Let $K =[0,1]$ and let $X \subset C^{\infty}(K)$ be the subspace of all functions vanishing on the end points of $K$. Show that the following seminorms are equivalent:
- $||D^nf||_1$
- $||D^nf||_2$
- $\sup\{|D^nf(x)|:x\in K\}$
Attempt:
Suppose $||D^\alpha f||_2 <\frac{1}{n}$ for all $\alpha \le n$.
We have that $||D^\alpha f||_1 \le ||D^\alpha f||_2||D^\alpha f||_2 < \frac{1}{n^2} < \frac{1}{n}$. And further:
$$\sup \{|D^{\alpha-1}f(x)|:x\in K\} \le \min\{|D^{\alpha-1}f(x)|\}+ V_0^1(D^{\alpha-1}f) < $$
$$<\frac{1}{(n-1)^2} + ||D^{\alpha} f||_1 <\frac{1}{n^2}+ \frac{1}{(n-1)^2}< \frac{1}{n-1}$$
For all $\alpha > 2$. But since $Df$ vanishes at least once (by roll theorem) we have $\sup\{|Df(x)|\}< V_0^1(Df) < \frac{1}{2}$ so the staement is true for all $\alpha \le n$ (where $V_0^1(-)$ stands for the total variation).
Lastly $||D^{\alpha-1} f||_2 \le \sup \{|D^{\alpha-1}f(x)|:x\in X\} < \frac{1}{n-1}$.
Denoting: $$A_n=\{||D^\alpha f||_2 < \frac{1}{n}: \forall {\alpha \le n} \}$$
$$B_n=\{||D^\alpha f||_1 < \frac{1}{n}: \forall {\alpha \le n} \}$$
$$C_n=\{\sup\{|D^\alpha f(x)|:x\in K\} < \frac{1}{n}: \forall {\alpha \le n} \}$$
We have shown that $A_{n-1} \subset C_{n-1} \subset B_n \subset A_n$ thereby showing that the topologies induced by the seminorms coincide.
Was that right? Did i interpret the question correctly?
We have$$||f||_1^n =\int_K |f^{(n)} (u)| du =\int_K 1\cdot |f^{(n)} (u)| du \leq \left(\int_K 1^2 du \right)^{\frac{1}{2}} \left(\int_K |f^{(n)} (u)|^2 du \right)^{\frac{1}{2}} =||f||^n_2$$ and $$f^{(n)} (t) =\int_0^t f^{(n+1)}(u) du \leqslant \int_K |f^{(n+1)}(u)| du =||f||^{n+1}_1$$ hence $$||f||^n_{\infty} \leq ||f||^{n+1}_1$$ and $$||f||^n_2 =\sqrt{\int_K |f^{(n)} (u)|^2 du}\leqslant ||f||^n_{\infty}$$ so we obtain $$||f||_1^n\leqslant ||f||^n_2 \leqslant ||f||^n_{\infty} \leqslant ||f||^{n+1}_1$$