10
$\begingroup$

If $f$ is a linear function (defined on $\mathbb{R}$), then for each $x$, $f(x) – xf’(x) = f(0)$. Is the converse true? That is, is it true that if $f$ is a differentiable function defined on $\mathbb{R}$ such that for each $x$, $f(x) – xf’(x) = f(0)$, then $f$ is linear?

  • 0
    @user1551: Well, yes, and no. The term "linear function" has two distinct, but related meanings, as explained in the Wikipedia article on "linear function". I'm using the meaning of "first degree polynomial", which is called a "linear function" because of its graph. But I appreciate your pointing out the ambiguity of the term, which I was unaware of. Anyway, here's the link to the Wikipedia article: http://en.wikipedia.org/wiki/Linear_function2011-08-23

5 Answers 5

7

Let $g(x)=f(x)-f(0)$, then g(x)=xg'(x) for all $x$. If $g(x)\neq0$, this means that $ \frac{\mathrm{d}}{\mathrm{d}x}\log(g(x))=\frac{1}{x}\tag{1} $ Solving $(1)$ equation yields $ \log(g(x))=\log(k)+\log(x)\tag{2} $ for some constant $k$ for either $x\in\mathbb{R}^+$ or $x\in\mathbb{R}^-$ (when $x<0$ or if $g(x)<0$ use $\log(-x)=\log(x)+\pi i$). Equation $(2)$ shows that if $g(x_0)\neq 0$ for some $x_0\neq0$, $g(x)\neq0$ for any $x$ with the same sign as $x_0$. Exponentiating equation $(2)$ and replacing $f$ yields $ f(x)-f(0)=kx\tag{3} $ Thus, $f$ is linear on $\mathbb{R}^+$ and on $\mathbb{R}^-$. However, the slope of $f$ can be different on $\mathbb{R}^+$ than it is on $\mathbb{R}^-$.

Now that I think about it, since $f$ is differentiable at $0$, we have \lim_{x\to0+}\frac{f(x)-f(0)}{x}=f'(0)=\lim_{x\to0-}\frac{f(x)-f(0)}{x}\tag{4} Thus, the slopes on $\mathbb{R}^+$ and $\mathbb{R}^-$ have to be the same.

As Andres Caicedo pointed out, and I have tried to account for, logs of negative numbers are problematic, as is $g(x)=0$. $g(x)=0$ is a solution, but then $f(x)=f(0)$ is still linear.

  • 0
    Yes, on second thought I realized that some cleanup was needed, although the basic idea was good. When I came back in here to point this out, I see that the cleanup has been addressed:)2011-08-23
4

Suppose $f$ is differentiable and satisfies your equation, f(x)-xf'(x)=f(0).

[Edit: This is a revised answer. The previous one was terribly flawed.]

Note that \displaystyle \frac{f(x)-f(0)}x=f'(x) for any $x\ne0$. This shows that f' is differentiable, except perhaps at 0. Differentiating this equation, we obtain f''(x)=\displaystyle \frac{xf'(x)-(f(x)-f(0))}{x^2}=0.

If f'' is continuous (at 0) it follows that f' is constant and therefore $f$ is linear.

In any case, the above shows that $f$ is linear in $(-\infty,0)$ and in $(0,\infty)$, and we are given that it is differentiable at 0.

But then $f$ is linear: Say that $f(x)=ax+b$ for $x<0$ and $f(x)=cx+d$ for $x>0$. Then f'(x)=a for $x<0$ and f'(x)=b for $x>0$ and f'(0) exists. Since derivatives satisfy the intermediate value theorem (see for example Rudin "Principles of Mathematical Analysis" Theorem 5.12), it follows that we must have a=f'(0)=b. Since differentiability implies continuity, it follows that $\displaystyle b=\lim_{x\to 0^-}f(x)=f(0)=\lim_{x\to0^+}f(x)=d$.

  • 0
    Hehe. Fixed now. @robjohn: Thanks for noticing the problem.2011-08-22
4

If $f\in C^2$ (it is twice differentiable, and f'' is continuous), then the answer is yes; I don't know if it's necessarily true without this hypothesis.

If f(x)-xf'(x)=f(0) for all $x$, then f(x)=xf'(x)+f(0), so that f'(x)=f'(x)+xf''(x) 0=xf''(x) This shows that f''(x)=0 for all $x\neq0$, but because f'' is continuous this forces f''(x)=0 everywhere. Thus f' must be constant, and thus $f$ must be linear.

  • 2
    No, $C^1$ is not needed. For $x\not=0$, the defining equation gives $f'(x)=(f(x)-f(0))/x$. Hence $f''(x)$ exists. Following the argument of Chonoles, we have $f''(x)=0$, so $f$ is affine on both $(0,\infty)$ and $(-\infty,0)$. Now the existence of $f'(0)$ implies that the two branches of $f$ must be collinear.2011-08-23
2

The following argument uses not much machinery.

Suppose that $f(0)=b$ and f'(0)=m. Let $g(x)=f(x)-(mx+b)$. Then $g(0)=0$ and g'(0)=0. Note that g(x)-xg'(x)=[f(x)-(mx+b)] -x(f'(x)-m)=0=g(0). So xg'(x)-g(x)=0 for all $x$.

For $x \ne 0$, let $h(x)=g(x)/x$. Then for any $x \ne 0$, we have h'(x)=\frac{xg'(x)-g(x)}{x^2}=0.

It follows that $h(x)$ is a constant $p$ on $(0, \infty)$, and a constant $n$ on $(-\infty,0)$.
Thus $g(x)=px$ on $(0,\infty)$ and $g(x)=nx$ on $ (-\infty,0)$.

But g'(x)=0. So $\lim_{x\to 0+} \frac{px-0}{x-0}=0.$ It follows that $p=0$. In the same way we can show that $n=0$. So $g(x)$ is identically $0$, and therefore $f(x)=mx+b$.

1

Suppose that $y=f(x)$ is a differentiable function that satisfies f(x)-xf'(x)=f(0) for all $x\in\mathbb{R}$. Then $y=f(x)$ satisfies the linear differential equation y'=\frac{y-k}{x}, where $k=f(0)$. The homogeneous system y'=\frac{y}{x} has solutions $y_\lambda=\lambda x$, for any $\lambda\in\mathbb{R}$. Also, the equation y'=(y-k)/x has a solution $y_p=x +k$. Thus, all the solutions of our differential equation are of the form $y=y_\lambda+y_p$ for some $\lambda\in\mathbb{R}$, i.e., $y=\mu x + k,$ where $\mu=\lambda+1\in\mathbb{R}$. Hence, all functions $f(x)$ with said property are linear.

Note: Technically, the existence and uniqueness theorem for linear differential equations tells us that the differential equation y'=f(x,y)=(y+k)/x has a unique solution $f_1(x)$ in $(0,\infty)$ and also a unique solution $f_2(x)$ in $(-\infty,0)$, because $f(x,y)$ has a discontinuity at $x=0$. The argument above says that $f_1(x)=\mu_1 x+k$ and $f_2(x)=\mu_2 x + k$ are linear. Since we are assuming that $f$ is differentiable, we must have $\mu_1=\mu_2=\mu$ and $f(x)=\mu x + k$.