These two norms of $C^k[a,b]$ are equivalent.

Question

In $C^k[a,b]$ we can define $$ \|f\|_* := \|f\|_\infty + \left\|f^{(k)}\right\|_\infty $$

and $$ \|f\|_{**} := \sum_{j=0}^k \left\|f^{(j)}\right\|_\infty$$

It is obvious that $\|f\|_*\leq \|f\|_{**}$. However, how can I show that $K\|f\|_*\geq \|f\|_{**}$ for $K>0$? Any hint?

Answer 1

First, let's consider $[a,b]=[0,1]$. From Using second derivative to find a bound for the first derivative, we know that

If $f:[0,1]\to\mathbb R$ satisfies $f(0)=f(1)=0$ and $\|f''\|_\infty \le 1$, then $\|f'\|_\infty \le \frac12. \tag{A1}\label{firstineq}$

We now remove the assumptions on $f$. If $\|f''\|_\infty=0$, then it is easy to see that $\|f'-f(1)+f(0)\|_\infty = \frac12 \|f''\|_\infty$. Otherwise, $\|f''\|_\infty>0$, and by using $\tilde f = \frac{f - f(0)-(f(1)-f(0))x}{\|f''\|_\infty}$, we have $$ \|f'-f(1)+f(0)\|_\infty \le \frac12 \|f''\|_\infty,$$ so that $ \|f'\|_\infty \le 2\|f\|_\infty + \frac12\|f''\|_\infty$, and so $$ \|f\|_{**} = \|f\|_* + \|f'\|_\infty \le 3 \|f\|_*.$$

Lets record the more general version of \ref{firstineq} above.

If $f:[0,1]\to\mathbb R$, then $ \|f'\|_\infty \le 2\|f\|_\infty + \frac12 \|f''\|_\infty \label{2nd}\tag{A2}$

The constants $2,1/2$ are not necessarily the most useful. Indeed, trying an induction for 3 derivatives leaves us at $\|f''\|\le2\|f'\|+\frac12\|f'''\|\le 4\|f'\| + {\color{red}{2 \times \frac12}}\|f''\| + \frac12 \|f'''\|$; if only $2\times\frac12<1$, this would be a proof for $k=3$. Luckily, the constant can change if we change the domain of $f$. Intuitively, if the domain is bigger, then knowing that $f''$ is small doesn't give such a tight bound on $f'$. So lets consider a function on $[a,b]$. As the supremum norm is translation invariant, we can still take $a=0$. With $f:[0,b]\to\mathbb R$, define $$ \tilde f:[0,1] \to \mathbb R, \tilde f(x) = f(bx).$$ Applying \ref{2nd} gives $$ \|\tilde f'\|_\infty \le 2\|\tilde f\|_\infty + \frac12 \|\tilde f''\|_\infty $$ By chain rule, $\|\tilde f^{(n)}\|_\infty = b^n\|f^{(n)}\|_{\infty;b}$, where $\|f\|_{\infty;b} :=\sup_{x\in[0,b]} |f(x)|$, giving

If $f:[0,b]\to\mathbb R$, then $$ \| f'\|_{\infty;b} \le \frac2b \| f\|_{\infty;b} + \frac{b}{2} \| f''\|_{\infty;b} \label{3rd}\tag{A3}$$

Finally, note that for $f:[0,b]\to\mathbb R$, we can use the bound for functions on $[0,\delta b]\to\mathbb R$, $0<\delta\le 1$ to change the value of the constant further, since $$\sup_{x\in[0,b]} = \sup_{t\in[0,b]}\sup_{x\in[t,t+\delta b]\cap [0,b]}$$ which gives $$ \|f'\|_{\infty;b} = \sup_{t\in[0,b]}\| f'(\cdot+t)\|_{\infty,\delta b} \le \frac2{\delta b} \| f\|_{\infty;b} + \frac{\delta b}{2} \| f''\|_{\infty;b}. $$ This is the version that allows induction to work:

If $f:[0,b]\to\mathbb R$, then for any $\delta \in (0,1]$, $$ \|f'\|_{\infty;b} \le \frac2{\delta b} \| f\|_{\infty;b} + \frac{\delta b}{2} \| f''\|_{\infty;b}. \tag{A4}\label{final}$$

Let me illustrate the $k=3$ case: first, you remove $\|f'\|_{\infty;b}$ by $$ \|f'\|_{\infty;b} \le \frac2b\|f\|_{\infty;b}+\frac{b}2\|f''\|_{\infty;b} \\ $$ Then for $\|f''\|_{\infty;b}$, $$\|f''\|_{\infty;b} \le \frac2b \|f'\|_{\infty;b} + \frac{b}2 \|f'''\|_{\infty;b} \le \frac4{\delta b^2} \|f\|_{\infty;b} + \delta \|f''\|_{\infty;b} + \frac{b}2 \|f'''\|_{\infty;b} $$ By using $\delta=\frac12$, we obtain a bound $$\|f'\|_{\infty;b} + \|f''\|_{\infty;b} \le C \|f\|_*,$$ where $C=C(b)$. In general, the induction statement to use is

$\mathfrak P(n)$: For any $0\le k< n$, there exists $C=C(k,n,b)$ such that for any $\beta\in(0,b]$, and any $f:[0,b]\to\mathbb R$, $$ \|f^{(k)}\|_{\infty;b} \le C\beta^{-k} \|f\|_{\infty;b} + C \beta^{n-k} \|f^{(n)}\|_{\infty;b}.$$

I took this induction proof and the above main idea from Giovanni Leoni's great book "A First Course In Sobolev Spaces"(2nd edition pg. 202, section on interpolation inequalities.) There, it is proven in greater generality and I highly encourage reading it there.

