MHB Find max(abs(a)+abs(b)+abs(c)), and a possible f(x)

[Albert1]

$given\,\,f(x)=ax^2+bx+c$
$if \,\,for\,\,all\,\,x \in [0,1]\,\, we\,\, have\left | f(x) \right |\leq1$
$find \,\, max(\left | a \right |+\left | b \right |+\left | c\right |)$
$and \,\, a\,\, possible \,\, f(x)$

 

[juantheron]

Given $f(x) = ax^2+bx+c$ and $0 \leq x\leq 1$ and $|f(x)|\leq 1$Put $x=0$ in $f(x)\;,$ We get $f(0) = c$ and put $x=1\;,$ We get $f(1) = a+b+c$and put $\displaystyle x=\frac{1}{2}\;,$ We get $\displaystyle f\left(\frac{1}{2}\right) = \frac{a}{4}+\frac{b}{2}+c\Rightarrow 4f(1) = a+2b+4c$So Using $\triangle$ Inequality::$\displaystyle |a| = \mid 2f(1)-4f\left(\frac{1}{2}\right)+2f(0)\mid \leq 2|f(1)|+4\mid f\left(\frac{1}{2}\right)\mid+2|f(0)|\leq 8$$\displaystyle |b| = \mid-f(1)-4f\left(\frac{1}{2}\right)-3f(0)\mid \leq |f(1)|+4\mid f\left(\frac{1}{2}\right)\mid+3|f(0)|\leq 8$

$|c|\leq 1$So we get $|a|+|b|+|c| \leq 8+8+1 \leq 17$

 

[Albert1]

Given $f(x) = ax^2+bx+c$ and $0 \leq x\leq 1$ and $|f(x)|\leq 1$Put $x=0$ in $f(x)\;,$ We get $f(0) = c$ and put $x=1\;,$ We get $f(1) = a+b+c$and put $\displaystyle x=\frac{1}{2}\;,$ We get $\displaystyle f\left(\frac{1}{2}\right) = \frac{a}{4}+\frac{b}{2}+c\Rightarrow 4f(1) = a+2b+4c$So Using $\triangle$ Inequality::$\displaystyle |a| = \mid 2f(1)-4f\left(\frac{1}{2}\right)+2f(0)\mid \leq 2|f(1)|+4\mid f\left(\frac{1}{2}\right)\mid+2|f(0)|\leq 8$$\displaystyle |b| = \mid-f(1)-4f\left(\frac{1}{2}\right)-3f(0)\mid \leq |f(1)|+4\mid f\left(\frac{1}{2}\right)\mid+3|f(0)|\leq 8$

$|c|\leq 1$So we get $|a|+|b|+|c| \leq 8+8+1 \leq 17$

Spoiler

one example of $f(x)$ : $a=8,b=-8,c=-1,\,\, we \,\, get \,\, f(x)=8x^2-8x-1$
$which \,\,max(|a|+|b|+|c|) =8+8+1 =17\,\,meets$
also for all $0 \leq x\leq 1 ,$ $|f(x)|\leq 1$

 

[Albert1]

Spoiler

one example of $f(x)$ : $a=8,b=-8,c=-1,\,\, we \,\, get \,\, f(x)=8x^2-8x-1---(*)$
$which \,\,max(|a|+|b|+|c|) =8+8+1 =17\,\,meets$
also for all $0 \leq x\leq 1 ,$ $|f(x)|\leq 1$

a typo in $(*):$

Spoiler

$a=8,b=-8,c=1$
$f(x)=8x^2-8x+1$
$which \,\,max(|a|+|b|+|c|) =8+8+1 =17\,\,meets$
also for all $0 \leq x\leq 1 ,$ $|f(x)|\leq 1$