Prove that a map is continuous Let $r:[0,1]\to\mathbb R$ be a continuous function and let $u_\lambda$ be the unique solution of the Cauchy Problem:
$$\begin{cases}u''(t)+\lambda r(t)u(t)=0,\quad\forall t\in [0,1],\\ u(0)=0,\quad u'(0)=1.\end{cases}$$
It is well known that $(u_{\lambda_n})$ converges uniformly on $[0,1]$ to $u_\lambda$ whenever $\lambda_n\to\lambda.$ Define the map $\tau:\mathbb R\to\mathbb R$ by setting
$$\tau(\lambda):=\inf\{t\in(0,1]\mid u_\lambda(t)=0\},$$ with the convention $\tau(\lambda)=1$ if $u_\lambda(t)\neq 0$ for every $t\in (0,1]$. Prove then that $\tau$ is continuous.
Edit:
ok so i've been trying to work on Robert hint. I am trying to prove that if $\tau(\lambda_0)=\tau_0<1$ then for some $\varepsilon>0$ small i must have $u_{\lambda_0}(\tau_0+\varepsilon)<0$ , but i seem to go nowhere farther from some silly tryings using mean value theorem or stuff like that.. Where am i missing the key point? 
 A: Hint: if $\tau(\lambda_0) = \tau_0  < 1$ and $\epsilon > 0$ is small, there is $\delta > 0$ such that $u_\lambda(t) > \delta$ if $\epsilon \le t \le \tau_0 - \epsilon$, while $u_\lambda(\tau_0 + \epsilon) < 0$.  
A: A few more hints:


*

*Using the uniqueness of the solution to the Cauchy problem, observe that if $u_\lambda(\tau(\lambda)) = 0$, then there exists some $\epsilon > 0$ such that $u_\lambda|_{(\tau(\lambda)-\epsilon,\tau)} > 0$ and $u_\lambda|_{(\tau,\tau+\epsilon)} < 0$. (Why can't $u$ locally have the same sign on the two sides of the zero?)

*If $u$ is a continuous function such that $u(\tau-\epsilon) > 0$ and $u(\tau+\epsilon) < 0$, what can you say about $u$?

*Since $r(t)$ is continuous, what can you say about $u_\lambda'(t)$, for $t$ sufficiently close to 0? What does this tell you about $u_\lambda$ in, say, $(0,\epsilon)$? 

*If $u_{\lambda}|_{[a,b]} > 0$, what can you say about $u_{\eta}|_{[a,b]}$ for $\eta$ sufficiently close to $\lambda$?


To put every thing together, let $\eta$ be a small perturbation of $\lambda$, and fix a very small $\epsilon$ depending on $\lambda$ and $r$. Use 4 to show that $u_\eta$ cannot vanish on $[\epsilon,\tau(\lambda)-\epsilon]$. Use 3 to show that $u_\eta$ cannot vanish on $(0,\epsilon)$. And use 1 & 2 to show that $u_\eta$ must vanish between $(\tau(\lambda)-\epsilon, \tau(\lambda)+\epsilon)$. 

Okay, more on point 1 by request. 


*

*The function $u \equiv 0$ is a solution to the differential equation (ignoring boundary conditions). So by uniqueness of solutions, if $\exists t_0$ such that $u_\lambda(t_0) = u_\lambda'(t_0) = 0$, $u_\lambda \equiv 0$ and cannot satisfy the boundary condition $u_\lambda'(0) = 1$. Hence for a solution to the ODE with the requisite boundary conditions, at any point where $u_\lambda(t_0) = 0$ we must have that $u_\lambda'(t_0) \neq 0$. 

*The usual regularity theory for ODEs guarantee that $u_\lambda'$ is continuous. 

