# approximately linear functions — more

Suppose $f,g$ are continuous functions from $\mathbb R$ to $\mathbb R$, with the property that $$f(x)+f(y)=g(x+y)$$ for all $x,y$. Taking $x=y=z/2$ implies that $g(x)=2f(x/2)$ so that the above condition becomes $$f(x)+f(y)=2f((x+y)/2).$$ This is known as Jensen's functional equation, and it implies that $f$ is linear.

There's also a generalization of Jensen's equation (I've seen it in work of Rassias, but it could be earlier): if $|f(x)+f(y)-2f((x+y)/2)|\leq\epsilon$ (and assuming WLOG that $f(0)=0$), then there is a linear function $L$ such that $|f(x)-L(x)\|\leq \epsilon$.

What I am interested in a generalization of all this: Suppose there are independent random variables $X,Y$ such that $$E[(f(X)+f(Y)-g(X+Y))^2]\leq\epsilon.$$ Is it possible to say anything about $f$ being (appropriately) approximately linear?

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I've removed the soft-question tag, as this doesn't really fit the bill. –  j.c. Jul 8 '10 at 22:23
Is this true for some class of $X$ and $Y$ or just two particular random variables? If it's two particular random variables then the condition certain doesn't imply anything without further knowledge of the distributions: X and Y could just be particular constants, and the inequality would tell you very little about f globally. –  Noah Stein Jul 9 '10 at 16:50

Comment : I think you could give more precision in the formulation of your question because you already have most of the answer.

In particular:

1-Is the generalization of Jensen equation you give valid on Hilbert Space ? (seems to be true, see http://www.emis.de/journals/JIPAM/images/075_02_JIPAM/075_02_www.pdf)

2-What assumption do you make on the random variables? do they have the same distribution ? are their distribution "equivakent" (i.e mutually absolutly continuous)

Answer: If (the answer to my question 1 is yes and if) $X$ and $Y$ have the same distribution $P$, because $L_2(P)$ is a hilbert space then you are done. Here, approximate linearity will mean that there exists a linear form $L$ such that $E_P[(f(X)-L(X))^2]<\epsilon$.

If $X$ and $Y$ have different distribution (say $P_X$ and $P_Y$) I see an easy case: when $E_{P_Y}[( dP_X/dP_Y )^2] ) < c$ ($\chi^2$ divergence between distributions bounded). Indeed, in this case you can work in $L_2(P_Y)$ using cauchy swartz inequality. Otherwise the question needs to be clarifyed (i.e what do you mean by "approximatly linear"? : in what banach space do you choose to work ? ).

Hope this helps !

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