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#1 h Jan 11, 2010, 8:13 PM • 2 Y

Y by Adventure10, Mango247

Find all differentiable functions $f;\mathbb{R}\to\mathbb{R}$ such that, for all real numbers $a,b,t$ with $0<t<1$ , $t^2f(a)+(1-t^2)f(b)\geq f(ta+(1-t)b)$

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#2 h Mar 18, 2025, 1:36 PM

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Let us write the above function equation according to:

$t^2f(a)+(1-t^2)f(b) = f(ta+(1-t)b) + g(a+b)$ (i)

for some differentiable function $g:\mathbb{R} \rightarrow [0,\infty)$ . Differentiating (i) with respect to $a$ and $b$ respectively yields:

$t^2f'(a) = tf'(ta+(1-t)b) + g'(a+b)$ (ii);

$(1-t^2)f'(b) = (1-t)f'(ta+(1-t)b) + g'(a+b)$ (iii)

and subtracting (iii) from (ii) produces:

$t^2f'(a) - (1-t^2)f'(b) = (2t-1)f'(ta + (1-t)b)$ (iv).

Differentiating (iv) with respect to $a$ and $b$ respectively yields:

$t^2f''(a) = t(2t-1)f''(ta+(1-t)b)$ (v);

$-(1-t^2)f''(b) = (1-t)(2t-1)f''(ta+(1-t)b)$ (vi)

and equating (v) with (vi) produces:

$tf''(a) = -(1+t)f''(b) \Rightarrow f''(a) = A \Rightarrow f(a) = Aa^2+Ba + C$ (vii)

for arbitrary $A,B,C \in \mathbb{R}$ . Substitution of (vii) into (i) next gives us:

$t^2(Aa^2+Ba+C) + (1-t^2)(Ab^2+Bb + C) = A[ta+(1-t)b]^2+B[ta+(1-t)b]+C + g(a+b)$ ;

or $Bat^2 -Bbt^2 = Bat-Bbt+2Aabt(1-t) + g(a+b)$ ;

or $g(a+b) = [B(a+b)+2b(Aa-B)](t^2-t)$ (viii).

Given that $t \in (0, 1)$ , we require $B(a+b)+2b(Aa-B) \le 0$ for all $a,b \in \mathbb{R} \Rightarrow A,B = 0$ . Therefore, $\textcolor{red}{f(x)=C}$ is the desired solution.

This post has been edited 9 times. Last edited by Mathzeus1024, Mar 18, 2025, 4:14 PM

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#3 h Yesterday at 12:47 PM

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@ #2: It is not clear why $t^2f(a)+(1-t^2)f(b) - f(ta+(1-t)b)$ should depend only on $a+b$ , as you claimed. Also, it seems that you assume that $f$ is twice differentiable, which is not given.
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