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June 23

Is there any general non-obvious sufficient condition, for a given differentiable injection $f(x)$ to satisfy that $f(x)/x$ (for all $x\ne 0$) is not injective?

Here are some simple examples of a differentiable injection $f$, for which $f(x)/x$ (for all $x\ne 0$) is not injective:

1. $f:x\mapsto A\cdot x,$ defined on the set of real numbres, for any real $A\ne 0.$
2. $f:x\mapsto A\cdot x^B+C,$ defined on the set of real numbres, for any real $A\ne 0,$ and for any odd number $B\ne 1,$ and for any real $C.$
3. $f:x\mapsto A\cdot x^B+C,$ defined on the set of positive numbres, for any real $A\ne 0,$ and for any natural $B\ne 1,$ and for any positive $C$
4. $f:x\mapsto {\log }_B(x),$ defined on the set of positive numbres, for any positive base $B\ne 1.$
5. $f:x\mapsto A\cdot \arcsin \sqrt{x},$ defined on the interval $[0,1],$ for any real $A\ne 0.$

By a "general" sufficient condition, I exclude any of the sufficient examples mentioned above. They are really sufficient (and non-obvious), yet not general, but rather special cases.

By a "non-obvious" sufficient condition, I exclude any obvious sufficient condition like the following (general) one: "$f:x\mapsto xg(x)$ for some differentiable function $g$ that is not injective while $f$ is".

2A06:C701:427F:6800:8CE8:BDC9:AFA0:45F4 (talk) 09:26, 23 June 2023 (UTC)

Few notes to make:

1. Assuming that $f(x)$ is $C^1$, $f(x)/x$ exists either properly or as a limit across the real line if and only if $f(0)=0$. The only if direction is obvious since $f(0)\ne 0$ would make $\underset{x\to 0}{\lim }f(x)/x$ not exist. The if direction results from L'Hôpital's rule, guaranteeing that $\underset{x\to 0}{\lim }f(x)/x=f^{\prime }(0)$.

2. If we do assume that $f(0)=0$, and if we furthermore assume that $f$ is $C^2$, the derivative of $f(x)/x$ exists everywhere (again either properly or as a limit), as by L'Hôpital's rule, $\underset{x\to 0}{\lim }(x{f}^{\prime }(x)-f(x))/x^2=f^{″}(0)/2$.

3. In general, assuming that $g(x)$ is differentiable, $g(x)$ is also injective if and only if $g^{\prime }(x)$ is always nonnegative or nonpositive, and only $0$ at isolated points (I will call this property A.)

4. Combined together, if we assume that $f(x)$ is a $C^2$ injection with $f(0)=0$, then it really boils down to $f^{\prime }(x)$ having property A, but $x{f}^{\prime }(x)-f(x)$ not having property A.

GalacticShoe (talk) 16:59, 23 June 2023 (UTC)

Few notes to make:

• Thanks ever so much.
• Re. the end of your first section: By the last mathematical expression $f^{\prime }(x),$ appearing to the right of the identity sign, you have probably meant $\underset{x\to 0}{\lim }{f}^{\prime }(x),$ haven't you?
• Re. the end of your second section: By the last mathematical expression $f^{″}(x)/2,$ appearing to the right of the identity sign, you have probably meant $\underset{x\to 0}{\lim }{f}^{″}(x)/2,$ haven't you?
• Re. your fourth section: Even though your general sufficient condition does not cover my fourth example above (because the logarithmic function is not defind for zero), nor does it cover my third example (for a similar reason), your condition is still a general (non-obvious) sufficient one, as required (BTW, practically speaking, I need all of this for functions not defined for zero. I forgot to mention that in my original question).

2A06:C701:7453:7D00:8F6:E1A9:A503:D522 (talk) 19:44, 24 June 2023 (UTC)

In regards to the second and third question, that's my mistake; it should be $f^{\prime }(0)$ and $f^{″}(0)/2$, I will edit my answer accordingly. GalacticShoe (talk) 20:59, 24 June 2023 (UTC)

You should have also added the condition that $f^{\prime }$ is continuous, shouldn't you? 2A06:C701:7453:7D00:8F6:E1A9:A503:D522 (talk) 21:09, 24 June 2023 (UTC)

Good point, let me change to continuous differentiability to make it work in general. GalacticShoe (talk) 22:29, 24 June 2023 (UTC)

You may want to precede 'lim' with a backslash. This would make the 'lim' a symbol, rendered in the upright font with appropriate spacing, instead of a blob of italic, varable-like letters. Compare Template:Nowrap → $\lim x$ to Template:Nowrap → $limx$. --CiaPan (talk) 20:24, 25 June 2023 (UTC)

Will do, thanks for the heads up! GalacticShoe (talk) 00:19, 26 June 2023 (UTC)