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I am somewhat new to functional analysis (and this site, so please constructively chastise me if I commit any faux pas on here). I am one chapter into Kreyszig (Intro.to Func.Anal.) and I am already having trouble. We are asked to show that

$d(x,y)=\int_{a}^{b}|x(t)-y(t)|dt$

is a metric on continuous functions on the interval [a,b]. I believe that I can do almost all the checks in my sleep, except for one:

if $d(x,y)=0$, then $x=y$

(Notice I said IF and not IFF, the other direction is necessary as part of the problem, but is super trivial.)

I started a proof by contradiction by assuming that $x\neq y$, say for at least one t (my intuition being that for discontinuous functions this would be the minimum difference between two functions that weren't equal).

Of course our assumption is that x and y are in the space C[a,b], thus we do have continuity, therefore if the functions differ at at least one t, then there must be a "measurable" amount of t's for which the functions differ, which then would mean that our integral must be nonzero, contradiction, qed. However, I have no idea how to mathematically show all those steps, and I feel like the whole purpose of Kreyszig's "INTRO" to Func.Anal. is to teach how to prove this kind of thing by the END of the book, not asking us to know it in the first chapter... Am I over thinking this? Did I miss something very straightforward?

Manny
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3 Answers3

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Hint: using continuity, if $y(t) \neq x(t)$ there is an interval $[c,d]$ with $c<t<d$ such as $$\forall tt\in [c,d] : |x(tt) - y(tt)| > \frac12 |x(t) - y(t)| $$

mookid
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Hint : if $f \, : \, [a,b] \, \rightarrow \, \mathbb{R}^{+}$ is continuous on $[a,b]$ and $\displaystyle \int_{a}^{b} f(t) \, dt = 0$ then $f \equiv 0$ on $[a,b]$.

This will help you prove that :

$$ \forall (f,g) \in \mathcal{C}([a,b])^{2}, \; \big( \, d(f,g) = 0 \, \big) \; \Rightarrow \; \big( \, f = g \, \big). $$

pitchounet
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  • I really like this response, shame on me for not recalling the corollary regarding the $f \equiv 0$. Any chance you could point me to an elegant proof of that one? – Manny Oct 15 '14 at 09:32
  • @Manny : See, for example : http://math.stackexchange.com/questions/541179/if-integral-is-zero-and-function-is-continuous-and-non-negative-then-what-about – pitchounet Oct 15 '14 at 09:43
  • That was great. Thank you. – Manny Oct 15 '14 at 09:50
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Yor proof is correct. Just note that if you have a point $t\in[a,b]$ such that $x(t)\neq y(t)$ then there is a whole neighbourhood $[c,d]$ on which $x$ and $y$ differ. Thus $|x(t)-y(t)|>0$ on said interval and: $$\int_c^d|x(t)-y(t)|\,\mathrm{d}t>0$$

Lolman
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