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am reading a long proof and I don’t get if this is true:

Let |||| be the Euclidean metric. If I show a function f is continuous. Will =||()−()||

be continuous? I am guessing yes because ℎ=()−()

is continuous.

So if g()is continuous

is $a(x)=||g(x)|| $ also continuous ? This should be very ba

Edit: is the metric a contour function? Assume $d$ is a metric on a set, so will the metric be continuous

Dhdh
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1 Answers1

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It won't let me comment so I can't really have a conversation to work out the proper answer. It depends on what kind of norm you are using. If you are referring to an operator norm between two normed F-vector spaces, then typically one refers to bounded linear operators which are continuous, this can include $n \times n$ matrices, derivatives (within stricter constraints), integral transformations and much more.

If you are referring to a norm on function spaces rather than operator spaces, this other question may help

Why are norms continuous?

So in other words, it depends on the context, but if you're looking through material in a class there are conventional theorems you can rely on. A metric immediately follows from a norm, that is, for any normed F vector space, $|| x-y ||$ generates a metric topology and so you can use $||x-y||$ as the metric.