Let $f(x)=x^2-x$, $x\in[-1,1]$. Find $\lVert f\rVert_2$.

Some other norms on $C[a,b]$:
$\lVert f\rVert_1=\int_a^b \lvert f(x)\rvert~\mathrm{d}x$.
$\lVert f\rVert_2=\left(\int_a^b \lvert f(x)\rvert^2~\mathrm{d}x\right)^{1/2}$.

I know the answer to this problem is $\sqrt{\frac{2}{3}}$, but am unable to solve it; the only way I get the answer is if I ignore the square. Can I get some help please?

  • 3
    $\begingroup$ Hint: $(x^2-x)^2=x^4-2x^3+x^2$. $\endgroup$
    – Lorago
    Commented Jan 1, 2023 at 12:59
  • 4
    $\begingroup$ Since you are only interested in $\|f\|_2$, what's the point of defining $\|f\|_1$? $\endgroup$ Commented Jan 1, 2023 at 13:12
  • $\begingroup$ How positive are you that the correct solution is $\sqrt{\frac{2}{3}}$? $\endgroup$ Commented Jan 1, 2023 at 13:20
  • $\begingroup$ 100% correct, that’s the answer given. $\endgroup$ Commented Jan 1, 2023 at 19:26
  • $\begingroup$ @Lorago already done this, it was first method does not provide the answer $\endgroup$ Commented Jan 1, 2023 at 19:28

1 Answer 1


Following your definition of $\left\Vert f\right\Vert_2$ one can simply insert $f\left(x\right)=x^2-x$ into the formula and integrate the sum term by term:

$$ \begin{aligned} \left\Vert f \right\Vert_2 &= \sqrt{\int_{-1}^1\left\vert x^2 - x\right\vert^2 dx}\\ &= \sqrt{\int_{-1}^1x^4-2x^3+x^2dx}\\ &= \sqrt{\left. \frac{1}{5}x^5 - \frac{1}{2}x^4 + \frac{1}{3}x^3 \right\rvert_{-1}^1}\\ &= \sqrt{\frac{1}{5}-\frac{1}{2}+\frac{1}{3}-\left(-\frac{1}{5}-\frac{1}{2}-\frac{1}{3}\right)}\\ &= \sqrt{\frac{2}{5}+\frac{2}{3}} = \sqrt{\frac{16}{15}} = \frac{4}{\sqrt{15}} \end{aligned} $$

  • $\begingroup$ answer as per text is sqrt(2/3) $\endgroup$ Commented Jan 2, 2023 at 0:26
  • 1
    $\begingroup$ Maybe you incorrectly transcribed the original definition in your question? $\endgroup$
    – Nerrit
    Commented Jan 2, 2023 at 0:47
  • $\begingroup$ No, I did not. But I feel the answer provided is incorrect, I have tried various options but by no means it is sqrt(2/3). for an abs function it is only positive in the region of -1<=x<0 for the function provided. The problem is they did not provide the solution in details. $\endgroup$ Commented Jan 2, 2023 at 2:14

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