The base cases $n=2,3$ are above (and $k=0,n$ is not hard). Suppose $\mathfrak P(n)$ is true, and we hope to prove $\mathfrak P(n+1)$. Then for $2\le k\le n$, the induction hypothesis gives for $\delta\in(0,1]$, $$ \|f' \|_{\infty;b} \le C(\delta \beta)^{-1} \|f\|_{\infty;b} + C (\delta \beta)^{k-1} \|f^{(k)}\|_{\infty;b},$$ and applying the induction hypothesis to $f'$, $$ \|f^{(k)} \|_{\infty;b} \le C\beta^{-(k-1)} \|f'\|_{\infty;b} + C \beta^{n-(k-1)} \|f^{(n+1)}\|_{\infty;b}.$$ Combining gives $$ \|f^{(k)} \|_{\infty;b} \le C\delta^{-1} \beta^{-k} \|f\|_{\infty;b} + C\delta ^{k-1} \|f^{(k)}\|_{\infty;b} + C \beta^{n+1-k} \|f^{(n+1)}\|_{\infty;b}.$$ This gives the result once we take $\delta$ small enough that $C\delta ^{k-1} < 1$.

We are left with $k=1$. In this case, estimate $f'$ by $$ \|f' \|_{\infty;b} \le C(\delta \beta)^{-1} \|f\|_{\infty;b} + C (\delta \beta)^{n-2} \|f^{(n-1)}\|_{\infty;b},$$ and then estimate $f^{(n-1)}$ by $$ \|f^{(n-1)} \|_{\infty;b} \le C\beta^{-(n-2)} \|f'\|_{\infty;b} + C \beta^{n-(n-2)} \|f^{(n+1)}\|_{\infty;b}.$$ This gives $$ \|f' \|_{\infty;b} \le C(\delta\beta)^{-1} \|f\|_{\infty;b} + C\delta^{n-2} \|f^{'}\|_{\infty;b} + C \delta^{n-2}\beta^{n} \|f^{(n+1)}\|_{\infty;b}.$$ By choosing $C\delta^{n-2}<1$, the proof is finished.

Answer 2

Finally I found and answer:

Firstly, it is obvious that $||f||_* \leq ||f||_{**}$.

In the other hand, it is sufficient to prove it for $C^k[0,1]$. We can write \begin{equation} f(x) = f(c) + \underbrace{f^{(1}(x)(x - c) + \cdots + \frac{f^{(k-1}(c)}{(k-1)!}(x-c)^{k-1}}_{P_{k,c}(x)} + \int_{(c,x)} f^k(t)\frac{(x-t)^{k-1}}{(k-1)!}\text{d} t \end{equation}

for every $c,x\in[0,1]$ (note that, because of that, this bound holds for $(x-c)\in [-1,2]$). Taking norms, $$ |P_{k,c}(x)| \leq 2 ||f||_\infty + ||f^{(n}||_\infty \leq 2||f||_* $$

Note: If you take $[a,b]$ instead of $[0,1]$ you will have $|P_{n,c}(x)| \leq 2 ||f||_\infty + k(a,b,n)||f^{(n}||_\infty$ with $k(a,b,n)$ a constant.

Lemma: Let $P(t)\in \mathcal{P}^k(I) = \{p(t) = \sum_{j=0}^l c_j t^j: l\leq k, a_j\in \mathbb{R}, t \in I\}$ for I a closed interval. If $|P(t)| < C$ for every $t\in I$, then its coefficients are also bounded.

Proof: the norms $$ ||p||_1 = \max_{0\leq j\leq k}\{c_j\}, \quad ||p||_2 = \sup_{t\in I}|p(t)| $$ are equivalent because $\mathcal{P}^k(I)$ is a normed and finite dimensional space.$\square$

In our case, $P_{k,c}\in \mathcal{P}^k([-c,1-c])$. Thus if we bound it, the constant will depend on $c$. To avoid this situation, we can suppose either $c>1/2$ or $c<1/2$. Without lost of generality, suppose $c>1/2$. Then, $(x-c)\in J = [-1,1/2]$.

Hence, since $||P_{k,c}||_2 \leq 2||f||_*$, then, there exists an $\alpha > 0$ such that $||P_{k,c}||_1 \leq 2\alpha||f||_*$, and so $||f^{(j}||_\infty \leq 2\alpha||f||_*$ for all $1\leq j\leq k$. There are not any problems related to the difference

Finally, $||f||_{**}\leq (2\alpha(k-1)+1)||f||_{*}$.