1
$\begingroup$

I have been stuck on trying to prove this without using proof by contradiction. Prove $g''(x)>0$ if $g(x) = xf(x), 1>x>0, f(x)>0, f'(x)>0,$ $\lim_{x \to 0} f'(x) =c>0$, and $f''(X)$ exists. I have tried proving it by showing the elasticity of $g$ with respect to $x$ is greater than 1, i.e. $\frac{d \ln g}{d \ln x} >1$, but I'm not sure this result guarantees $g''(x)>0$.

Proof by contradiction: Suppose $x \in [0,1]$ and $f: [0,1] \rightarrow (0,\infty)$ where $f \in C^2$ and $f'(x) >0.$ Define a function $g$ such that $g(x) = xf(x)$.

Suppose $g''(x)<0$ and define a function $h(x) = x^2f'(x).$ It follows $h'(x) = xg''(x)<0.$ Then for $t<1,$ we have $h(t)>h(1)$, which can also be written as $tf'(t)>\frac{f'(1)}{t}.$ Thus $$ f(1) - f(x) = \int_x^1 f'(t)dt \geq \int_x^1 t f'(t)dt > \int_x^1 \frac{f'(1)}{t}dt= - f'(1) \ln x.$$

The above statement can be written as $f(x) < f(1) + f'(x) \ln x, $ which implies $ \lim_{x \rightarrow 0} f(x) = -\infty, $ which contradicts the assumption that $f(x)>0 $ for all $x \in [0,1].$

$\endgroup$
2
+50
$\begingroup$

You can't, because it isn't.

We have $$ \begin{align*} g'(x) &= xf'(x) + f(x) \\ g''(x) &= xf''(x)+f'(x)+f'(x) = xf''(x)+2f'(x). \end{align*} $$

Now, let

$$ f(x) := -\frac{c}{3}x^3+cx+d. $$

For $d$ sufficiently large, we have $f(x)>0$ for $0<x<1$. We have

$$ f'(x) = -cx^2+c, $$

so $f'(x)>0$ and $\lim_{x\to 0}=c$. Next,

$$ f''(x) = -2cx. $$

Thus,

$$ g''(x) = xf''(x)+2f'(x) = -2cx^2-2cx^2+2c = -4cx^2+2c, $$

which is negative for $x\to 1$.


The key thing you need to look at is your equation

$$ g''(x) = xf''(x)+2f'(x). $$

Thus, to control $g''$, you need to control both $f'$ and $f''$ across the entire interval $0<x<1$. Controlling $f'$ on the entire interval but $f''$ only pointwise (as in a limit for $f''(x)$ as $x\to 0$) is not enough. (If you add another condition on $\lim_{x\to 1} f''(x)$, I am confident we can create another counterexample that fails somewhere else on the interval.) Derivatives of well-behaved functions can behave quite badly indeed.


Where does your proposed proof break down? You switch quantifiers in the middle without noticing. Your function $h$ indeed satisfies $h'(x)<0$, but not for all $x$, but only for some $x$ near $1$. Therefore, your integral inequality

$$ \int_x^1tf'(t)\,dt \geq\int_x^1\frac{f'(1)}{t}\,dt $$

does not necessarily hold.

Bottom line: explicitly write out quantifiers and make sure your deductions hold for the quantifiers you actually have.


The following is my original answer to the question pre-edit, where the condition $\lim_{x\to 0}f'(x)=c>0$ was not yet imposed.

You can't, because it isn't.

Now, let $$ f(x) := -x-\frac{1}{x}+c, $$ which is positive for $0<x<1$ if $c$ is large enough. We have $$ \begin{align*} f'(x) = & -1+\frac{1}{x^2} \\ f''(x) = &-\frac{2}{x^3}, \end{align*} $$ so $f'(x)>0$ for $0<x<1$, and $f''(x)$ exists, but $$ g''(x) = xf''(x)+2f'(x) = -\frac{2}{x^2}-2+\frac{2}{x^2} = -2 <0. $$


Your proposed proof (without the condition added later) breaks down at the very end, where you write that

$f(x) < f(1) + f'(x) \ln x, $ which implies $ \lim_{x \rightarrow 0} f(x) = -\infty $

This implication does not hold unless you can bound the behavior of $f'(x)$ as $x\to 0$. If you can't, $f'(x) \ln x$ can go elsewhere than to $-\infty$ as $x\to 0$. For instance, if $f'(x)=x$, then $\lim_{x\to 0}x\ln x=0$ by L'Hôpital's rule.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Great work. Thanks! Can you please help me see why this proof by contradiction is wrong then? Please see the original question. $\endgroup$ – dlnB May 11 at 19:29
  • 1
    $\begingroup$ I'll happily take a look (promising nothing) if you could include a link to the original question. $\endgroup$ – Stephan Kolassa May 11 at 19:30
  • 1
    $\begingroup$ No problem, I edited with a new proposal. $\endgroup$ – Stephan Kolassa May 11 at 20:06
  • 1
    $\begingroup$ If $\lim_{x\to 0} f'(x)=c>0$ then yes, it looks good (from my cursory look). $\endgroup$ – Stephan Kolassa May 11 at 20:13
  • 2
    $\begingroup$ I edited the answer to address this. In general, I would appreciate it if you didn't change the question substantially after it has been answered, because that invalidates answers. (Edits for clarity are fine.) Much better to ask a new question and link the two. $\endgroup$ – Stephan Kolassa May 20 at 7:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.