Confirm a Metric on the set of Continuous function on [0,1]

Question

I was hoping I could get some resolution on this problem of mine. I had to create a metric on the set of continuous functions on $[0,1]$ and I came up with this $$d(f,g) = \sup\left\{\left| \int_a^b(f-g)(x)\, \mathrm dx\right| : a,b\in\mathbb{R}\cap[0,1]\right\}$$

I know that a better solution would be to put the absolute values inside the integral enclosing $f-g$, but I didn't think of that at the time.

Anyway, I'm rather convinced that this isn't a metric as my proof is a little handwavy, (mainly the part of $d(f,g)=0 \implies f=g$), but my professor is somewhat convinced it is a metric regardless of my proof.

I'd be grateful if anyone could perhaps confirm whether or not this is indeed a metric, and if possible, give me tips on how to improve my proof if necessary.

Thanks a ton. The proof for the difficult part is below (The other parts for a metric are more or less trivial)

Proof for the part in question:

Answer 1

I think the problem with the proof is the conclusion from $\int_u^v (f(x)-g(x))dx =0$ to existence of some $x_0\in[u,v]$ such that $f(x_0) \neq g(x_0)$. How do you derive this in detail?

Answer 2

$f \neq g$ on $[p; q]$ doesn't imply $\int_p^q (f - g)(x)\, \mathrm dx \neq 0$. It can be fixed: if, for example, $f(x_0) > g(x_0)$, then $f(x) > g(x)$ in some neighborhood of $x_0$. And integral of strictly positive function isn't zero.

Also, formally in cases $x_0 = u$ or $x_0 = v$ you will not have $(x_0 - \varepsilon; x_0 + \varepsilon) \subseteq [u; v]$, and if $x_0 \in \{0, 1\}$ you can't even integrate over such interval, but it is easy to fix by taking one-side neighborhood.

Answer 3

Alternatively, you can prove this this way:

If $d(f,g)=0$, then $\int_0^t f(x)dx = \int_0^t g(x)dx$ for all $t$. Then, differentiating with respect to $t$, we get $f(t)=g(t)$.