# $f'+\lambda f$ acts as $f'$ without integral

## Problem

Suppose that $f(x)$ is a differentiable function and $f'(x)$ is the derivative of $f(x)$. Without aids of integral, can we prove that $f'(x)+\lambda f(x)$ has intermediate property?

## Intermediate property

A (real) function $f(x)$ having intermediate property means that if $a,b\in f(\Bbb R)$ and $a<c<b$, then $c\in f(\Bbb R)$, where $f(\Bbb R)=\{\;f(x):x\in\Bbb R\;\}$

## With aids of integral

Let $g(x)=f'(x)+\lambda f(x)$. For $f(x)$ is continuous, we have $f(x)$ is Riemann-integrable. Let $G(x)=f(x)+\int_0^xf(t)dt$, we have $G'(x)=g(x)$; therefore, we can apply Darboux's theorem to $G(x)$, and we've done.

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@AndréNicolas That's not true, every function on the reals is the sum of two with IVP (c.f. the Wiki page for Darboux's Theorem). – Rofler Aug 16 '12 at 8:34
@Rofler I don't know about whether the sum of an arbitrary Darboux function and a continuous function is a Darboux function – Frank Science Aug 16 '12 at 8:35
Thanks, I even knew that (not from old real knowledge, but from recent post on MSE). – André Nicolas Aug 16 '12 at 8:38
@FrankScience ams.org/journals/proc/1976-060-01/S0002-9939-1976-0417350-1/… seems to prove that an arbitrary sum of a Darboux and continuous function can fail IVP pretty bad. – Rofler Aug 16 '12 at 8:44
@Rofler Thanks. Now it's not high time for me to read that. I'm just working on a problem in basic calculus, which results in the stated problem. I think using integral is not appropriate because in that chapter, integral is not introduced! Therefore it's better to find a more elementary proof, or skip it, and use another way to solve the original problem. – Frank Science Aug 16 '12 at 8:46

Let $a\lt b$, $g(t)=f'(t)+\lambda f(t)$, and $z$ in the interval between $g(a)$ and $g(b)$. One wants to show that there exists $x$ in $[a,b]$ such that $g(x)=z$. If $\lambda=0$, this is the usual case. Assume that $\lambda\ne0$.

One can assume without loss of generality that $g(a)\lt z\lt g(b)$. Consider the function $h$ defined by $h(t)=\mathrm e^{\lambda t}(f(t)-z/\lambda)$. Then $h'(t)= \mathrm e^{\lambda t}(g(t)-z)$, hence $h'(a)\lt0\lt h'(b)$. The usual intermediate value theorem for derivatives applied to $h'$ ensures that $h'(x)=0$ for some $x$ in $(a,b)$. Thus, $g(x)=z$.

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Pretty good! Thanks! – Frank Science Aug 16 '12 at 23:53
Can we deal with $\sum_{k=0}^m a_kf^{(k)}(x)$ using the same trick? – Frank Science Aug 16 '12 at 23:59