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I am doing the following exercise from my functional analysis course.

Prove that $$d(\{x_n\},\{y_n\}):=\sum_{n\geq 1} 2^{-n} \dfrac{|x_n-y_n|}{1+|x_n-y_n|}$$ is a metric on $\mathbf{R}^{\mathbf{N}}$.

Is this some kind of known norm? I tried looking on the internet and I found similar expressions with the name F-norm. Any information on this norm is helpful.

Currently, I am stuck proving the triangle inequality for this metric. This must follow from the triangle inequality for $|\cdot|$.

I think it is equivalent to proving that

$$\frac{|x_n-z_n|}{1+|x_n-z_n|}\leq \frac{|x_n-y_n|}{1+|x_n-y_n|}+\frac{|y_n-z_n|}{1+|y_n-z_n|},$$

which is in turn equivalent to

$$1+\frac{1}{1+|x_n-z_n|}\geq \frac{1}{1+|x_n-y_n|}+\frac{1}{1+|y_n-z_n|}.$$

Reformulating, I am trying to prove that if $a,b,c\geq 0$ and $a\leq b+c$, then $1+\frac{1}{1+a}\geq \frac{1}{1+b}+\frac{1}{1+c}$. I am not really good at inequalities, can someone give a hint?

rae306
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  • Also:https://math.stackexchange.com/q/297818/42969, https://math.stackexchange.com/q/2466122/42969, https://math.stackexchange.com/q/2948450/42969. – Martin R Mar 15 '20 at 11:31
  • Also: https://math.stackexchange.com/q/360989/42969, https://math.stackexchange.com/q/686792/42969, https://math.stackexchange.com/q/685504/42969 – all found with Approach0 – Martin R Mar 15 '20 at 11:35

1 Answers1

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Since $a\leq b+c$, it's enough to show

$$1+\frac{1}{1+b+c}\geq \frac{1}{1+b}+\frac{1}{1+c}$$

or:

$$1+b+c+1\geq \frac{1+b+c}{1+b}+\frac{1+b+c}{1+c}=1+\frac{c}{1+b}+1+\frac{b}{1+c}$$

or

$$b+c\geq \frac{c}{1+b}+\frac{b}{1+c}$$

Can you see why this holds?

$$\frac{c}{1+b}\leq \frac{c}{1+0}=c$$

LHF
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