Counterexample for the stronger statement of Riesz's lemma 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 $d(x,y)<1$?
 A: 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
$$\|x - y\|_\infty \le \max_t|f_n(t)| \, M \le ( 1+ \frac1n) \, M$$.
