Given a function $g(x)$ which has derivative $g'(x)$ for every x and satisfies $g'(0)=2$ and $g(x+y) = e^y g(x)+ e^x g(y)$ for all x and y. Find g.

Question

Given a function $g(x)$ which has derivative $g'(x)$ for every x and satisfies $g'(0)=2$ and $g(x+y) = e^y g(x)+ e^x g(y)$ for all x and y. Find g.

Actually I have seen the solution attached to the problem and I agree to it surely. The issue is that I am not able to figure out what why does my solution works.

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g(0)=0

My solution: Put $x=y$,

$g(2x) = 2e^xg(x)$

Till now everything is good, I don't know where things go wrong from here on.

I thought of telescoping and using continuity of the function at last. $\frac{g(x)}{g(\frac{x}{2^n})}=e^{x[1-\frac{1}{2^n}]}.2^n$

My next step was to apply limit n tends to inf both sides.

$\lim_{n \to \infty} (\frac{g(x)}{g(\frac{x}{2^n})}) = \lim_{n \to \infty} e^{x[1-\frac{1}{2^n}]}.2^n$

$g(x)=e^x \lim_{n \to \infty} g(\frac{x}{2^n})2^n$

If this would have been determinate, I would have been satisfied and happy. Now stuck here, I thought of undergoing my routine procedure of creating $\lim_{n \to \infty} \frac{g(\frac{x}{2^n})}{2^{-n}}$ to apply L'hopital rule.

Now the uncertainty in my mind is how do I differentiate the function in numerator. It is a function of x and n both and I have to differentiate wrt n.

I did use substitution $t=x/2^n$ and then happily differentiated f(t) w.r.t $t$. ( I still had doubts about taking x out of the limit)

But I am somehow still unable to get a solid understanding of what I did or something. Can't we differentiate (or maybe partially differentiate) $f(x,n) $ w.r.t. n? I am a bit perplexed.

Answer 1

It turns out you can avoid L'Hôpital's rule entirely.

Firstly, the identity $g(2x)=2e^xg(x)$, applied to $x=0$, means that $g(0)=0$. Now, let's start from $$g(x)=e^x\lim_{n\to\infty} g\left(\frac x{2^n}\right)2^n.$$ Write $t_n=x/2^n$; we have $$g(x)=e^x\lim_{n\to\infty}g(t_n)(x/t_n)=xe^x\lim_{n\to\infty}\frac{g(t_n)}{t_n}.$$ This limit looks rather like the expression defining $g'(0)$. Indeed, we have $$2=g'(0)=\lim_{t\to 0}\frac{g(t)-g(0)}t=\lim_{t\to 0}\frac{g(t)}t.$$ (This limit exists since $g'(0)$ is well-defined.) Now, the sequence $\{t_n\}$ tends to zero as $n\to\infty$. This means that, since the limit in the above line exists, $$\lim_{n\to\infty}\frac{g(t_n)}{t_n}=\lim_{t\to 0}\frac{g(t)}t=2.$$ We conclude that $g(x)=2xe^x$ for all $x$.