Prove the equivalence of these norms

I found the following exercise (Smirnov, Linear Analysis page 43 ex. 12): Show that the following "norms" in the vector space $\mathscr C^1 ([a,b]) $ are in fact norms and that they are equivalent. $$p_1 (f) = |f(a)| + \int_a^b |f'(t)|\ dt $$ $$p_2(f) = \int_a^b |f(t)| \ dt \ + \int_a^b |f'(t)|\ dt $$

I showed that they were norms without great trouble, but the equivalence part took me a long time. I showed that there are constants $C_1, C_2 >0$ such that: $$C_1 p_1 (f) \le p_2(f) \le C_2 p_1 (f)$$ The constants I used were: $$C_1 = \min(\frac{1}{2}, |1-\frac{b-a}{2}|) $$ $$C_2 = b-a +1$$

I think they are valid but the proof that they are, and finding them in the first place, took a lot more time than expected, especially $C_1$, and they are not strict inequalities at all. I will show how I found $C_2$:

$ \int_a^b |f(t)| \ dt \le M(b-a)$, where $M$ is the maximum value of the function in that interval, and $\int_a^b |f'(t)|\ dt \ge M-|f(a)|$. It's interesting to note that both equalities can be simultaneous only for constant functions.

Then, subtracting the integral of the derivative on both sides, we have: $$p_2(f)- \int_a^b |f'(t)|\ dt \le M(b-a) \ \ \ \ \text {and}\ \\ \ C_2 |f(a)| + (C_2-1) (M-|f(a)|) \le C_2 p_1 (f) - \int_a^b |f'(t)|\ dt$$ Now it's enough to find a $C_2$ such that: $$M(b-a) \le (C_2-1)M + |f(a)| $$ and $C_2 = b-a +1$ works perfectly.

The proof for $C_2$ is simple enough, but just the verification that $C_1$ works takes a lot of work, and my process for getting there seemed a bit too arbitrary. So, my questions are: Is my value for $C_2$ the minimum possible, and how to get to a possible value of $C_1$ (not necessarily the best one) without having to spend five minutes on algebric manipulations just to check it?

Solution 1:

Your $C_2$ is optimal. Let $$f(t) = \begin{cases} n \cdot (t - a),\ t \in \left[a, a + \frac{1}{n} - \frac{1}{10\cdot n^2}\right] \\ 1,\ t > a + \frac{1}{n}\\ \text{something smooth with derivative at most 1},\ t \in \left(a + \frac{1}{n} - \frac{1}{10\cdot n^2}, a + \frac{1}{n}\right] \end{cases}$$

Then we have $p_1(f) \approx 1$ and $p_2(f) \approx b - a + 1$, which means we can't decrease $C_2$. We can't actually achieve equality with such $C_2$ (except for $f(t) = 0$), because, as you noted, both inequalities you used are non-strict only if $f$ is constant, but that would mean $p_2(t) = (b - a) \cdot p_1(t)$, without extra $1$.

For simplicity, assume $f(a) = 1$ (if $f(a) \neq 0$ replace $f(t)$ with $\frac{f(t)}{f(a)}$, and if $f(a) = 0$ then we have our inequality with $C_1 = 1$).

If $f(t) > \frac{1}{2}$ for all $t \in [a, b]$, we have our inequality with $C_1 = \min\left(\frac{1}{2}, \frac{b - a}{2}\right)$.

Otherwise, consider point $c > a$ s.t. $f(c) = \frac{1}{2}$ and $f(t) > \frac{1}{2}$ if $t \in [a, c]$.

We have $\int_a^c |f'(t)|\ dt \leq \left|\int_a^c f'(t) dt\right| = \frac{1}{2}$, so $3\int_a^b |f'(t)|\ dt \geq \int_a^b |f'(t)|\ dt + 1 = p_1(f)$, so we have our inequality with $C_1 = \frac{1}{3}$.