In the space of functions $\mathcal{C}^1([0, 1])$ we define two norms:
$$N_1 = \bigg( \int_0^1 |f(x)|^2dx \bigg)^{1/2} + \bigg( \int_0^1 |f'(x)|^2dx \bigg)^{1/2} \\ N_2 = |f(0)| + \bigg( \int_0^1 |f'(x)|^2dx \bigg)^{1/2}$$
Determine if the two norms are equivalent.
This is part of an assignment. The answer is "yes", but I don't see it: the main reason being that while the second half of the norms are the same, I believe I can vary the first half arbitrarily with an appropriate choice of functions – thus there should be no way for me to determine appropriate constants such that:
$$ C_1 N_1 \le N_2 \le C_2 N_1 $$
I've tried with a family of functions $f(x)=e^{ax}, a\in \Bbb{R}$. The two norms yield:
$$||f||_1 = \sqrt{\frac{e^{2a}-1}{2a}} + \sqrt{\frac{a(e^{2a}-1)}{2}} \\
||f||_2 = 1 + \sqrt{\frac{a(e^{2a}-1)}{2}}$$
It looks like I can make the first term arbitrarily big or small with the choice of $a$, thus invalidating any constant. Am I missing something?
Best Answer
In your example with $f(x)=e^{ax}, a\in \Bbb{R}$, the terms are equivalent as $a\to 0$ and as $a\to \infty$ hence this does not give a counter-example.
Here are some hints to prove the equivalence of the norms.
Notice that $$f(x)^2\leqslant 2(f(x)-f(0))^2+f(0)^2 $$ and by the fundamental theorem of calculus, $f(x)-f(0) =\int_0^xf'(t)dt$ hence $$f(x)^2\leqslant 2\left(\int_0^x\left\lvert f'(t)\right\rvert dt\right)^2+f(0)^2\leqslant 2\left(\int_0^1\left\lvert f'(t)\right\rvert dt\right)^2+f(0)^2.$$
Using Cauchy-Schwarz inequality, we get $$f(x)^2\leqslant 2 \int_0^1\left\lvert f'(t)\right\rvert ^2dt +f(0)^2 .$$
For the opposite direction, an other use of the fundamental theorem of calculus shows that for all $x\in[0,1]$, $$ f(x)^2=2\int_0^x f'(t)f(t)dt+f(0)^2.$$ Therefore, $$ f(0)^2\leqslant f(x)^2+2\int_0^x \left\lvert f'(t)f(t)\right\rvert dt\leqslant f(x)^2+2\int_0^1 \left\lvert f'(t)f(t)\right\rvert dt $$ Integrate this equality with respect to $x$ on $[0,1]$ and use the fact that $2\left\lvert f'(t)f(t)\right\rvert\leqslant f'(t)^2+f(t)^2 $ to conclude.