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My Attempt: If Variable is same then $$(f+g)x=f(x)+g(x)$$ it's always true. Thanks in Advance..

G. Chiusole
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Amit Tikhe
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  • Yes you're right – sai-kartik Mar 17 '20 at 06:22
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    Yes, this is the definition of function addition. We define a new function $f + g$ by the above rule, i.e. the value of $f + g$ at $x$ is the sum of the values of $f$ and $g$ at $x$. – user759562 Mar 17 '20 at 06:24
  • I would like to add that this definition may not be consistent, depending on the functions. Consider two functions $f,g : \mathbb{N} \to {0,1}$. $f(x)+g(x)$ may have no sense. – nicomezi Mar 17 '20 at 06:27
  • Yes, again depending on the context however, but I would adapt the notation accordingly. Writing $f+g$ strongly suggest something close to the relationship given by the OP. @sai-kartik – nicomezi Mar 17 '20 at 06:29
  • Makes sense @nicomenzi. Would rather stick to same variables to avoid confusion – sai-kartik Mar 17 '20 at 06:31
  • I am asking in general... Is there is any contradict ex. for this..? – Amit Tikhe Mar 17 '20 at 06:55
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    There isn't any contradiction to this. This is how function addition is defined. If you attempt to add two functions which don't have the same variable though, I'm not so sure this definition will still hold . – sai-kartik Mar 17 '20 at 07:04

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The equation $(f+g)(x)=f(x)+g(x)$ is how addition of functions is generally defined (where “generally” here is more of a hedge word; I'm not aware of any other definition in use, however I don't want to completely exclude the possibility). It is called “pointwise addition”.

More exactly, if $A$ is an additive structure (that is, a set on which addition is defined), then given two functions $f:X\to A$ and $g:Y\to A$, then their sum is the function $(f+g):(X\cap Y)\to A$ with $(f+g)(x) = f(x)+g(x)$.

celtschk
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Whenever it makes sense to talk about $f + g$, it's true. Specifically, it makes sense when:

  • The domains of $f$ and $g$ to be the same set $A$,
  • The codomains of $f$ and $g$ to be the same set $B$,
  • The operation $+$ makes sense on $B$.

By the third point, I mean that $B$ is a set like $\Bbb{R}$, or $\Bbb{C}$, or $M_{2 \times 2}(\Bbb{R})$, or just some other set with some sensible addition defined on it. All we require is that the set be closed under the $+$ operation, i.e. $b_1 + b_2 \in B$ for all $b_1, b_2 \in B$.

For example, if $f, g : \Bbb{C} \to \Bbb{C}$ such that $f(z) = \overline{z}$ and $g(z) = |z|$, then $f + g$ makes sense, and we define $f + g : \Bbb{C} \to \Bbb{C}$ by $$(f + g)(z) = f(z) + g(z) = \overline{z} + |z|.$$

If, on the other hand, we had $f : \Bbb{R} \to \Bbb{R}$ and $g : M_{2 \times 2}(\Bbb{R}) \to \Bbb{R}$ defined by $f(x) = 2x$ and $g(M) = \operatorname{Tr}(M) - 6$, then $f + g$ would not make sense, since $f$ and $g$ have different domains. We could not say $(f + g)(x) = f(x) + g(x)$, because it's not clear whether $x$ is a real number, or a two-by-two real matrix.

So yeah, whenever $f + g$ makes sense, what you wrote is true.

user759562
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