1
$\begingroup$

$C_b([0,\infty])$ is the space of all bounded, continuous functions.

Let $||f||_a=(\int_{0}^{\infty}e^{-ax}|f(x)|^2)^{\frac{1}{2}}$

First I want to prove that it is a norm on $C_b([0,\infty])$. The only thing I have problems with is the triangle inequality, I do not know how to simplify

$||f+g||_a=(\int_{0}^{\infty}e^{-ax}|f(x)+g(x)|^2)^{\frac{1}{2}}$

The second thing I am interested in is how to show that there are constants $C_1,C_2$ such that $||f||_a\le C_1||f||_b$ and $||f||_b\le C_2||f||_a$ so for $a>b>0$ the norms $||.||_a$ and $||.||_b$ are not equivalent.

$\endgroup$
  • $\begingroup$ for the triangle inequality, just expand the |f+g|^2, then use Cauchy inequality. $\endgroup$ – lee Dec 2 '12 at 12:56
  • $\begingroup$ Ok thanks, and what about the non-equivalence of the norms? $\endgroup$ – Voyage Dec 2 '12 at 13:03
1
$\begingroup$

For the first one: Use

$$e^{-\alpha \cdot x} \cdot |f(x)+g(x)|^2 = \left|e^{-\frac{\alpha}{2} \cdot x} \cdot f(x)+ e^{-\frac{\alpha}{2} \cdot x} \cdot g(x) \right|^2$$

and apply the triangel inequality in $L^2$.

Concerning the second one: For $a>b>0$ you have

$$e^{-a \cdot x} \cdot |f(x)|^2 \leq e^{-b \cdot x} \cdot |f(x)|^2 $$

i.e. $\|f\|_a \leq \|f\|_b$. The inequality $\|f\|_b \leq C \cdot \|f\|_a$ does not hold in general. To see this let $c \in (b,a)$ and

$$f_n(x) := \min\{n,e^{c \cdot x}\} (\in C_b)$$

Then $\|f_n\|_a < \infty$,$\|f_n\|_a \to c<\infty$, but $\|f_n\|_b \to \infty$.

$\endgroup$
1
$\begingroup$

For triangular inequality, write, using Cauchy-Bunyakovsky-Schwarz, \begin{align}\lVert f+g\rVert_a^2&=\int_0^{+\infty}e^{-ax}(|f(x)|^2+|g(x)|^2+2f(x)g(x))dx\\ &\small\leqslant\int_0^{+\infty}e^{-ax}(|f(x)|^2+|g(x)|^2)dx+2\left(\int_0^{+\infty}e^{-ax}|f(x)|^2dx\right)\left(\int_0^{+\infty}e^{-ax}|g(x)|^2dx\right)\\ &=\left(\lVert f\rVert_a^2+\lVert g\rVert_a^2\right)^2. \end{align} Let $0<a<b$. As $e^{-bx}\leqslant e^{-ax}$, we have for all $f\in C_b([0,+\infty))$ that $\lVert f\rVert_b\leqslant \lVert f\rVert_a$.

For non equivalence, take $f_n$ a function which is $e^{b/2x}$ on $[0,n]$, $0$ for $x\geqslant n+1$ and linear in $[n,n+1]$.

$\endgroup$
1
$\begingroup$

For the triangle inequality, note $\|f\|_a=\|e^{-ax/2}f\|_2$, so you can apply Minkowski.

To see that the norms are not equivalent (and so such constants cannot both exist), let $$ f_n(x)=\begin{cases}e^{bx/2},&\text{ if } x\in[0,n] \\ \text{line},&\text{ if } x\in[n,n+e^{-bn/2}] \\ 0,&\text{ if }x>n+e^{-bn/2}\end{cases} $$ Then $$ \|f_n\|_a\leq\left(\int_0^ne^{(b-a)x}\,dx +1\right)^{1/2}\leq\left(\frac1{b-a}+1\right)^{1/2}, $$ $$ \|f_n\|_b\geq\left(\int_0^n 1\right)^{1/2}=\sqrt n $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.