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Prove that for every $x$, we have $\Delta[f(x)+g(x)]=\Delta f(x)+ \Delta g(x)$. Thanks in advance.

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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Julian Kuelshammer Nov 11 '12 at 13:43
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What is $\Delta(f(x))$ for you? – Sigur Nov 11 '12 at 13:43
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Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Julian Kuelshammer Nov 11 '12 at 13:44
The laplacian $\Delta$ is a linear differential operator. Don't you understand my answer? Well, try to explain what you are asking, please. – Siminore Nov 11 '12 at 14:26

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u can solve this by using forward differences w.k.t D=delta (for my convinent) Df(x)=f(x+h)-f(x) LHS let a(x)=f(x)+g(x) Da(x)=a(x+h)-a(x) now substitute for a(x) D[f(x)+g(x)]=f(x+h)+g(x+h)-(f(x)+g(x))...................1 now take rhs of equ Df(x)=f(x+h)-f(x)......i Dg(x)=g(x+h)-g(x)......ii add i &ii lhs=rhs

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Since Notyathing said "derivative question", it's likely $\Delta$ is the Laplacian (or perhaps some other differential operator). It does not represent general differences as in your answer. – Mark S. Nov 11 '12 at 15:50

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