Let $f(t)=t^2$. Find the best approximation to $f(t)$ over $[0,1]$.

Consider the Banach space $(C[0,1],\|\cdot\|_\infty)$. Then the best constant approximation, in my understanding, should be $g(t)\equiv 1$, since


In $\left(L^1[0,1], d_1\right)$ we have


So the best approximation in this case is $g(t) \equiv \frac13$.

In $\left(L^2[0,1], d_2\right)$ we have


So the best approximation in this case is $g(t) =\frac{1}{\sqrt{5}}$.

Please let me know if my understanding is correct.

Also, it is noted that the result for $L^2[0,1]$ can be obtain in at least two ways. What could a second way possibly be? I'd appreciate a hint on this.

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    $\begingroup$ Are they not asking for a constant $c$ that minimizes $\| f(t) - c \|$? $\endgroup$ – user365239 Nov 26 '17 at 1:11
  • $\begingroup$ @user365239 Thanks for your comment. Would this approach be correct? $\inf\limits_{c\in[0,1]} \|t^2-c\|_\infty=\inf\limits_{c\in[0,1]} |1-c|=1$. $\endgroup$ – sequence Nov 26 '17 at 1:51
  • $\begingroup$ You need to find $c$ that actually attains $\inf\limits_{c\in[0,1]} \|t^2-c\|_\infty$ (that can be denoted $\arg\inf\limits_{c\in[0,1]} \|t^2-c\|_\infty$) -- yes, that's correct. But why did you replace it with $\inf\limits_{c\in[0,1]} |1-c|$? That is wrong. Note that $t^2$ ranges from $0$ to $1$, so depending on the values of both $t$ and $c$ the infimum (or minimum) is NOT necessarily attained at $t=1$. $\endgroup$ – zipirovich Nov 26 '17 at 2:18
  • $\begingroup$ @zipirovich So if $t^2$ stays unaffected, how can we come up with a $c$ that is a constant number and yet is the minimum distance from $t^2$? In my understanding, $c=0$. Is this correct. Also, can you please clarify what the notation of $arg$ means in this case? $\endgroup$ – sequence Nov 26 '17 at 2:39
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    $\begingroup$ $\arg$ in this case means the input value that yields the desired maximum or minimum. For example, $\max\limits_{t\in[0,2]}t^3=8$, the maximal possible output value, and $\arg\max\limits_{t\in[0,2]}t^3=2$, the input value of $t$ that yields that maximum. $\endgroup$ – zipirovich Nov 26 '17 at 3:31

In all three cases, you found the norm of the function $f(t)=t^2$ in the respective space instead of addressing the actual question. But the norm $\|f\|$ is not necessarily the best constant approximation to the function $f$. As was already pointed out in the comments, you need to minimize $\|f(t)-c\|$ in each case.

Let's consider $C[0,1]$ with the supremum norm $\|h\|_{\infty}=\sup\limits_{t\in[0,1]}|h(t)|$. You correctly found the norm of $f(t)=t^2$ as $\|f\|_{\infty}=1$, but that wasn't the question. What you need is to find $$\inf_{c\in\mathbb{R}}\|f-c\|_{\infty}=\inf_{c\in\mathbb{R}}\left(\sup_{t\in[0,1]}|f(t)-c|\right).$$

Let's look at some examples first. If $c=0$, then $\|f-c\|_{\infty}=\sup\limits_{t\in[0,1]}|t^2-0|=1$, attained at $t=1$. If $c=1$, then $\|f-c\|_{\infty}=\sup\limits_{t\in[0,1]}|t^2-1|=1$, attained at $t=0$. But both are far from the best (minimal) value.

Note that the values of $f(t)=t^2$ on $t\in[0,1]$ range from $0$ to $1$. Then the triangle inequality tells us that the norm $\|f-c\|_{\infty}$ can't be less than $1/2$. Indeed, if $\|f-c\|_{\infty}=\sup\limits_{t\in[0,1]}|f(t)-c|<\frac{1}{2}$, then $|f(t)-c|<\frac{1}{2}$ for all ${t\in[0,1]}$, and $|f(1)-f(0)|\le|f(1)-c|+|c-f(0)|<\frac{1}{2}+\frac{1}{2}=1$, contradicting the fact that $f(1)-f(0)=1^2-0^2=1$. So we can't do better than $1/2$. But if you think for a moment, you should realize that we can actually achieve $1/2$, i.e. we can find a constant $c$ such that $\|f-c\|_{\infty}=\sup\limits_{t\in[0,1]}|t^2-c|=\frac{1}{2}$. Roughly speaking, the idea is that in the supremum norm the best $c$ lies right in the middle of the range.

With the integral norms $L^1[0,1]$ or $L^2[0,1]$, the exact approach to finding the right $c$ will be different, but at least the underlying idea is the same: find a value of $c$ such that the calculated value of the norm $\|f-c\|$ will be minimal possible (compared to what we'd get with any other $c$).

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  • $\begingroup$ Thank you. But isn't it true that if $\sup\limits_{t\in [0,1]} |t^2-c|=\frac12$ then $c$ could take any value in the interval $[0,1]$? Say, if $t=\frac{1}{\sqrt2}$ then $c=0$. So there are infinitely many possible $c$. Thus we probably have the set $W=\left\{c\in [0,1]: \| t^2 - c \|_\infty =\frac12\right\}$. Moreover, while your argument proves that the norm cannot be less than $\frac12$, it doesn't prove that it cannot be less than $\frac{1}{1.9}$. $\endgroup$ – sequence Nov 26 '17 at 5:40
  • $\begingroup$ I'm afraid you have a serious logical misconception here. $c$ can NOT depend on $t$; in other words, you can NOT choose a value of $t$ and then choose a value of $c$ for it. Instead, once you pick a value of $c$, then with this $c$ you consider ALL values of $t$ from the interval $t\in[0,1]$, calculate $|t^2-c|$ for all of these $t$ (but still with the same $c$!), and then of all those output values find the biggest one. $\endgroup$ – zipirovich Nov 26 '17 at 5:47
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    $\begingroup$ "But isn't it true that if $\sup\limits_{t\in[0,1]}|t^2−c|=\frac{1}{2}$ then $c$ could take any value in the interval $[0,1]$?" <-- No, it isn't true: $\sup\limits_{t\in[0,1]}|t^2−c|=\frac{1}{2}$ is only true for certain $c$ (in fact, there's only one such $c$), but not for any $c$. Read my answer again, as two counterexamples are provided there: if $c=1$, then $\sup\limits_{t\in[0,1]}|t^2−c|=1\neq\frac{1}{2}$, and if $c=0$, then $\sup\limits_{t\in[0,1]}|t^2−c|=1\neq\frac{1}{2}$. $\endgroup$ – zipirovich Nov 26 '17 at 5:50

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