Determine if two norms are equivalent.

Question

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?

Answer 1

Eventually I came up with my own answer to the question, which is hopefully correct. Recall that: $$ f(x)-f(0) = \int_0^xf'(t)dt $$ So we start by proving: $$|f(0)| = \bigg|f(x) + \int_x^0 f'(t)dt\bigg| \le |f(x)|+\bigg|\int_0^x f'(t)dt\bigg| \le |f(x)|+\int_0^x |f'(t)|dt$$ We may extend the rightmost integral all the way up to 1 and since the integrand is a positive function, the integral is nondecreasing in x. $$|f(0)| \le |f(x)|+\int_0^1 |f'(t)|dt$$ Now integrating the expression and using monotonicity, leaving the real numbers the same since we're integrating on an interval of unitary measure: $$|f(0)| \le \int_0^1|f(x)|dx +\int_0^1 |f'(x)|dx$$ In general, by Hölder's inequality: $$||f||_{L^1([0, 1])} \le ||f||_{L^2([0, 1])}\\ \implies \int_0^1|f(x)|dx+\int_0^1 |f'(x)|dx \le \Big(\int_0^1|f(x)|^2dx\Big)^{1/2} +\Big(\int_0^1 |f'(x)|^2dx\Big)^{1/2} $$ And this implies finally that: $$ |f(0)| \le \Big(\int_0^1|f(x)|^2dx\Big)^{1/2} +\Big(\int_0^1 |f'(x)|^2dx\Big)^{1/2}\\ \implies N_2 \le 2N_1 $$

For the other direction: $$ (a+b)^2 \le 2a^2 + 2b^2 \\ \implies |f(x)|^2 \le \bigg|2(f(0))^2 + 2\Big(\int_0^x f'(t)dt\Big)^2\bigg| \le 2\Big(|f(0)|^2 + \bigg|\int_0^x f'(t)dt\bigg|^2) \le 2\Big(|f(0)|^2 + \big(\int_0^x |f'(t)|dt\big)^2\Big) $$ So we're using the same tricks as before (most of the absolute values are redundant, but I'll keep them). Then we use an inequality for square roots that holds for positive $a, b$: $$ (a+b)^{1/2} \le a^{1/2} + b^{1/2} $$ So we will integrate the above over the interval after extending the integral over to 1, then apply the square root inequality: $$\int_0^1|f(x)|^2dx \le 2\Big(|f(0)|^2 + \big(\int_0^1 |f'(t)|dt\big)^2\Big) \\ \implies \Big(\int_0^1|f(x)|^2dx\Big)^{1/2} \le \sqrt{2}|f(0)| + \sqrt{2}\int_0^1 |f'(x)|dx \le \sqrt{2}|f(0)| + \sqrt{2}\Big(\int_0^1 |f'(x)|^2dx\Big)^{1/2}$$ So we determine that: $$N_1 \le (\sqrt{2} + 1)N_2$$

Answer 2

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.

1. 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 $$f(x)^2\leqslant 2 \int_0^1\left\lvert f'(t)\right\rvert ^2dt +f(0)^2 .$$

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.