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Let $X$ be a vector space and $f_k$ be semi-norms on $X$ such that for every $x\neq 0$ there is a $k_0\in \Bbb N$ satisfying $f_{k_0}(x)>0$. Then $$d(x,y)=\sum_{k\in \Bbb N} 2^{-k} \frac{f_k(x-y)}{1+f_k(x-y)}$$ defines a metric on $X$.

I proved everything except the triangle inequality and it's thoroughly confusing me. We must somehow use triangle inequality for the semi-norms but I can't figure out how to make it work. Could you kindly help me out?

Daniel Fischer
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Casey Harrison
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1 Answers1

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Hint: for each $k$ (and arbitrary $z\in X$) we have $$f_k(x-y) \leq f_k(x-z) + f_k(z-y).$$ Combining this with the observation that the function $x\mapsto \frac{x}{1+x}$ is monotonically increasing, we get (for each seminorm $f_k$):

$$\frac{f_k(x-y)}{1+f_k(x-y)} \leq \frac{f_k(x-z)+f_k(z-y)}{1+f_k(x-z)+f_k(z-y)}.$$

Now, since $f_k$ is nonnegative, it is not hard to conclude that the right-hand side satisfies

$$\frac{f_k(x-z)+f_k(z-y)}{1+f_k(x-z)+f_k(z-y)} = \frac{f_k(x-z)}{1+f_k(x-z)+f_k(z-y)} + \frac{f_k(z-y)}{1+f_k(x-z)+f_k(z-y)} \\ \leq \frac{f_k(x-z)}{1+f_k(x-z)} + \frac{f_k(z-y)}{1+f_k(z-y)}.$$

Thus, we get $$\frac{f_k(x-y)}{1+f_k(x-y)} \leq \frac{f_k(x-z)}{1+f_k(x-z)} + \frac{f_k(z-y)}{1+f_k(z-y)},$$ which should give you the desired triangle inequality.

bakula
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