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Given $f_1,f_2: X \to X$ where $X$ is a metric space and the mappings are continuous and bounded (and therefore Lipschitz). Does it necessarily follow that $$(f_1+f_2)(x) = f_1(x)+f_2(x)$$ for some $x \in X$?

I've seen the property that $(x+y)(t) = x(t) + y(t)$ a couple of times in my life in calculus or in linear algebra sometimes. Are there times when it doesn't hold?

Does the above question hold if $X$ is a normed space?

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What's $f_1+f_2$? Is it $f_1 \circ f_2$? – Kaster Feb 14 '13 at 7:02
Adding functions, not taking compositions – Hawk Feb 14 '13 at 7:05
How do you define that sum? – Kaster Feb 14 '13 at 7:07
Sorry I am kinda lost here. Could you show me an example of your question? – Hawk Feb 14 '13 at 7:08
So am I. Can you tell me, please, how exactly you construction $f_1+f_2$ maps some number $x \in X$ to some other number $y \in X$? Functions are defined by their mapping. – Kaster Feb 14 '13 at 7:10
up vote 1 down vote accepted

What is the definition of $f+g$?

Given two maps $f,g:X\to A$, where $A,X$ are sets such that there is a 'sum' $+_A$ in $A$, $f+g$ is defined by $(f+g)(x)=f(x)+_Ag(x)$, for all $x\in X$.

Note that I used two different symbols to denote a sum, that's because they are two different sums. The sum $+$ in $f+g$ is a binary operation defined in the set of all maps from $X$ to $A$, $A^X$, while the sum $+_A$ is a binary operation defined in $A$.

Almost always mathematicians will write $(f+g)(x)=f(x)+g(x)$, but that's essetially incorrect because the same symbol $+$ is used to denote two different things. What they mean is what I wrote above.

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Do you think question might serve as an answer? :) – Kaster Feb 14 '13 at 7:04
@Kaster Yes. ${}{}{}$ – Git Gud Feb 14 '13 at 7:05
So, in this case question is rhetorical, because identity is valid by definition, since $+_A \equiv +$ in $X$. – Kaster Feb 14 '13 at 7:23
@Kaster It's not rethorical. I believe the OP doesn't understand the definition of $f+g$. – Git Gud Feb 14 '13 at 7:25
What if we assume it is the same binary operation? – Hawk Feb 14 '13 at 7:30

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