Proving that if $f'=cf$ on some interval then $f=ke^{cx}$ on the interval.

Question

Suppose that on some interval the function $f$ satisfies $f'=cf$ for some number $c$.

(a) Assuming that $f$ is never $0$, prove that $|f(x)|=le^{cx}$ for some number $l\gt 0$. It follows that $f(x)=ke^{cx}$ for some $k$.

(b) Show that this result holds without the added assumption that $f$ is never $0$. Hint: Show that $f$ can't be $0$ at the endpoint of an open interval on which it is nowhere $0$.

(c) Suppose that $f'=fg'$ for some $g$. Show that $f(x)=ke^{g(x)}$ for some number $k$.

My work:

(a) is easy. For (b), first assume that $f$ does not vanish on the entire interval. Then $f(x_0)\neq 0$ for some point. Hence, by continuity there is some open interval around $x_0$ for which the function is nonzero, and hence by (a), $f(x)=ke^{cx}$. Now as the hint suggests, I want to prove that $f$ can't be $0$ at the endpoint of an open interval on which it is nowhere $0$. Again by continuity, the endpoint of this interval would be $w:=\sup\{x\gt x_0 : f(x)\neq 0\}$ and $z:=\inf\{x\lt x_0 : f(x)\neq 0\}$.

Here is where I'm struggling. I think that this $f$ can't be $0$ on both of these endpoints because $f$ is either always positive or always negative on this interval and since it is also continuous on the endpoints, it can't suddenly vanish at the point. However, I cannot precisely show this. How can I rigorously prove this idea?

Finally, for (c), solving differential equations I get $|f|=de^g$ for some number $d$. Again, to show that $f=ke^g$ on the interval, I need to show that $f$ is never $0$. However, unlike the situation in (b), I think since $g$ is also a function, it can rapidly go to $\infty$ as $x$ goes to the endpoint of the same proposed interval, and then $f$ can actually go to zero. How can I show that like (b), $f$ is never zero on the entire interval?

I would greatly appreciate any help for the above two questions.

Answer 1

Update: I am not sure if OP is the downvoter. But I guess OP wants to complete his ongoing proof of this statement.

If $f$ is continuous and differentiable on $[a, b]$ and $f(x) \neq 0$ for all $x \in (a, b)$ then $f(a) \neq 0 \neq f(b)$.

Clearly this statement is wrong and counter examples abound. Check $f(x) = x(1 - x)$ on $[0, 1]$.

However the result holds for the specific function in question. Switching to the notations used by OP we have $f(x) = ke^{cx}$ on some interval $I$ and $z, w$ with $z < w$ are the end-points of this interval. Also $f$ is continuous on $[z, w]$ and $f(x) \neq 0$ for all $x \in (z, w)$. This means that $k \neq 0$. Since $f(x) = ke^{cx}$ for all $x \in (z, w)$ it follows by continuity that $$f(z) = \lim_{x \to z^{+}}f(x) = \lim_{x \to z^{+}}ke^{cx} = ke^{cz} \neq 0$$ Similarly we can show that $f(w) \neq 0$. Nothing more than continuity of $f$ is required here.

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Note that the question in part a) has a simple solution only when we know the properties of exponential and logarithmic functions. Without knowing anything about these functions it is bit hard to establish a) and this has been done here.

Although it is not necessary to know whether $f$ vanishes or not (and I don't know why you want to deal with vanishing/non-vanishing of $f$) you can do it easily as follows.

For Part b) we have $f'(x) = cf(x)$ for all $x \in I$. If $c = 0$ then $f$ is a constant and hence if it vanishes at one point it vanishes for all $x \in I$. So let's assume that $c \neq 0$. Let $a \in I$ be such that $f(a) \neq 0$. Without any loss of generality we can assume that $a$ is an interior point of $I$ (because if $a$ were an end point then by continuity we could find an interior point nearby where $f$ is non-zero). We will show that $f(x) \neq 0$ for all $x \in I$. Consider the function $g(x) = f(a + x)f(a - x)$. Clearly since $a$ is interior point of $I$ the function $g$ is defined in some interval of type $[0, p)$. We can see that \begin{align} g'(x) &= f'(a + x)f(a - x) - f(a + x)f'(a - x)\notag\\ &= cf(a + x)f(a - x) - cf(a + x)f(a - x)\notag\\ &= 0\notag \end{align} so that $g$ is a constant and thus $f(a + x)f(a - x) = g(x) = g(0) = \{f(a)\}^{2} > 0$. It follows that both $f(a + x), f(a - x)$ are non-zero for all $x \in [0, p)$. If $(a + p)$ (or $(a - p)$) is an interior of $I$ then we can continue the same argument by replacing $a$ with $(a + p)$ (or $(a - p)$) and thereby extend the region where $f$ does not vanish. Using the same argument repeatedly we can show that $f$ does not vanish at any point in $I$.

For part c) (which is similar to part a))we can consider the function $h(x) = f(x) / \exp(g(x))$. We have \begin{align} h'(x) &= \frac{\exp(g(x))f'(x) - f(x)g'(x)\exp(g(x))}{\exp(2g(x))}\notag\\ &= \frac{f(x)g'(x)\exp(g(x)) - f(x)g'(x)\exp(g(x))}{\exp(2g(x))}\notag\\ & = 0\notag \end{align} and therefore $h(x) = k$ where $k$ is some constant number. It thus follows that $f(x) = k\exp(g(x))$. Note that this proof would be difficult (like proof of part a)) if we don't know anything about exponential and logarithmic functions.

Note: We don't need to know about vanishing / non-vanishing of $f$ (provided we know the properties of exponential and logarithmic functions). Thus for part a) we use $h(x) = f(x)/\exp(cx)$ and get $h'(x) = 0$ so that $h(x) = k$ and $f(x) = k\exp(cx)$.

Answer 2

Considering $f'/f$ is fine for conjecturing the result, but dividing by $f$ makes too much complication. So, after guessing $f(x)=k\cdot e^{g(x)}$, define $$ k(x) = f(x) \cdot e^{-g(x)} $$ instead. (That function is conjectured to be constant.) Then take its derivative: $$ k'(x) = f'(x) \cdot e^{-g(x)} - f(x) \cdot e^{-g(x)}g'(x) = (f'(x) - f(x)g'(x)) \cdot e^{-g(x)} = 0. $$ Therefore, $k(x)$ is constant.

Answer 3

For part (b) I'll look at a special case that I think captures all that is needed. Let $I=(0,b).$ Suppose $f$ is differentiable and nonzero on $I,$ with $f'(x) = cf(x), x \in I,$ where $c$ is a nonzero constant. Then $\lim_{x\to 0^+}f(x) \ne 0.$

Proof: The statements below pertain to all of $I.$ WLOG, $f>0.$ We then have

$$(\ln f)'(x) = f'(x)/f(x) = c.$$

Thus the derivative of $\ln f(x)- cx$ is $0,$ which implies $\ln f(x)- cx = d$ for some constant $d.$ This gives $f(x) = e^{cx+d},$ and the conclusion follows.