I'm self-studying from Erwin Kreyszig's Introductory Functional Analysis with Applications, and I'm having trouble understanding a specific part of a proof in Section 1.5-9. Here, they prove that the set of all continuous real-valued functions on $J=[0,1]$ equipped with the metric:
$ d(x,y)=\int\limits_0^1 |x(t)-y(t)| dt $
is not a complete metric space. I've reproduced the two pages containing the proof.
My issue is understanding the last part of the proof, the part where it reads thus:
"Hence $d(x_m,x) \rightarrow 0$ would imply that each integral approaches zero (understood this part) and, since $x$ is continuous, we should have:
$x(t)=0 \text{ if } t \in [0,\frac{1}{2})$
$x(t)=1 \text{ if } t \in (\frac{1}{2}, 1]$
But this is impossible for a continuous function (The part I do not understand)."
I understand that $\int\limits_0^{1/2}|x(t)| dt=0$ implies that $x(t)=0 \text{ if } t \in [0,1/2)$. but what about the second implication $x(t)=1 \text{ if } t \in (\frac{1}{2}, 1]$. Why would $x(t)=1$ in the entire interval $(1/2, 1]$ when there is a term $\int\limits_{1/2}^{a_m} |x_m(t)-x(t)| dt$ in the middle?
I'd love some clarification, in case I am blanking out on something obvious.
Thanks in advance for your help.
The term in the middle doesn't matter, it goes to zero. You have $$ \int_{a_m}^1|1-x(t)|\,dt\to0. $$ Suppose that $x(t_0)\ne1$ for some $t_0\in(1/2,1)$. By continuity there exists an interval $(c,d)\subset(1/2,1)$ with $t_0\in(c,d)$ and $\delta>0$ such that $|1-x(t)|>\delta$ on $(c,d)$. By choosing $m$ big enough, we can guarantee that $a_m<c$. Then $$ \int_{a_m}^1|1-x(t)|\,dt\geq\int_c^d|1-x(t)|\,dt\geq\delta(d-c), $$ and you have a contradiction since the left-hand-side goes to zero with $m$.