Is the following statement provable?

Question

The statement is: $f$ is a real fucntion on $\mathbb R$. Then if $f'(x)=f(x)$ and $f(0)=1$, then $f(x)\neq 0.$

Answer 1

Here's another solution — this one doesn't require knowing about the exponential function.

We're given that $f$ is differentiable on $\mathbb{R}$, $f'(x)=f(x)$ for all $x \in \mathbb{R}$, and $f(0)=1.$

Assume that there is some $a \in \mathbb{R}$ such that $f(a) = 0$. We must have $a \ne 0$ since $f(0)=1.$

Note that, since $f$ is differentiable, it must be continuous.

Claim 1: There exists $b \ne 0$ such that $f(b)=0$ and there are no zeros of $f$ between $0$ and $b$.

Proof of Claim 1: If $a \gt 0$, take $b$ to the the greatest lower bound of $\lbrace x | x \gt 0$ and $f(x)=0\rbrace$. If $a \lt 0$, take $b$ to the the least upper bound of $\lbrace x | x \lt 0$ and $f(x)=0\rbrace$. In either case, $f(b)=0$ by the continuity of $f$. Notice that $b$ can't equal $0$, since $f(0)=1$. There are no zeros of $f$ between $0$ and $b$ by the choice of $b$, proving Claim 1.

Since $f$ is continuous and $f(0)=1$, there exists $d$ with sign opposite that of $b$ such that there are no zeros of $f$ between $d$ and $0$; hence there are no zeros of $f$ between $d$ and $b$. Let $I$ be the open interval between $d$ and $b$.

Claim 2: For every $x \in I$, $f(x/2)^2=f(x)$.

Proof of Claim 2: Define $g(x)=\frac{f(x/2)^2}{f(x)}$. The function $g$ is defined on $I$ since $f$ has no zeros there. Differentiating $g$ and using the fact that $f'=f$, we find that $g'(x)=0$ for $x \in I$. It follows that $g$ is constant on $I$. But $0 \in I$ and $g(0)=1$, so $g$ must be identically $1$ on $I$, proving Claim 2.

Now we use continuity again. If $b \gt 0$, then $$0 = f(b) = \lim_{x \to b^{-}} f(x) = \lim_{x \to b^{-}} f(x/2)^2 = f(b/2)^2.$$ If $b \lt 0$, then $$0 = f(b) = \lim_{x \to b^{+}} f(x) = \lim_{x \to b^{+}} f(x/2)^2 = f(b/2)^2.$$ In either case, $f(b/2)=0,$ contradicting the fact that $f$ has no zeros between $0$ and $b.$

Answer 2

Well,it should be clear that what we actually have here is a differential equation with 3 initial conditions 1) For every $x\in\mathbb R$, f'(x) = f(x).

2)f(0)=1.

3) For every $x\in \mathbb R$, $f(x)\neq 0$

We need to test all 3 conditions on any candidate function. So let's integrate both sides:

$\int f'$ = $\int f $ $\rightarrow$ f(x) +c = $\int f $.

The 2 initial conditions are really what solves the problem. The only closed form real valued function that satisfies all 3 of these conditions is f(x) = $e^{x}$. Verify as follows: f'(x) = 1* $e^{x}$ = $e^{x}$. Also f(0) = $e^{0}=1$. Also, for every $x\in \mathbb R$, $e^{x}\neq 0$.

So the real valued function that satisfies all 3 conditions is f(x) = $e^{x}$ and that ends the proof.

Answer 3

Here's a good way of seeing this, knowing basic facts about the exponential function:

Assume that $f$ is differentiable on $\mathbb{R}$, $f'(x) = f(x)$ for all $x \in \mathbb{R}$, and $f(0)=1$.

Then $g(x) = \mathrm{e}^{-x} f(x)$ is differentiable, since it's the product of differentiable functions.

Applying the product rule, $g'(x) = \mathrm{e}^{-x} f'(x) - \mathrm{e}^{-x} f(x).$ But this equals $\mathrm{e}^{-x}(f'(x)-f(x))$, which is $0$ because of the assumption that $f'(x)=f(x)$.

Since $g'(x)$ is identically $0$, $g(x)$ must be a constant function. We have $g(0) = \mathrm{e}^{-0} f(0) = f(0) = 1$; since $g$ is a constant function, $g(x)=1$ for all $x$.

By the definition of $g$, we now know that $\mathrm{e}^{-x} f(x) = 1$ for all $x$. It follows that $f(x)$ can never be $0$, since something times $0$ can't equal $1$, completing the proof.

By the way, how do we know that $g'$ being identically $0$ implies that $g$ is a constant function? Apply the mean value theorem: Suppose there were two distinct numbers $a$ and $b$ such that $g(a) \neq g(b)$. Without loss of generality, $a<b$ (if not, switch $a$ and $b$). Then there exists $c$ such that $a < c < b$ and $g'(c) = \frac{g(b)-g(a)}{b-a}$, which is non-zero, contradicting the fact that $g'$ is identically $0$.

Answer 4

Obviously the statement is provable and moreover it is provable without any knowledge of $e^{x}$.

Consider the function $g(x) = f(x)f(-x)$ and then we can see that $$g'(x) = f'(x)f(x) - f(x)f'(-x) = f(x)f(-x) - f(x)f(-x) = 0$$ and hence $g$ is constant. Then $g(x) = g(0) = f(0)f(0) = 1$. Thus $f(x)f(-x) = 1$ for all $x$. It follows that $f(x)$ is non-zero for all $x$.

Using similar approach you should try to prove the following:

If $f'(x) = f(x)$ for all real $x$ then either $f(x) = 0$ for all values of $x$ or $f(x) \neq 0$ for all values of $x$.