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#1 h Nov 19, 2013, 9:50 PM • 2 Y

Y by Adventure10, Mango247

Let $f:R->R$ a function with the following properties
1) $f(x+y)<= f(x)+f(y)$ for any $x,y\in R$
2) $f(x)<=e^x-1$ ,for any $x\in R$
show that f has primitives on R

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#2 h Yesterday at 11:32 AM

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Let us express Properties (1) and (2) according to:

(1) $f(x) + f(y) = f(x+y)+g(x+y)$ (for $g:\mathbb{R} \rightarrow [0,\infty)$ );

(2) $e^{x}-1 = f(x) + h(x)$ (for $h:\mathbb{R} \rightarrow [0, \infty)$ )

where $f,g,h$ are each differentiable over the reals. We can differentiate Property (1) with respect to $x$ and $y$ to obtain:

$f'(x)=f'(x+y)+g'(x+y)=f'(y) \Rightarrow f(x) = Ax+B$ ;

for arbitrary $A,B \in \mathbb{R}$ . We next obtain:

$Ax+B+Ay+B = A(x+y)+B + g(x+y) \Rightarrow g(x+y) = B$ ;

of which we require $B \ge 0$ . Looking at Property (2) we obtain:

$e^{x}-1 = Ax+B+h(x) \Rightarrow e^{x}-1-Ax-B = h(x) \ge 0 \Rightarrow e^{x}-1 \ge Ax + B$ (i).

If $B>0$ , then we violate (i) (regardless of the value for $A$ ). This gives us the necessary condition $B=0$ . The tangent line for $y=e^{x}-1$ at $(x,y)=(0,0)$ is $y=x \Rightarrow A=1$ (which satisfies (i) for all $x \in \mathbb{R}$ ). Hence, $\textcolor{red}{f(x)=x}$ is the desired function which has a primitive over all $\mathbb{R}$ .

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#3 h Yesterday at 12:26 PM • 1 Y

Y by Filipjack

@ #2: It is not given that $f$ is differentiable (indeed, then there would be nothing to show since every continuous, and in particular every differentiable, function admits primitives). Also, it is not clear why $f(x)+f(y)-f(x+y)$ should depend only on $x+y$ , as you claimed.
Solution

This post has been edited 1 time. Last edited by Gryphos, Yesterday at 12:29 PM

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