If $f(x–y)= f(x)·g (y)–f (y)·g(x)$ and $g(x – y)=g(x)·g(y)+f(x)·f(y)$ for all $x, y \in R$. Find following for the following given conditions.

Question

If $f (x – y) = f (x) · g (y) – f (y) · g (x)$ and $g (x – y) = g (x) · g (y) + f (x) · f (y)$ for all $x, y \in R$. If right hand derivative at $x = 0$ exists for $f (x)$. Find derivative of $g (x)$ at $x = 0$.

My attempt is as follows:-

$$\lim_{h\to0^{+}}=\dfrac{f(h)-f(0)}{h}$$

Putting $x=0,y=0$ in the first functional equation

$$f(0)=f(0)g(0)-f(0)g(0)$$ $$f(0)=0$$

Putting $x=0,y=0$ in the second functional equation

$$g(0)=g^2(0)$$ $$g(0)=0 \text { or } g(0)=1$$

Case $1$: $g(0)=0$

Putting $x=h,y=0$ in the first equation

$$f(h)=f(h)g(0)-f(0)g(h)$$

$$f(h)=0$$

So $$\lim_{h\to0^{+}}=\dfrac{f(h)-f(0)}{h}=0=f'(0^+)$$

Putting $x=h,y=0$ in the second equation

$$g(h)=g(h)g(0)+f(h)f(0)$$ $$g(h)=0$$

$$g'(0)=\lim_{h\to0}=\dfrac{g(h)-g(0)}{h}=0$$

Case $2$: $g(0)=1$

Putting $x=0,y=h$ in the first equation

$$f(0-h)=f(0)g(h)-f(h)g(0)$$ $$f(-h)=-f(h)$$

$$g(0-h)=g(0)g(h)+f(0)f(h)$$ $$g(-h)=g(h)$$

$$\lim_{h\to 0}\dfrac{g(h)-g(0)}{h}$$

Put $x=\dfrac{h}{2},y=-\dfrac{h}{2}$ in the second functional equation

$$g(h)=g\left(\dfrac{h}{2}\right)g\left(-\dfrac{h}{2}\right)+f\left(\dfrac{h}{2}\right)f\left(-\dfrac{h}{2}\right)$$

$$g(h)=g\left(\dfrac{h}{2}\right)g\left(\dfrac{h}{2}\right)-f\left(\dfrac{h}{2}\right)f\left(\dfrac{h}{2}\right)\tag{1}$$

Putting $x=y$ in the second functional equation

$$g(0)=g(x)^2+f(x)^2$$

$$g(h)=1-f\left(\dfrac{h}{2}\right)f\left(\dfrac{h}{2}\right)-f\left(\dfrac{h}{2}\right)f\left(\dfrac{h}{2}\right)\tag{1}$$

$$\lim_{h\to 0}\dfrac{-2f^2\left(\dfrac{h}{2}\right)}{h}$$

$$\lim_{h\to 0}\dfrac{-2hf^2\left(\dfrac{h}{2}\right)}{4\left(\dfrac{h}{2}\right)^2}$$

$$\dfrac{-1}{2}\cdot\left(\lim_{h\to 0}h\right)\cdot\lim_{h\to 0}\dfrac{f^2\left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)^2}$$

$$\dfrac{-1}{2}\cdot\left(\lim_{h\to 0}h\right)\cdot f'(0)^2=0$$

Answer 1

Although it has been asked, I would like to present an answer different from that of the given ones.

It apparently has something to do with complex numbers.

We define a function $h: \Bbb R \rightarrow \Bbb C$ by $h(x) = g(x) + if(x)$.

The two identities for $f, g$ then translates to just one identity: $$h(x - y) = h(x)\overline{h(y)},\forall x, y \in \Bbb R.\tag{*}$$

If the function $h$ is identically zero, then of course both $f$ and $g$ are identically zero.

Suppose now that there exists $a\in \Bbb R$ such that $h(a) \neq 0$.

Setting $y = x - a$ in (*), we get $h(a) = h(x)\overline{h(x - a)}$. Hence $h(x)\neq 0$ for all $x\in \Bbb R$.

Setting $y = \frac x 2$ in (*), we get $h(\frac x 2) = h(x)\overline{h(\frac x 2)}$, hence $h(x) = h(\frac x 2)/\overline{h(\frac x 2)}$. In particular, we have $|h(x)| = 1$ for all $x \in \Bbb R$.

We may then rewrite (*) as: $h(x - y) = h(x)h(y)^{-1}, \forall x, y \in \Bbb R$. Changing $x$ to $x + y$, we get: $$h(x + y) = h(x)h(y), \forall x, y \in \Bbb R.\tag{**}$$

Hence $h$ is a group homomorphism from $\Bbb R$ to the circle $\{z\in \Bbb C: |z| = 1\}$. In particular, we have $h(0) = 1$, which implies $g(0) = 1$ and $f(0) = 0$; and also $h(-x) = h(x)^{-1} = \overline{h(x)}$, which implies $g(x) = g(-x)$ and $f(x) = -f(x)$ for all $x\in\Bbb R$.

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At this point, the original question can be answered.

Since the right derivative $\lim_{x \rightarrow 0+}\frac{f(x)}x$ exists, and $f$ is odd, we know that there exists $\alpha > 0$ and $b > 0$ such that for any $x$ with $|x| < b$, we have $|f(x)| \leq \alpha|x|$.

Shrinking $b$ if necessary, we may assume that $|f(x)| < \frac 1 2$ whenever $|x| < b$. It follows that, for such $x$, we have $h(x) = e^{i\theta}$ with $\theta \in (-\frac\pi 6, \frac\pi 6) \cup (\frac{5\pi} 6, \frac{7\pi} 6)$. However, if $\theta \in (\frac{5\pi} 6, \frac{7\pi} 6)$, then by (**), we should have $h(\frac x 2)^2 = h(x) = e^{i\theta}$, which implies that $h(\frac x 2) = e^{i\tau}$ with $\tau \in (\frac{5\pi}{12}, \frac{7\pi}{12}) \cup (\frac{-7\pi}{12}, \frac{-5\pi}{12})$. This then leads to $|f(\frac x 2)| > \sin(\frac{5\pi}{12}) > \frac 1 2$, a contradiction. Hence we have $\theta \in (-\frac\pi 6, \frac\pi 6)$. In particular, it follows that $g(x) > 0$ whenever $|x| < b$.

Therefore, for any $x$ with $|x| < b$, we have $$|g(x) - 1| = |1 - \sqrt{1 - f(x)^2}| = \left|\frac{f(x)^2}{1 + \sqrt{1 - f(x)^2}}\right|\leq |f(x)|^2 \leq \alpha^2 |x|^2.$$

Hence the limit $\lim_{x\rightarrow 0}\frac{|g(x) - 1|}x$ is clearly $0$, i.e. we have $g'(0) = 0$.