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Let $f:\mathbb{R}\to\mathbb{R}$ be a differentiable function s.t. $f’$ is continuous. Suppose $f’\left(\frac{1}{2}\right)=0$, prove that there is $c\in\left(0,\frac{1}{2}\right)$ s.t.

$$f'(c)=2c(f(c)-f(0))$$

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Consider the function

$$

g(x)=e^{-x^2}(f(x)-f(0))\tag{1}

$$

Taking the derivative, we get

$$

g'(x)=e^{-x^2}\left(f'(x)-2x(f(x)-f(0))\right)\tag{2}

$$

Notice that

$$

g(0)=0\tag{3}

$$

and

$$

g(1/2)=e^{-1/4}(f(1/2)-f(0))\tag{4}

$$

and

$$

g'(1/2)=-e^{-1/4}(f(1/2)-f(0))\tag{5}

$$

If $g(1/2)=0$, then by $(3)$ and the Mean Value Theorem, there is a $c\in(0,1/2)$ so that $g'(c)=0$.

If $g(1/2)\gt0$, then there is an $x_0$ in $(0,1/2)$ so that $g'(x_0)=2(g(1/2)-g(0))\gt0$. Furthermore, by $(4)$ and $(5)$, $g'(1/2)\lt0$. By the intermediate value theorem, there is a $c\in(x_0,1/2)$ so that $g'(c)=0$.

If $g(1/2)\lt0$, then there is an $x_0$ in $(0,1/2)$ so that $g'(x_0)=2(g(1/2)-g(0))\lt0$. Furthermore, by $(4)$ and $(5)$, $g'(1/2)\gt0$. By the intermediate value theorem, there is a $c\in(x_0,1/2)$ so that $g'(c)=0$.

Thus, there is a $c\in(0,1/2)$ so that $g'(c)=0$, which by $(2)$, gives us

$$

f'(c)-2c(f(c)-f(0))=0\tag{6}

$$

This is a partial answer.

If $f'(x) = 0$ for all $x \in (0, 1/2)$ then the proof is trivial because then $f(x)$ is a constant and the relation $f'(x) = 2x\{f(x) – f(0)\}$ holds for any $x \in (0, 1/2)$.

If we assume that $f'(0) \neq 0$ then the proof is easy. In this case by the virtue of continuity of $f'(x)$ there will be an interval of type $(0, a)$ where the function $f'(x)$ maintains its sign. Also by continuity of $f'(x)$ there will be a first value $\alpha \in (0, 1/2]$ such that $f'(\alpha) = 0$ and $f'(x)$ is of constant sign in $[0, \alpha)$. Hence in each case the function $G(x)$ given by $$G(x) = \dfrac{f'(x)}{{\displaystyle x\int_{0}^{x}f'(t)\,dt}} = \dfrac{f'(x)}{x\{f(x) – f(0)\}}$$ is well defined for $(0, \alpha]$. Clearly it is continuous in $(0, \alpha]$ and $G(\alpha) = 0$. Also as $x \to 0+$ the function $G(x) \to \infty$. Hence by IVT there is a $c \in (0, \alpha) \subset (0, 1/2)$ for which $G(c) = 2$ and this completes the proof.

If $f'(0) = 0$ then we have bit more difficulty. In this case it may happen that there is an interval of type $(0, a)$ in which $f'(x)$ is non-zero and therefore maintains its sign. Then the proof proceeds as before. We only need to note that the ratio $$\dfrac{f'(x)}{{\displaystyle x\int_{0}^{x}f'(t)\,dt}} = \dfrac{f'(x)}{x^{2}f'(\xi)}$$ where $\xi \in (0, x)$ tends to $\infty$ as $x \to 0+$ because by continuity of $f'(x)$ at $x = 0$ the ratio $f'(x)/f'(\xi)$ remains bounded.

There may be a scenario when $f'(0) = 0$ and every interval of type $(0, a)$ contains a zero of $f'(x)$ (e.g. $f'(x) = x\sin\{(1/x) – 2\}, f'(0) = 0$). This is the only remaining case where I have not been able to establish the claim.

From the partial solution above I note two things:

1) The $1/2$ in $f'(1/2) = 0$ can be replaced by an positive $a$ with $f'(a) = 0$.

2) In the conclusion $f'(c) = 2c\{f(c) – f(0)\}$ the constant $2$ can be replaced by any positive number $K$.

We have supposed $f'(\frac 12)=0$ and $f$ is differentiable over $\Bbb R$ and $f’$ is continuous over $\Bbb R$. By continuity of $f'(x)$, noting the inequality $2c^2\le\frac12\forall c\in(0,\frac12)$ and by the known value $f'(\frac12)=0$, for every $d\in (0,\frac 12)$ there exists a value $c\in(d,\frac12)$ such that $2c^2f'(d)=f'(c)$.

Further, we can take the Mean Value Theorem and set $e\in(0,c)$ such that $f'(e)=\frac{f(c)-f(0)}{c-0}$. Since $d$ can take on any value in the interval $(0,\frac 12)$, and since $e\in(0,c)\subseteq (0,\frac12)$, there must exist a $c,d$ such that $f'(d)=f'(e)$, which means that

$$f'(c)=2c^2f'(d)=2c(f(c)-f(0))$$

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