Prove this is a metric, what else should I consider?

Question

Let $C_b(\mathbb{R})$ be the space of the bounded continuous functions with values in $\mathbb{C}$ defined in $\mathbb{R}$ ($f:\mathbb{R}\rightarrow\mathbb{C}$) prove that: with $x\in \mathbb{R}, h\in[0,1]$ $$sup|\int_{x}^{x+h}(f(t)-g(t))dt|$$ is a metric.

I tried using the fundamental theorem of calculus to do the integral, i've got to $|F(x+h)-F(x)-G(x+h)+G(x)|$

I'd say since the absolute value is defined (or whatever norm is this) it automatically is positive and $0$ iff $f$ and $g$ are the same function. also symmetry is easy since you can multiply for $|-1|$, and finally triangular inequality is also part of the norm, is this correct?

Well in my imagination this works for $\mathbb{R}$, but this functions are made for $\mathbb{C}$ and they are bounded, when do I have to consider this?

and finally: what is that $b$ in $C_b$ ?

Answer 1

Let's verify the triangular inequality: Given, $f,g,k\in C_b(\mathbb{R})$, \begin{align*} d(f,k)&=\sup_{x,h}\left|\int_x^{x+h}f(t)-k(t)dt\right|\\ &=\sup_{x,h}\left|(\int_x^{x+h}f(t)-g(t)dt)+(\int_x^{x+h}g(t)-k(t)dt)\right|\\ &\leq^\star\sup_{x,h}\left(\left|\int_x^{x+h}f(t)-g(t)dt\right|+\left|\int_x^{x+h}g(t)-k(t)dt\right|\right)\\ &\leq^{\star\star}\left(\sup_{x,h}|\int_x^{x+h}f(t)-g(t)|\right)+\left(\sup_{x,h}|\int_x^{x+h}g(t)-k(t)|\right)\\ &=d(f,g)+d(g,h) \end{align*} (You should be careful with the inequalities $\star$ and $\star\star$: The first one uses the fact that if $A$ and $B$ are functions and $A\leq B$, then $\sup_yA(y)\leq\sup_y B(y)$, and the second one uses the fact that $\sup_y(A(y)+B(y))\leq\sup_y a(y)+\sup_y B(y)$.)

Symmetry is just as you noticed.

Now suppose that $f\neq g$, and let's show that $d(f,g)>0$. Since $f\neq g$, we have $f(x_0)\neq g(x_0)$ for some $x_0$. WLOG, let's say that $f(x_0)>g(x_0)$. Let $\delta=f(x_0)-g(x_0)$. Then there exists an interval $(x_0-h,x_0+h)$ ($h>0$) around $x_0$ such that $f(t)>g(t)+\delta/2$ for every $t$ in this interval. We can moreover assume that $h<1$. Then, since the distance is defined by a supremum, we have $$d(f,g)\geq|\int_{x_0}^{x_0+h}f(t)-g(t)dt|=\int_{x_0}^{x_0+h}f(t)-g(t)dt\geq \int_{x_0}^{x_0+h}\delta/2dt=h\delta/2>0.$$

Answer 2

The $b$ in $C_b$ is for "bounded".

The fact that $d(f,g)=0$ implies $f=g$ is not that "obvious", in my opinion (not saying that it's hard). If $d(f,g)=0$, it follows that, for all $x$, $$ \int_x^{x+h} (f-g)=0. $$ The limit as $h\to0$ of this expression is $f(x)-g(x)=0$ (here is where we use continuity).

For the triangle inequality, $$ \left|\int_x^{x+h} f-g\right|=\left|\int_x^{x+h} f-k+k-g\right| =\left|\int_x^{x+h} f-k+\int_x^{x+h}k-g\right| \\ \leq \left|\int_x^{x+h} f-k\right|+\left|\int_x^{x+h}k-g\right|\\ \leq d(f,k)+d(k,g), $$ and now since the rightmost expression is an upper bound for the leftmost one, $$ d(f,g)\leq d(f,k)+d(k,g). $$