder inequality implies
$$ \vert\vert u\vert \vert_{L^1} \le c_1 \vert \vert u \vert \vert_{L_2} \le c_2 \vert \vert u\vert\vert_{L^\infty}.$$
Now to functional analysis: You can show that a linear map $T:( E,\vert\vert\cdot\vert\vert_E) \rightarrow ( F,\vert\vert\cdot\vert\vert_F)$ between normes spaces is continuous if and only if it is bounded, i.e. there is a constant $c>0$ with
$$ \vert\vert Tu\vert\vert_F \le c \vert\vert u \vert\vert_E.$$
From the previous  inequalities you can simply conclude that the identity is continuous as a map
$$ X_1 \rightarrow X_3 \quad \text{and}\quad X_3 \rightarrow X_2 \quad \text{and}\quad X_1 \rightarrow X_2$$
This solves a) and c) of your question.
If b) was true you would also have an inequality of the form
$$\vert\vert u \vert\vert_{L^\infty} \le c_3 \vert\vert u \vert\vert_{L^1}.$$
To show there is no such inequality you can use the sequence of functions suggested by Mathematician 42.
