I have got this sequence of function defined on the space $\Re_{[0,1]}$ with the integral metric. $n\in\mathbb{N}$ and $M\leq n$, $M=\sup f_n$

How can we know: $$\int_{0}^{\frac{1}{n^2}} (n-M) dx+\int_{\frac{1}{n^2}}^{\frac{1}{M^2}} \left(\frac 1x -M\right)dx \leq \int \left| f-f_n\right| dx$$ Given the fact the book presents this:

\begin{align}f:[0,1]&\to\mathbb{R}\\ x&\rightarrow\begin{cases}\frac{1}{\sqrt{x}}&\text{if }x\geq\frac{1}{n^2}\\ n&\text{otherwise}\end{cases}\end{align}

\begin{align}d(f,f_n)&=\int_0^1\left|f-f_n\right|\\ &\geq \int_0^{\frac{1}{M^2}}\left|f-f_n\right|\\ &\geq \int_{0}^{\frac{1}{n^2}}n-M\;dx+\int_{\frac{1}{n^2}}^{\frac{1}{M^2}}\frac{1}{\sqrt{x}}-M\;dx\\ &=\frac 1M - \frac 1n \end{align}

  • $\begingroup$ You might want to look at the formatting of your post - there's a few oddities cropping up! :) $\endgroup$
    – lioness99a
    Mar 16, 2017 at 13:11
  • $\begingroup$ If you know how to improve it you are welcome! $\endgroup$ Mar 16, 2017 at 13:32
  • $\begingroup$ Thank you I was not capable of that. Sorry if I was not polite in my last comment! $\endgroup$ Mar 16, 2017 at 14:53
  • $\begingroup$ How does my edit look? Feel free to update if I've got anything wrong $\endgroup$
    – lioness99a
    Mar 16, 2017 at 14:54
  • 1
    $\begingroup$ I didn't downvote it, but I can see why someone might have: The question is very hard to understand. $\endgroup$
    – zhw.
    Mar 20, 2017 at 6:58

1 Answer 1


$$ \int_0^{\frac1{M^2}} |f - f_n| ~\mathrm{d}x = \int_0^{\frac1{n^2}} |f - f_n| ~\mathrm{d}x + \int_{\frac{1}{n^2}}^{\frac1{M^2}} |f - f_n| ~\mathrm{d}x $$ seeing that $M \leq n$ and so $\frac{1}{M^2} \geq \frac{1}{n^2}$.

In the first integral, $f(x) = n$ by definition since $x \in (0, \frac1{n^2})$. By assumption $f_n \leq M$. So $f - f_n \geq n - M$ and hence $|f - f_n| \geq n-M$, seeing as $n-M$ is non-negative.

In the second integral, $f(x) = \frac{1}{\sqrt{x}}$ by definition as $x\in (\frac{1}{n^2}, \frac{1}{M^2})$. Since $x \leq \frac{1}{M^2}$, we have that $\sqrt{x} \leq \frac{1}{M}$ and $f(x) \geq M$. So $f(x) - M \geq 0$. Again since $f_n \leq M$ we have that $f - f_n \geq f - M \geq 0$ so that $|f - f_n| \geq \frac{1}{\sqrt{x}} - M$.

  • $\begingroup$ Thank you for answering. What I cannot understand is precisely what you do not point out. Why is $f-f_n\geqslant n-M$? Should not be $f-f_n\leqslant n-M$ instead, since $f_n\leqslant M$ and $f_n=n$ when $x \in [0,1/n^2]$ $\endgroup$ Mar 24, 2017 at 12:21
  • 1
    $\begingroup$ @PedroGomes: $f_n \leq M$ so $- f_n \geq -M$ so $f + (-f_n) \geq f + (-M) = n - M$ $\endgroup$ Mar 24, 2017 at 16:42
  • $\begingroup$ Thank you! The points are yours $\endgroup$ Mar 24, 2017 at 18:12

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