For each $f\in C[0,1]$ set $$\|f\|_1 = \left(\int_0^1 |f(x)|^2 dx\right)^{1/2},\quad\quad \|f\|_2 = \left(\int_0^1 (1+x)|f(x)|^2 dx\right)^{1/2}$$

Then prove that $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent norms on $C[0,1]$.

So we want to show that for positive real numbers $a,b$ that $$a\|f\|_2 \leq \|f\|_1 \leq b\|f\|_2$$

Since $\|f\|_2^2 = \int_0^1 |f(x)|^2 dx + \int_0^1 x|f(x)|^2 dx= \|f\|_1^2+ \int_0^1 x|f(x)|^2 dx$ and because we know that $\int_0^1 x|f(x)|^2 dx\geq 0$ we have that:

$$\|f\|_1 \leq 1\times \|f\|_2$$

Now we want to find some $a\gt 0$ such that:

$$a\cdot\|f\|_2\leq \|f\|_1 $$

This I am not sure how to do.


In this case $1+x \leq 2$ so $\| f \|_2 \leq \sqrt{2} \| f \| _1$.

  • $\begingroup$ Does $|f(x)|^2$ get treated as a real number or something? If not wouldn't we be concerned about how $x$ effects $f(x)$ and hence integration by parts? $\endgroup$ – Functional Analysis Sep 7 '15 at 8:19
  • $\begingroup$ $\|f\|_2^2 = \int_0^1 (1+x)|f(x)|^2\mathrm{d}x \leq \int_0^1 2|f(x)|^2\mathrm{d}x=2\|f\|_1^2$ Since we have $1+x \leq 2$ for all $x \in (0,1)$. $\endgroup$ – Nigel Overmars Sep 7 '15 at 8:24
  • $\begingroup$ @NigelOvermars What does $|f(x)|$ actually mean? $\endgroup$ – Functional Analysis Sep 7 '15 at 8:38
  • $\begingroup$ $|f(x)| = f(x)$ if $f(x) \geq 0$ and $|f(x)| = -f(x)$ if $f(x) < 0$. I.e., the distance between $f(x)$ and $0$. $\endgroup$ – Nigel Overmars Sep 7 '15 at 8:53
  • $\begingroup$ @NigelOvermars Thanks. I think I get it now. Any $g(x)$ out the front, if taken to the maximum value it could attain on the interval will contribute strictly more than it would if you let it range over the interval. $\endgroup$ – Functional Analysis Sep 7 '15 at 8:58

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