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user44143
user44143

This is again too long for a comment:

Let $a=-f'(0),\ b=-f'(1),\ c=f''(0),\ d=-f''(1)$ to show the narrow range of polynomials that might work. Since $f$ is convex on $[0,p]$ and concave on $[p,1]$ we also knowWe let $f''(p)=0$$a=-f'(0),\ b=-f'(1),\ c=f''(0),\ d=-f''(1)$.

So, ifIf $f$ is a polynomial, then $f''(x)$ is of the form $$f''(x)=(p-x)\left(\frac{c(1-x)}{p}+\frac{dx}{1-p}+x(1-x)q(x)\right)$$ for some polynomial $q(x)$ such that $f''(x)/(p-x)$ is positive on $[0,1]$. In particular, the factor of $p-x$ comes from the requirement that $f''(p)=0$, which in turn is the consequence of $f$ being convex on $[0,p]$ and concave on $[p,1]$.

Then there are only four more remaining conditions on $f''$ to solve the problem:

$$-f'(p) + \int_0^p f''(x)\, dx = a$$

$$f'(p) + \int_p^1 f''(x)\, dx = -b$$

$$\int_p^1 \left(f'(p) + \int_p^u f''(x)\, dx\right)\, du = 0$$

$$\int_0^p \left(f'(p) - \int_u^p f''(x)\, dx\right)\, du = 0$$ \begin{align} -f'(0)=a&:\ \ \ \ \ \ -f'(p) + \int_0^p f''(x)\, dx = a\\ -f'(1)=b&:\ \ \ \ \ \ \ \ \ \ f'(p) + \int_p^1 f''(x)\, dx = -b\\ f(0)=f(p)&:\ \int_p^1 \left(f'(p) + \int_p^u f''(x)\, dx\right)\, du = 0\\ f(p)=f(1)&:\ \int_0^p \left(f'(p) - \int_u^p f''(x)\, dx\right)\, du = 0 \end{align}

So one approach is towe could try $q(x)$ of the form $$q(x)=r x^i + s (1-x)^j + tx(1-x)$$ and solve the four equations above for $r,s,t$ and $f'(p)$.

The integrals are easy. Solving the equations is easy. Ensuring the positivity of $f''(x)/(p-x)$ is difficult. But if we can choose $i$ and $j$ to make $r,s,t$ positive, that will be enough. The trick is finding how $i$ and $j$ can depend on $a,b,c,d$ to get this positivity, or getting something similar for another form of $q(x)$.

This is again too long for a comment:

Let $a=-f'(0),\ b=-f'(1),\ c=f''(0),\ d=-f''(1)$. Since $f$ is convex on $[0,p]$ and concave on $[p,1]$ we also know $f''(p)=0$.

So, if $f$ is a polynomial, then $f''(x)$ is of the form $$f''(x)=(p-x)\left(\frac{c(1-x)}{p}+\frac{dx}{1-p}+x(1-x)q(x)\right)$$ for some polynomial $q(x)$ such that $f''(x)/(p-x)$ is positive on $[0,1]$.

Then there are only four more remaining conditions on $f''$ to solve the problem:

$$-f'(p) + \int_0^p f''(x)\, dx = a$$

$$f'(p) + \int_p^1 f''(x)\, dx = -b$$

$$\int_p^1 \left(f'(p) + \int_p^u f''(x)\, dx\right)\, du = 0$$

$$\int_0^p \left(f'(p) - \int_u^p f''(x)\, dx\right)\, du = 0$$

So one approach is to try $q(x)$ of the form $$q(x)=r x^i + s (1-x)^j + tx(1-x)$$ and solve the four equations above for $r,s,t$ and $f'(p)$.

The integrals are easy. Solving the equations is easy. Ensuring the positivity of $f''(x)/(p-x)$ is difficult. But if we can choose $i$ and $j$ to make $r,s,t$ positive, that will be enough. The trick is finding how $i$ and $j$ can depend on $a,b,c,d$ to get this positivity, or getting something similar for another form of $q(x)$.

This is a comment to show the narrow range of polynomials that might work. We let $a=-f'(0),\ b=-f'(1),\ c=f''(0),\ d=-f''(1)$.

If $f$ is a polynomial, then $f''(x)$ is of the form $$f''(x)=(p-x)\left(\frac{c(1-x)}{p}+\frac{dx}{1-p}+x(1-x)q(x)\right)$$ for some polynomial $q(x)$ such that $f''(x)/(p-x)$ is positive on $[0,1]$. In particular, the factor of $p-x$ comes from the requirement that $f''(p)=0$, which in turn is the consequence of $f$ being convex on $[0,p]$ and concave on $[p,1]$.

Then there are only four more remaining conditions on $f''$ to solve the problem: \begin{align} -f'(0)=a&:\ \ \ \ \ \ -f'(p) + \int_0^p f''(x)\, dx = a\\ -f'(1)=b&:\ \ \ \ \ \ \ \ \ \ f'(p) + \int_p^1 f''(x)\, dx = -b\\ f(0)=f(p)&:\ \int_p^1 \left(f'(p) + \int_p^u f''(x)\, dx\right)\, du = 0\\ f(p)=f(1)&:\ \int_0^p \left(f'(p) - \int_u^p f''(x)\, dx\right)\, du = 0 \end{align}

So we could try $q(x)$ of the form $$q(x)=r x^i + s (1-x)^j + tx(1-x)$$ and solve the four equations above for $r,s,t$ and $f'(p)$.

The integrals are easy. Solving the equations is easy. Ensuring the positivity of $f''(x)/(p-x)$ is difficult. But if we can choose $i$ and $j$ to make $r,s,t$ positive, that will be enough. The trick is finding how $i$ and $j$ can depend on $a,b,c,d$ to get this positivity, or getting something similar for another form of $q(x)$.

Source Link
user44143
user44143

This is again too long for a comment:

Let $a=-f'(0),\ b=-f'(1),\ c=f''(0),\ d=-f''(1)$. Since $f$ is convex on $[0,p]$ and concave on $[p,1]$ we also know $f''(p)=0$.

So, if $f$ is a polynomial, then $f''(x)$ is of the form $$f''(x)=(p-x)\left(\frac{c(1-x)}{p}+\frac{dx}{1-p}+x(1-x)q(x)\right)$$ for some polynomial $q(x)$ such that $f''(x)/(p-x)$ is positive on $[0,1]$.

Then there are only four more remaining conditions on $f''$ to solve the problem:

$$-f'(p) + \int_0^p f''(x)\, dx = a$$

$$f'(p) + \int_p^1 f''(x)\, dx = -b$$

$$\int_p^1 \left(f'(p) + \int_p^u f''(x)\, dx\right)\, du = 0$$

$$\int_0^p \left(f'(p) - \int_u^p f''(x)\, dx\right)\, du = 0$$

So one approach is to try $q(x)$ of the form $$q(x)=r x^i + s (1-x)^j + tx(1-x)$$ and solve the four equations above for $r,s,t$ and $f'(p)$.

The integrals are easy. Solving the equations is easy. Ensuring the positivity of $f''(x)/(p-x)$ is difficult. But if we can choose $i$ and $j$ to make $r,s,t$ positive, that will be enough. The trick is finding how $i$ and $j$ can depend on $a,b,c,d$ to get this positivity, or getting something similar for another form of $q(x)$.