Show that $\| v\|_{L^2(0,b)}=o(\| u\|_{L^2(0,b)})$ as $b \to \infty$

Question

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.

Answer 1

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 $b \to \infty$.

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