I'll prepend the above question 1. and 2. by a question 0.,
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QUESTION 0. In what sense can we talk about a mathematical notion, say a function $\,F,\,$ being the most important among all of them?
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One may talk about the ubiquity of $\, F,\, $ or about its engineering-like logical necessity, etc.
A time ago I've asked this question at MO in the context of education. I asked about the crucial place/function which is a central point for learning mathematics. You aim at it, and then you go from there in all possible directions.
Frankly speaking, I asked about two stages. The first one was the field of complex numbers $\,\Bbb C.\,$ And I had also my own answer for the 2nd stage hiding, waiting for reactions from others. Sure enough, one (or more?) participants mentioned the complex exponential function -- of course complex since
it had to follow the field of complex numbers.
My own answer was nearly equivalent, I consider that this most important function is the complex logarithm.
Now back to the OP's questions.
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QUESTION 1. Do you think, or is it common knowledge to mathematicians, that the exponential function is indeed the most important in all of mathematics?
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Certainly, some people would say that the zeta function is the most important. It encodes the secrets of prime numbers so well.
I would consider the partition function of the statistical physics at least as important as any other. It's the fundamental ingredient of the fundamental law of statistical mechanics. Richard Feynman said:
- ... the entire subject is either the slide-down from this summit ... or the climb-up ...*
(of course, the exponential function serves the partition function faithfully).
Thus, to answer Q1, it is NOT common knowledge, and for two reasons. (A) people simply don't know or don't care to think about such issues; (B) some mathematicians may have a different opinion, it all depends.
Nevertheless, if we talk about learning mathematics and having a wide open road for doing research than indeed the complex logarithm would be it.
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QUESTION 2. If so, why is this the case? What could make it more important than all the conceivable and inconceivable functions (linking arbitrary pairs of sets) out there?
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This question is really two questions: internal (atomic) and external (the world view). Namely: (A) What's so wonderful about the making of function $\,\log\,$ (about its structure/construction) that this construction is so helpful around mathematics? (B) What are the crucial achievements of function $\, \log\, $ and how deeply $\, \log\, $ is intertwined with nearly all mathematics?
(A) The internal answer, from the inside of $\,\exp/\log,\,$
is that they connect tightly the two most important elementary operations, the complex addition, and multiplication -- this is a true miracle (without which mathematics wouldn't be the same).
(B) First, we need to talk about the field of complex numbers
$\,\Bbb C\,$ -- it provides the full scope for $\,\exp/\log.\ $ In particular, $\,C\,$ is truly Euclidean plane geometry. It's a great shame that children are not taught Euclidean Geometry via complex numbers, they would master the complex numbers and Euclidean Geometry by the end of the 5th great, no sweat (well, poor teachers, but that's a different story).
Now, $\ \exp/\log\ $ are the very first transcendental function, they
are not algebraic (you can't obtain them by the five arithmetic operations), and they form a straight bridge between algebra and analysis; you can say
-- between finiteness and infinity.
The logarithmic function is at the start of algebraic topology. That's where the true wonders of topology and its complexity shine. You may recall the Eilenberg's theorem about dissecting $\, Bbb\,$ which is a concert played on two instruments, $\, \exp\,$ and $\,\log.$
Of course, these two functions appear anywhere you look at, e.g. algebra, combinatorics, geometry, probability, statistical mechanics, ...
One could praise these two guys forever.
PS Statistical Mechanics is so rich and complete that if mathematicians were restricted to Statistical Mechanics only, it'd make hardly any difference, they wouldn't miss much.
