Triple Thunder!

1. For any function $f \in \mathcal F$ , we must have $|f(x) - f(y)| \le \ln\tfrac{x}{y}$ for all $x > y > 0$ . To see this, for any $n \in \mathbb{N}$ we put $h = \tfrac{x-y}{n}$ and see that $\begin{aligned} |f(x) - f(y)| & = \; \left| \sum_{j=1}^n \big(f(y+jh\big) - f\big(y+(j-1)h\big)\right| \\ & \le \; \sum_{j=1}^n \left|\big(f(y+jh\big) - f\big(y+(j-1)h\big)\right| \; \le \; \sum_{j=1}^n \frac{h}{y + (j-1)h} \\ & \le \; \frac{h}{y}\sum_{j=1}^n \frac{1}{1 + (j-1)\frac{h}{y}} \; = \; \frac{h}{y} + \sum_{j=1}^{n-1} \frac{1}{1 + j\frac{h}{y}} \\ & \le \; \frac{x-y}{ny} + \int_0^{(n-1)\frac{h}{y}} \frac{dt}{1+t} \; \le \; \frac{x-y}{ny} + \int_0^{\frac{x-y}{y}}\frac{dt}{1+t} \\ & = \; \frac{x-y}{ny} + \ln\tfrac{x}{y} \end{aligned}$ We have used here the fact that $(1+t)^{-1}$ is a decreasing function of $t$ , and hence that the Riemann sum is less than the matching integral. Letting $n \to \infty$ gives us the result. We also note that the function $f_0(x) = \ln x$ has the property that $|f_0(x+y) - f_0(y)| \; = \; \ln\left(1 + \tfrac{x}{y}\right) \; \le \; \tfrac{x}{y} \hspace{1cm} x,y > 0$ so that $f_0 \in \mathcal F$ and $|f_0(x) - f_0(y)| \; = \; \ln\tfrac{x}{y} \hspace{2cm} x > y > 0$ Thus the maximum value $A \ln2$ is achieved by choosing $f=f_0$ , and hence $A\ln 2 \; = \; \sum_{j=1}^{10}|f_0(2^{10}) - f_0(2^j)| \; = \; \sum_{j=1}^{10} \ln(2^{10-j}) \; = \; \sum_{j=0}^9 \ln 2^j \; = \; \ln 2^{45} \; = \; 45\ln2$ and so $A = 45$ .
2. The transformations $g(x) = e^{-x}f(x)$ for $x > 0$ and $h(x) = g(e^x)$ for $x \in \mathbb{R}$ lead to the facts that $h$ is differentiable on $\mathbb{R}$ and $h(x+y) = h(x)+h(y)$ for all real $x,y$ . Thus $h(x) = \alpha x$ for real $x$ , and hence $f(x) = \alpha e^x \ln x$ for $x > 0$ . Thus $f'(x) = \alpha e^x(\ln x + x^{-1})$ , so the requirement that $f'(1) = e$ tells us that $\alpha=1$ , so that $B = f(2) = e^2\ln2$ . It is worth noting that assuming that $f$ is continuous is enough to get this result.
3. Putting $y=1$ , we see that $2f(x) = f(x^2)$ . The equation now reads $f(uv) = f(u) + f(v)$ for $u,v > 0$ and (since $f$ is differentiable), $f(x) = \alpha \ln x$ for $x > 0$ . Since $f'(1)=1$ , we deduce that $\alpha=1$ , and so $f(x) = \ln x$ . Thus $C = \ln4 = 2\ln2$ . Again, we only need assume that $f$ is continuous.

This makes the answer $A+B+C = 45 + (e^2 + 2)\ln2 = \boxed{51.50799776}$ .

Relevant wiki: Limits of functions - Problem solving

$1$ . $|f(p)-f(q)|<=$ $\displaystyle \sum_{j=1}^{n}\frac{h}{q+(j-1)h}$ (why? try to break $f(p)-f(q)$ into a summation and then use the given property...you will get)( where $h=\frac{p-q}{n}$ ) $<=\frac{h}{q}\displaystyle \sum_{j=1}^{n}\frac{1}{1+(j-1)\frac{h}{q}}$ $=$ $\frac{h}{q}+$ $\displaystyle \sum_{j=1}^{n-1}\frac{1}{1+j\frac{h}{q}}$ $<=\frac{p-q}{nq}+ln\frac{p}{q}$ which tends to $ln\frac{p}{q}$ .From here conclude that the function will be $f(x)=ln x$ and hence answer is $45ln2$ $\&$ $A=45$ .

$2$ .Put $x=y=1$ to get $f(1)=0$ . $f'(x)$ = $\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}$ = $\lim_{h\to0} \dfrac{f(x.(1+\frac{h}{x}))-f(x)}{h}$ = $\lim_{h\to0} \dfrac{f(x)(e^h-1)+e^{h-1+x-\frac{h}{x}}(f(1+\frac{h}{x})-f(1)}{h})$ = $f(x)$ + $\frac{e^{x-1}.f'(1)}{x}$ .Now you should easily get that $\frac{1}{x}$ = $\frac{d(f(x))}{dx(e^{x})}$ .Hence $f(x)=e^xln|x|$ and $B=e^2ln2$ .

$3$ .Let $xy=u$ $\&$ $\frac{x}{y}=v$ . $2f(\sqrt{uv})$ = $f(u)+f(v)$ .Differentiate w.r.t $u$ ,we get $2f'(\sqrt{uv})$ * $\frac{\sqrt{v}}{2\sqrt{u}}$ = $f'(u)$ .Put $u=1$ and replace $v$ with $t^2$ .. $f'(t)=$ $\frac{1}{t}$ .So $f(x)=ln|x|$ and $C=ln4$ .