Two questions on continuous functions I am stuck on two questions :


If $f,g\in C[0,1]$ where $C[0,1]$ is the set of all continuous functions in $[0,1]$ then is the mapping $id:(C[0,1],d_2)\to (C[0,1],d_1)$ continuous ? where $id$ denotes the identity mapping.


where  $d_2(f,g)=(\int _0^1 |f(t)-g(t)|^2dt )^{\frac{1}{2}} $ and $d_1(f,g)=(\int _0^1 |f(t)-g(t)|dt )$
?

2.If $f\in L^2(\Bbb R)$ does it imply that $f\in L^1(\Bbb R)$.

My try:

1.The first question reduces to proving the fact that if $(\int _0^1 |f(t)-g(t)|^2dt )^{\frac{1}{2}} &lt;\infty$ does it imply that  $(\int _0^1 |f(t)-g(t)|dt )&lt;\infty$ which I am unable to prove.

I am also unable to conclude anything for the 2nd one.

Please give some hints
@marwalix got the first question.

For the second, consider the function $f(x) = \min\left\{1,\frac{1}{x}\right\}$.

$$\int_{\Bbb R} |f|^2 = \int_{-\infty}^{-1} \frac{dx}{x^2} + \int_{-1}^1 dx + \int_{1}^{\infty} \frac{dx}{x^2}= 1+2+1 = 4$$

but $\int_{\Bbb R}|f|$ doesn't converge.
Take $f(x)=\sqrt{x}$ one has $f\in L^2([0,1])$ and obviously $f\notin L^1([0,1])$ for the integral diverges at $0$.

Now consider a sequence of functions in $L^2$ that converges to the above $f$. If the identity were continuous the sequence should converge in $L^1$ but we have seen that $f\notin L^1$ so the identity is not continuous.