Your statements are hinting at something called convexity; however, your notation is a bit off: as it is now, the statements don't hold even for simple cases. I'm going to assume that you intended to talk about convexity and answer myself :)
A function is said to be convex if, for any two points $x$ and $y$ and any $\lambda\in [0,1]$, then
$$ f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y).$$
Note, in particular, that this means that the value of the function at the midpoint of $x$ and $y$ is smaller than the midpoint of the values of the function at $x$ and $y$:
$$ f\left(\frac{x+y}{2}\right) \leq \frac{f(x) + f(y)}{2}.$$
If a function is not convex it may be concave, if
$$ f(\lambda x + (1-\lambda)y) \geq \lambda f(x) + (1-\lambda)f(y).$$
If a function is both convex and concave, then it is linear. Note that there are functions that are neither convex nor concave!
In the examples you allude to, we have that the linear function is (clearly) linear, so
$$ f(\lambda x + (1-\lambda)y) = \lambda f(x) + (1-\lambda)f(y).$$
Let's look at the exponential function $f(x) = \mathrm{e}^x$; this is a convex function, so in this case we would have
$$ \mathrm{e}^{\lambda x + (1-\lambda)y} \leq \lambda \mathrm{e}^x + (1-\lambda)\mathrm{e}^y.$$
The general proof requires derivatives and Taylor expansions; however, for the midpoint, the proof is very easy. In this case we have to prove that
$$ \mathrm{e}^{\frac{x+y}{2}} \leq \frac{\mathrm{e}^x + \mathrm{e}^y}{2},$$
for all $x$ and $y$. Consider the following steps
$$ \mathrm{e}^{\frac{x+y}{2}} \leq \frac{\mathrm{e}^x + \mathrm{e}^y}{2}$$
$$ 2\mathrm{e}^{\frac{x+y}{2}} \leq \mathrm{e}^x + \mathrm{e}^y$$
$$ 2\mathrm{e}^{x/2}\mathrm{e}^{y/2} \leq \mathrm{e}^x + \mathrm{e}^y$$
$$ 0 \leq \mathrm{e}^x - 2 \mathrm{e}^{x/2}\mathrm{e}^{y/2} + \mathrm{e}^y$$
$$ 0 \leq \left(\mathrm{e}^{x/2} - \mathrm{e}^{y/2}\right)^2$$
Note that the right hand side is a square, so the inequality holds for all $x$ and $y$!
The logarithm function, on the other hand, is concave, so the reverse inequality holds: for all positive $x$ and $y$,
$$\log\left(\frac{x+y}{2}\right) \geq \frac{\log(x) + \log(y)}{2}.$$
We can easily prove this with the following chain of inequalities:
$$\log\left(\frac{x+y}{2}\right) \geq \frac{\log(x) + \log(y)}{2}$$
$$\frac{x+y}{2} \geq \exp\left(\frac{\log(x) + \log(y)}{2}\right)$$
$$\frac{x+y}{2} \geq \exp\left(\frac{\log(x)}{2}\right)\exp\left(\frac{\log(y)}{2}\right)$$
$$\frac{x+y}{2} \geq \exp\left(\log(\sqrt{x})\right)\exp\left(\log(\sqrt{y})\right)$$
$$\frac{x+y}{2} \geq \sqrt{x}\sqrt{y}$$
$$x+y \geq 2\sqrt{x}\sqrt{y}$$
$$x+y - 2\sqrt{x}\sqrt{y} \geq 0$$
$$\left(\sqrt{x} - \sqrt{y}\right)^2 \geq 0,$$
which, again, is true for all (positive) $x$ and $y$!
Best Answer
No, there's no connection between the injectivity (what you call "one-to-one") property and the various algebraic properties. The theorems just happen to be stated one after another in this particular book.
Knowing that the exponential and logarithmic functions are injective is just a useful fact to know. Really, knowing that any function $f$ is injective is useful if you plan on using $f$ a lot, because it allows you to cancel $f$, in other words, if in your calculations you should end up with:
$$f(x)=f(y)$$
You can immediately conclude that:
$$x=y$$
Which is only true if $f$ is injective, of course. For instance, $x^2$ is not injective, which is why $x^2=y^2$ does not imply $x=y$.