# Thread: [SOLVED] Metric Space and Inequality

1. ## [SOLVED] Metric Space and Inequality

Problem:
For any two functions f and g in C([0,1],R), define

a. Prove that this defines a metric on C([0,1],R)
b. Prove that the following inequality relating this metric and the uniform metric:

c. Compare the concepts of convergence of a sequence of functions in this metric and in the uniform metric.
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Attempt:
a. First, C([0,1],R) denotes that the function is continuous.
We need to prove that

i. if and only if f = g. If f = g, then

ii. d*(f,g) = d*(g,f) (symmetry): This is true since , thus

iii. Triangle Inequality:

Thus, d*(f,g) is a metric on C([0,1],R)

b.We need to prove that

I'm unsure where to start. I try to divide both sides by (b-a) which yields

Here, I use the Mean Value Theorem for integrals, since the function on [a,b] is continuous, then there exists a c such that

c. I'm not sure.

Thank you for your time.

2. Originally Posted by Paperwings
a. Prove that this defines a metric on C([0,1],R)
You almost proved this. You did not prove the "only if" part in the second part of being a metric space.
If, $\int_a^b |f(x)-g(x)| dx = 0$ then it means $f(x)-g(x)=0$ i.e. $f=g$.
Because if $f,g$ are continous then $|f-g|$ is continous.

To show $d^*(f,g) \leq (b-a)d(f,g)$. I think here $f,g\in \mathcal{C}([a,b],\mathbb{R})$. I am also assuming that you define $d(f,g) = \sup\{ |f(x)-g(x)| : x\in [a,b]\}$. Clearly, $|f(x)-g(x)|leq d(f,g)$ for all $x\in [a,b]$. And thus, $\smallint_a^b |f-g| \leq \smallint_a^b d(f,g) \implies d^*(f,g)\leq (b-a)d(f,g)$.

3. Thank you very much TPH. I also understand part (c) more clearly as well. Cheers.