Let $\alpha<0$ and let $u(x):= e^{\alpha x}$ for $x \geq 0$. I'm reading a paper which states that there are constants $d\_{j}, \beta\_{j} \in \mathbb{C}$ with $\beta\_j>0$ such that if we define
$$v(x) =\sum\_{j=1}^n d\_{j} e^{ \beta\_{j} (x-b)}$$
then $w^{(j)}(b)=0$ for $j=0,1, \ldots, n-1$ if $w(x):=v(x)+u(x)$, and
\begin{equation}\tag{1}
\| v\|\_{L^2(0,b)}=o(\| u\|\_{L^2(0,b)}) \ \ \ \mbox{as} \ \ b \to \infty
\end{equation}


I think that the constants $d\_j$ depend on $b$. I am struggling to prove (1). 


**My attempt**


I tried the following:
$$\|v\|\_{L^2(0,b)}^2 \leq n M\_b \int\_{[0,b]} \sum\_{j=1}^n e^{2 \beta\_{j}(x-b)}= nM\_b \sum\_{j=1}^n \frac{1-e^{-2 \beta\_{j}b}}{2 \beta\_{j}},$$
where $M\_b:=\max\{|d\_{j,b}|^2: j=1, \ldots,n\}$. But using that $\|u \|^2\_{L^2(0,b)}=(e^{\alpha b}-1)/\alpha$ I can not get (1).


Thanks for any help you can give me.



Show that $\| v\|_{L^2(0,b)}=o(\| u\|_{L^2(0,b)})$ as $b \to \infty$
Not sure if I understood correctly the question, but this is too long for a comment.


Fix $\alpha < 0$ and take any distinct $\beta\_j > 0$, $j = 1, 2, \ldots n$. Then there is a unique solution $d\_j$, $j = 1, 2, \ldots, n$, of the system of linear equations $$ \frac{d^i}{dx^i} \bigg|\_{x = 0} \biggl(\sum\_{j = 1}^n d\_j e^{\beta\_j x}\biggr) = -\alpha^i , \qquad i = 0, 1, \ldots, n - 1.$$ For $b > 0$ define $$ v(x) = \sum\_{j = 1}^n d\_j e^{\beta\_j (x - b) + \alpha b} $$ (note the additional factor $e^{\alpha b}$!). Then $w(x) = v(x) + e^{\alpha x}$ has derivatives of order $0, 1, \ldots, n - 1$ at $x = b$ all equal to zero, as desired. Furthermore, the $L^2$ norm of $e^{\alpha x}$ on $(0, b)$ converges to a constant as $b \to \infty$, while the $L^2$ norm of $v$ on $(0, b)$ decays as $e^{\alpha b}$ as $b \to \infty$.


Does this answer your question?


