# Prove that if $\phi'(x) = \phi(x)$ and $\phi(0)=0$, then $\phi(x)\equiv 0$. Use this to prove the identity $e^{a+b} = e^a e^b$.

I am given the following. hint Consider $f(x)=e^{-x} \phi(x)$.

I am unsure how to approach this problem.

Note that $f'(x)=0$ for all values of $x$ by construction. Hence $f(x)$ is constant. Given that $\phi(0)=0$ it must be true that $f(x)=0$ for all values of $x$ and hence that $\phi(x)=0$ for all values of $x$.

• Then take $\phi(x)=e^{a+x}-e^a e^x$ for the second part of the question. – JPi Feb 1 '14 at 21:22
• I see, forgot to use the hypothesis part of the question – grayQuant Feb 1 '14 at 21:23

If $$\phi'(t)=\phi(t),$$ then $$\mathrm{e}^{-t}\big(\phi'(t)-\phi(t)\big)=0.$$ But $$0=\mathrm{e}^{-t}\big(\phi'(t)-\phi(t)\big)=\left(\mathrm{e}^{-t}\phi(t)\right)'$$ and thus $\mathrm{e}^{-t}\phi(t)$ is a constant function, i.e., $$\mathrm{e}^{-t}\phi(t)=\mathrm{e}^{-0}\phi(0)=0.$$ Thus $\phi(t)=0$, for all $t$.

Then consider $\varphi(t)=\mathrm{e}^{a+t}-\mathrm{e}^a\mathrm{e}^t$. Clearly, $\varphi(0)=0$ and $\varphi'(t)=\varphi(t)$, and hence $\varphi(t)=0$.

If we carefully observe the answers given here, it will be found that the answers make use of three crucially important properties of $f(x) = e^{x}$:

1) $f'(x) = f(x)$

2) $f(0) = 1$

3) $f(x) \neq 0$ for all $x$

and then using these properties it is proven that $f(x + y) = f(x)f(y)$.

Justification of existence of a function $f(x)$ with the above properties is effectively equivalent to development of theory of exponential and logarithmic functions.

It is of interest to prove the result without using the properties of function $f(x) = e^{x}$. We proceed to provide such a solution as follows.

Case 1: Let us suppose that there is a constant $a$ such that $\phi(a) = k > 0$ and consider the function $\psi(x) = \phi(x + a)$. Clearly $\psi'(x) = \psi(x)$ and $\psi(0) = \phi(a) = k > 0$. We now claim that $\psi(x) > 0$ for all $x$. This will lead to a contradiction because $\psi(-a) = \phi(0) = 0$.

To establish our claim let's first consider positive values of $x$. If $\psi(x)$ vanishes for some positive $x$ then by continuity of $\psi(x)$ there is a least such positive value of $x$ for which $\psi(x) = 0$. Let this value be called $b$. Thus we have $\psi(b) = 0$ and $\psi(x) \neq 0$ for $x \in [0, b)$. Then by mean value theorem we have $\psi'(c) = (\psi(b) - \psi(0))/b = -k/b < 0$ for some $c \in (0, b)$. Clearly then $\psi(c) = \psi'(c) < 0$. Now $\psi(0) = k > 0, \psi(c) < 0$ and hence by intermediate value theorem $\psi(x)$ vanishes at least once in $(0, c) \subset (0, b)$ and this contradicts that $b$ is the least positive value of $x$ for which $\psi(x) = 0$. It follows that $\psi(x) \neq 0$ for all $x > 0$. Since $\psi(0) = k > 0$ it follows that $\psi(x) > 0$ for all $x > 0$ (if $\psi(x) < 0$ then by IVT $\psi(x)$ would vanish somewhere in between).

Now consider the function $h(x) = \psi(-x)\psi(x)$. Clearly $$h'(x) = \psi'(x)\psi(-x) - \psi(x)\psi'(-x) = \psi(x)\psi(-x) - \psi(x)\psi(-x) = 0$$ so that $h(x)$ is constant and hence $\psi(x)\psi(-x) = h(x) = h(0) = \psi(0)\psi(0) = k^{2} > 0$. It follows that $\psi(-x)$ has same sign as that $\psi(x)$. It now follows $\psi(x) > 0$ for all $x$.

