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DSM
DSM
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Too long to comment.

Here is a suggestion. Consider transforming the question to finding a $C^{\infty}$ nonan analytic non-positive convex function $f$ over [0,1] with (i) zeros at 0 and 1, (ii) the required first and second order derivatives and (iii unimodal at $p$. In that case, one can then multiply by $g(x) = \frac{2}{1+e^{K(x-p)}}-1$, $K$ sufficient large and positive, and see that for $h(x)=f(x)g(x)$, $h(p)=0$ and $$ h'' = f''g + 2f'g' + g''f $$ is non-negative (and hence conves) in $[0,p]$ and non-positive in $[p,1]$. Reason for signs of curvature: (i) for $x \in [0,p]$, $f''g$ is non-negative since $f''$ and $g$ are positive, $f'g'$ is non-negative since $f$ reaches minimum at $p$ and $g'$ is non-positive, and $g''f$ is non-negative since $g''$ is non-positive and $f$ is non-positive. The same can be reasoned out for concavity in $[p,1]$.

Too long to comment.

Here is a suggestion. Consider transforming the question to finding a $C^{\infty}$ non-positive convex function $f$ over [0,1] with (i) zeros at 0 and 1, (ii) the required first and second order derivatives and (iii unimodal at $p$. In that case, one can then multiply by $g(x) = \frac{2}{1+e^{K(x-p)}}-1$, $K$ sufficient large and positive, and see that for $h(x)=f(x)g(x)$, $h(p)=0$ and $$ h'' = f''g + 2f'g' + g''f $$ is non-negative (and hence conves) in $[0,p]$ and non-positive in $[p,1]$. Reason for signs of curvature: (i) for $x \in [0,p]$, $f''g$ is non-negative since $f''$ and $g$ are positive, $f'g'$ is non-negative since $f$ reaches minimum at $p$ and $g'$ is non-positive, and $g''f$ is non-negative since $g''$ is non-positive and $f$ is non-positive. The same can be reasoned out for concavity in $[p,1]$.

Too long to comment.

Here is a suggestion. Consider transforming the question to finding an analytic non-positive convex function $f$ over [0,1] with (i) zeros at 0 and 1, (ii) the required first and second order derivatives and (iii unimodal at $p$. In that case, one can then multiply by $g(x) = \frac{2}{1+e^{K(x-p)}}-1$, $K$ sufficient large and positive, and see that for $h(x)=f(x)g(x)$, $h(p)=0$ and $$ h'' = f''g + 2f'g' + g''f $$ is non-negative (and hence conves) in $[0,p]$ and non-positive in $[p,1]$. Reason for signs of curvature: (i) for $x \in [0,p]$, $f''g$ is non-negative since $f''$ and $g$ are positive, $f'g'$ is non-negative since $f$ reaches minimum at $p$ and $g'$ is non-positive, and $g''f$ is non-negative since $g''$ is non-positive and $f$ is non-positive. The same can be reasoned out for concavity in $[p,1]$.

Source Link
DSM
DSM
  • 1.2k
  • 7
  • 12

Too long to comment.

Here is a suggestion. Consider transforming the question to finding a $C^{\infty}$ non-positive convex function $f$ over [0,1] with (i) zeros at 0 and 1, (ii) the required first and second order derivatives and (iii unimodal at $p$. In that case, one can then multiply by $g(x) = \frac{2}{1+e^{K(x-p)}}-1$, $K$ sufficient large and positive, and see that for $h(x)=f(x)g(x)$, $h(p)=0$ and $$ h'' = f''g + 2f'g' + g''f $$ is non-negative (and hence conves) in $[0,p]$ and non-positive in $[p,1]$. Reason for signs of curvature: (i) for $x \in [0,p]$, $f''g$ is non-negative since $f''$ and $g$ are positive, $f'g'$ is non-negative since $f$ reaches minimum at $p$ and $g'$ is non-positive, and $g''f$ is non-negative since $g''$ is non-positive and $f$ is non-positive. The same can be reasoned out for concavity in $[p,1]$.