Identity function on $C[0,1]$

Question

Consider $C[0,1]$ with metrics $d_1$ and $d_2$ where $$d_1(f,g)=\int_0^1 \vert f(x)-g(x) \vert dx$$ and

$$d_2(f,g)=\Big(\int_0^1 \vert f(x)-g(x) \vert^2 dx\Big)^\frac{1}{2}$$

Let $id$ be the identity map of $C[0, 1]$ onto itself

Then $id:(C[0,1],d_2) \rightarrow (C[0,1],d_1)$ is continuous

Here's my try:

$\vert f(x)-g(x) \vert \in [0,1]$ implies $\vert f(x)-g(x) \vert^2 \leq \vert f(x)-g(x) \vert$

Hence $d_1(f,g) \leq d_2(f,g)$ and so the $id$ is Lipchitz of constant $1$, continuity follows!

Is this correct?

Answer 1

Your proof is not correct. The standard proof uses Hölder's / Cauchy-Schwarz inequality $$\left|\int_0^1 FG \, dx\right|\le \left(\int_0^1 F^2\, dx\right)^\frac12\left(\int_0^1 G^2\, dx\right)^\frac12$$ with $F=|f-g|$ and $G=1$.

Answer 2

Your proof is not correct, since it is not true in general that$$\bigl\lvert f(x)-g(x)\bigr\rvert^2\leqslant\bigl\lvert f(x)-g(x)\bigr\rvert.$$

Since $\operatorname{Id}$ is linear, in order to prove that it is continuous, all you need is to prove that there is a $k\in\mathbb{R}^+$ such that$$(\forall f\in\mathcal{C}[0,1]):d_1(f,0)\leqslant kd_2(f,0).$$But\begin{align}d_1(f,0)&=\int_0^1\bigl\lvert f(x)\bigr\rvert\,\mathrm dx\\&=\int_0^11\times\bigl\lvert f(x)\bigr\rvert\,\mathrm dx\\&\leqslant\sqrt{\int_0^11^2\,\mathrm dx}\sqrt{\int_0^1\bigl\lvert f(x)\bigr\rvert^2\,\mathrm dx}\\&=d_2(f,0).\end{align}