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Here is a counterexample for the stronger statement of Riesz's lemma and I don't understand it. Why for all $x$, such that $||x||=1$, there exists $y \in Y$, such that $d(x,y)<1$?

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Jaimi
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1 Answers1

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Take $x$ in the unit sphere of $X$. That is, $x \in C([0,1])$, $\|x\|_\infty \le 1$ and $x(0) = 0$. Then, $$M := \Big|\int_0^1 x(t) \, \mathrm{d}t \Big| < 1.$$

Now, take an arbitrary $y \in Y$. That is, $y \in C([0,1])$, $y(0) = 0$ and $\int_0^1 y(t) \, \mathrm{d}t = 0$. Let us show $$M \le \|x - y\|_\infty$$ by contradiction. If this would not be the case, we would have $|x(t) - y(t)| < M$ for all $t \in [0,1]$, hence we would get the contradiction $$\Big|\int_0^1 y(t) \, \mathrm{d}t\Big| \ge -\int_0^1 |x(t) - y(t) | \, \mathrm{d}t + M > 0.$$

It remains to construct $y_n \in Y$ with $\|x - y_n\|_\infty \to M$ as $n \to \infty$. To this end, for any $n$ choose a function $f_n$ with the following properties:

  • $f_n \in C([0,1])$ and $f(0) = 0$
  • $\int_0^1 f_n(t) \, \mathrm{d}t = 1$
  • $|f_n(t)| \le 1 + 1/n$ for all $t \in [0,1]$

It is clear that such functions exist. Now, we set $$y_n(t) = x(t) - f_n(t) \, \int_0^1 x(s) \, \mathrm{d}s.$$ It is easy to see that $y_n \in Y$ and we get $$\|x - y\|_\infty \le \max_t|f_n(t)| \, M \le ( 1+ \frac1n) \, M$$.

gerw
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    Thank you very much for your decision!

    I managed to simplify it a bit. If we have $x \in C([0,1])$, $||x|| \le 1$, $x(0)=0$ and $M:=\left|\int\limits_0^1 x(t) d(t)\right| < 1$, we shuld take n, such that $$ \frac{M}{1-M}<n. $$

    Then we get $f(t)$: [ f(t) = \begin{cases} \frac{(n+1)^2}{n}t, & x \in [0,\frac{1}{n+1}] \ \frac{n+1}{n}, & x \in (\frac{1}{n+1},\frac{n}{n+1}] \ -\frac{(n+1)^2}{n}t + \frac{(n+1)^2}{n}, & x \in (\frac{n}{n+1},1] \end{cases}. ].

    Let $y(t) = x(t) - M f(t)$. $y \in Y$ and $||x-y|| = ||Mf(t)|| = M \cdot \frac{n+1}{n}<1$. Hense, $d(x,Y)<1$.

     – Jaimi Dec 23 '15 at 02:33