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I am stuck on two questions :

  1. 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 $(\int _0^1 |f(t)-g(t)|^2dt )^{\frac{1}{2}} <\infty$ does it imply that $(\int _0^1 |f(t)-g(t)|dt )<\infty$ which I am unable to prove.

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

Please give some hints

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@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.

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  • $\begingroup$ Why does $\int |f|$ not converge ;is it because $\int_{-\infty }^{-1} \frac{1}{x}=\ln x|_{-1}^{-\infty}$ does not exist? $\endgroup$ – Learnmore Oct 18 '16 at 6:34
  • $\begingroup$ It's because $\int_1^n dx/x = \ln n$ is unbounded as $n \to \infty$. $\endgroup$ – Alexis Olson Oct 18 '16 at 6:39
  • $\begingroup$ Ok thank you very much $\endgroup$ – Learnmore Oct 18 '16 at 6:43
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The answer to the first is yes, by the Cauchy-Schwarz inequality (you simply have $d_1(f,g)\leq d_2(f,g)$). But it only works because the interval $[0,1]$ is finite. On ${\Bbb R}$ it does not hold (as already mentioned in other posts).

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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.

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    $\begingroup$ How is $\sqrt{x} \in L^2(\Bbb R)$? $\endgroup$ – Alexis Olson Oct 18 '16 at 5:35
  • $\begingroup$ Sorry I meant $L^2([0,1])$ will edit immediately $\endgroup$ – marwalix Oct 18 '16 at 5:38

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