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.