We have thus completed case 1.

Case 2: Let us now suppose that there is some constant $a$ for which $\phi(a) = k < 0$. In this case proceeding as before we can show that the function $\psi(x) = \phi(x + a)$ is such that $\psi(x) < 0$ for all $x$ and this contradicts the fact that $\psi(-a) = \phi(0) = 0$.

Since both the cases are ruled out we are left with the only alternative that $\phi(x) = 0$ for all $x$.

Summing it up, we can see that if there is a function $\phi(x)$ which satisfies $\phi'(x) = \phi(x)$ for all $x$ then $\phi(x)$ is of constant sign for all $x$ i.e. either $\phi(x)$ is identically zero for all $x$ or $\phi(x)$ is positive for all $x$ or $\phi(x)$ is negative for all $x$.

The simple constraint $\phi'(x) = \phi(x)$ thus controls the behavior of the function $\phi(x)$ in a very strong way (and fortunately for us, in a very positive way leading to theory of exponential and logarithmic functions).

Note that the reasoning in case 1 above also shows that a function $f(x)$ satisfying $f'(x) = f(x)$ and $f(0) = 1$ is such that $f'(x) = f(x) > 0$ for all $x$. Thus $f(x)$ is strictly increasing and positive and hence admits an inverse $g(x)$ which is also strictly increasing and then $g(1) = 0$ and by rule of derivative of inverse functions $g'(x) = 1/x$ so that $g(x) = \int_{1}^{x}(1/t)\, dt$ and then it is easy to show that $g(xy) = g(x) + g(y)$ and this relation leads to $f(x + y) = f(x)f(y)$ and thereby it justifies the existence of $f(x) = e^{x}$ as well as the property $e^{x + y} = e^{x}e^{y}$ is also established.

Update: There is another approach which is simpler than the one used above. We are to prove that if $\phi'(x) = \phi(x)$ for all $x$ and $\phi(0) = 0$ then $\phi(x) = 0$ for all $x$. Let us suppose for example that $\phi(a) \neq 0$ and consider the function $\psi(x) = \phi(a + x)\phi(a - x)$. Clearly we have $$\psi'(x) = \phi'(a + x)\phi(a - x) - \phi(a + x)\phi'(a - x) = \phi(a + x)\phi(a - x) - \phi(a + x)\phi(a - x) = 0$$ so that $\psi(x)$ is a constant function and hence $\psi(x) = \psi(0) = \phi^{2}(a) > 0$. It follows that $\phi(a + x)\phi(a - x) > 0$. Putting $x = a$ in this equation we get $\phi(2a)\phi(0) > 0$ which is contrary to the fact that $\phi(0) = 0$.

What we have established so far is that if a function $\phi(x)$ satisfies $\phi'(x) = \phi(x)$ for all $x$ and is such that it takes a non-zero value at some point then it does not vanish for any value of $x$. Then by continuity $\phi(x)$ is of constant sign. Thus if we assume $\phi'(x) = \phi(x)$ and $\phi(x)$ not identically zero then we can define another function $f(x) = \phi(x)/\phi(0)$. Then $f'(x) = f(x), f(0) = 1$ and $f(x) > 0$ for all $x$. If $y$ is a constant and we define $g(x) = f(x + y)/f(x)$ then $$g'(x) = \frac{f(x)f'(x + y) - f(x + y)f'(x)}{\{f(x)\}^{2}} = \frac{f(x)f(x + y) - f(x + y)f(x)}{\{f(x)\}^{2}} = 0$$ so that $g(x)$ is a constant and hence $g(x) = g(0) = f(y)$. We thus have $f(x + y) = f(x)g(x) = f(x)f(y)$ so that the functional equation of $f(x)$ is proved.