Is $(X,d)$ a metric space?

Question

$X = C_{[a,b]}^1$, $d(f,g) = \displaystyle\max_{x \in [a,b]} |f'(x) - g'(x)| + |f(a) - g(a)|$. Is $(X,d)$ a metric space?

My attempt: These conditions $d(f,g) \geq 0$, $d(f,g) = 0$, $d(f,g) = d(g,f)$ are trivial We condider the last condition for a metric, that is $d(f,h) \leq d(f,g) + d(g,h), \forall f,g,h \in X$.

$d(f,h) = \displaystyle\max_{x \in [a,b]} |f'(x) - h'(x)| + |f(a) - h(a)|$.

We have $|f'(x) - h'(x)| = |f'(x) - g'(x) + g'(x) - h(x)| \leq |f'(x) -g'(x)|+|g'(x) - h'(x)|$,

$|f'(x) - h'(x)| \leq |f'(x) -g'(x)|+|g'(x) - h'(x)| \Rightarrow \displaystyle\max_{x \in [a,b]} |f'(x) - h'(x)| \leq |f'(x) -g'(x)|+|g'(x) - h'(x)| \Rightarrow \displaystyle\max_{x \in [a,b]} |f'(x) - h'(x)| \leq \displaystyle\max_{x \in [a,b]} |f'(x) - g'(x)|+|g'(x) - h'(x)|$

Similarly we conclude that $|f(a) - h(a)| \leq |f(a) - g(a)|+|g(a) - h(a)| \Rightarrow \displaystyle\max_{x \in [a,b]} |f(a) - h(a)| \leq \displaystyle\max_{x \in [a,b]} |f(a) - g(a)|+|g(a) - h(a)|$

Hence, $\displaystyle\max_{x \in [a,b]} |f'(x) - h'(x)| + |f(a) - h(a)| \leq \displaystyle\max_{x \in [a,b]} |f'(x) - g'(x)| + |f(a) - g(a)| + \displaystyle\max_{x \in [a,b]} |g'(x) - h'(x)| + |g(a) - h(a)|$

Thus, $d$ is a metric on $X$. Is my proof true? Thank all!

Answer 1

Is it true that if two functions have the same (continuous) first derivative and take on the same value at a point then they are the same function? This assertion I would not consider trivial. The rest of the argument looks fine, though.

Answer 2

Instead of those explicit calculations, I would find it easier to to keep the various pieces of the construction separate by proving in general:

Lemma 1. If $(X,d_X)$ and $(Y,d_Y)$ are metric spaces then $$ d((x_1,y_2),(x_2,y_2)) = \max(d_X(x_1,x_2), d_Y(y_1,y_2)) $$ defines a metric on $X\times Y$.

Lemma 2. If $(Z,d_Z)$ is a metric space and $\varphi: W\to Z$ is any injective map, then $$ d_W(w_1,w_2) = d_Z(\varphi(w_1),\varphi(w_2)) $$ defines a metric on $W$.

Then apply these with $X=\mathcal C_{[a,b]}^0$ with the sup-norm, $Y=\mathbb R$, $Z=X\times Y$, and $\varphi(f) = (f',f(a))$.

(You would need to prove that $\varphi$ is injective, but that's a simple analysis exercise).