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 $x \in [0,1].$


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,

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


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

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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ No problem, I edited with a new proposal. $\endgroup$ – Stephan Kolassa May 11 at 20:06
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    $\begingroup$ 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
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    $\begingroup$ 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

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