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Find all closest lines $p(x)=ax+b$ to $f(x)=x^2$ in the $C^1[0,1]$ norm. Note that the best approximation is not unique

Attempt : Let $r(x)=x^2-ax-b$.

Then $\|r(x)\|_{C^1}=\max\{|r^{(i)}(x)| : 0 \le i \le 1 \}$

$\|r(x)\|_\infty=r(0)=|b|$ or $r\left(\frac{a}{2}\right)=\left|\frac{a^2}{4}+b\right|$ or $r(1)=|a+b-1|$

and $\|r^{(1)}(x)\|_\infty=r(0)=|a|$ or $r(1)=|a-2|$

at this moment I do not know what should I do, because $\|r(x)\|_\infty $change its value as $a$,and $b$ change. But when I subsitute some value on $a$ and $b$, $\|r(x)\|_{C^1}$ is always less than 1. So I quess If I find $ax+b$ which satisfies $r(x)=1$, it can be all closest line to $x^2$. It just quess. please give me a hint. Thank you

mechanodroid
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fivestar
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1 Answers1

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Just to make sure, the $C^1[0,1]$ norm in question is $\|f\|_{C^1} = \max\{\|f\|_\infty, \|f'\|_\infty\}$.

You have obtained:

$$\|r\|_{C^1} = \max\left\{|a|, |b|, |a-2|, \left|\frac{a^2}4 - b\right|, |a+b-1|\right\}$$

Let's plug in $a = b = 1$.

$$\|r\|_{C^1} = \max\left\{|1|, |1|, |-1|, \left|-\frac{3}4\right|, |1|\right\} = 1$$

So, the minimum value of $\|r\|_{C^1}$ must be $\le 1$.

$$g(a,b)=\max\left\{|a|, |b|, |a-2|, \left|\frac{a^2}4 - b\right|, |a+b-1|\right\} \le 1$$

We have:

$$|a| \le 1 \implies a \in [-1,1]$$ $$|a-2| \le 1 \implies a-2 \in [-1,1] \implies a \in [1,3]$$

Hence, $a \in [-1,1] \cap [1,3] = \{1\}$ so necessarily $a = 1$. Plugging this into $\|r\|_{C^1}$ we get:

$$\|r\|_{C^1} = \max\left\{1, |b|, \left|\frac{1}4 - b\right| \right\}$$

This is obviously minimal when $\|r\|_{C^1} = 1$ so:

$$|b| \le 1 \implies b \in [-1,1]$$ $$\left|\frac{1}4 - b\right| \le 1 \implies b - \frac14 \in [-1,1] \implies b \in \left[-\frac34, \frac54\right]$$

So, $b \in [-1,1] \cap \left[-\frac34, \frac54\right] = \left[-\frac34, 1\right]$

Therefore, the minimal distance is $1$, and it is obtained for $p(x) = x + b$, with $b \in \left[-\frac34, 1\right]$.

This is the graph of $g(a,b)$ over the region $[-3,3]^2$:

enter image description here

Jack D'Aurizio
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mechanodroid
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  • Thank you for your kind explanation. But there is some part that I cannot understand . It is "This is obviously minimal when ... so: ...". could you explain more detail...? – fivestar Nov 18 '17 at 18:09
  • @fivestar At that point we had $$|r|{C^1} = \max\left{1, |b|, \left|\frac{1}4 - b\right| \right}$$ so $|r|{C^1} \ge 1$. In particular, $\min_{a,b} |r|{C^1} \ge 1$. Furthermore, for any $b$ with $|b| \le 1$ and $\left|\frac{1}4 - b\right| \le 1$, such as $b = 0$, we have $|r|{C^1} = 1$, so $\min_{a,b} |r|{C^1} \le 1$. We conclude that $\min{a,b} |r|_{C^1} = 1$. – mechanodroid Nov 18 '17 at 18:56
  • ahh . I understand perfectly. Thank you every time. I will remember your efforts. – fivestar Nov 18 '17 at 18:59