the following is an exercise taken from the written exam of the functional analysis course that I am following
Let $f \, : \, [0,1] \times \mathbb{R} \to \mathbb{R}$ be a $C^1$ function that satisfies
$$ \lim_{|\xi|\to \infty}{\frac{f(x,\xi)}{\xi}} = 0 $$ uniformly with respect to $x \in [0,1]$. Prove that the problem $$ \begin{cases} &-u^{''} = f(x,u(x)) \\ &u(0) = u(1) = 0\end{cases} $$ has a unique classical solution
I managed to prove the existence of the solution but I didn't managed to prove the uniqueness of the solution. I would like to demonstrate uniqueness by not using tools external to the course, whcih are : Sturm-Liouville theory, a bit of Sobolev spaces, Nemitski operators, Fourier transform, Functional calculus and spectral theorem, Lax-Milgram theorem, Schauder-Fixed point theorem, Implicit function theorem for Banach spaces, Inversion theorem for Banach spaces and Lagrange multipliers for Banach spaces.
This is how I proved the existence.
PROOF OF EXISTENCE :
Consider the operator $\tilde{T} : L^2([0,1]) \to H^1_0{([0,1])}$ that maps $g \in L^2([0,1])$ in the unique solution $\tilde{T}(g) = v$ of the Sturm-Liouville problem $$\begin{cases} &-v^{''} = g \\&v(0) = v(1) = 0\end{cases}$$
one can prove that $\tilde{T}$ is continuos because multiplying by $\tilde{T}(g)$, integrating and using integration by parts we find
$$ \int_{0}^{1}{ g \cdot (\tilde{T}g) dx} = -\int_{0}^{1}{(\tilde{T}g)^{''}\cdot(\tilde{T}g)dx} = \int_{0}^{1}{ (\tilde{T}g)^2 dx}$$
Combining this with the Cauchy-Schwarz inequality we find ( from now on I will omit $([0,1])$ in the functional spaces for simplicity )
$$ ||(\tilde{T}g)' ||^2_{L^2} \leq ||(\tilde{T}g)' ||_{L^2}||g ||_{L^2}$$
dividing we find
$$ ||(\tilde{T}g)' ||_{L^2} \leq ||g ||_{L^2}$$
which proves the continuity by the Poincarè inequality.
Now let $i \; : \; H^1_0 \to L^2$ be the inclusion operator, we know it is compact thus $T = i \circ \tilde{T}$ is compact
Consider the Nemitski non-linear operator $L^2 \ni u \mapsto Gf = f(x,u(x)) \in L^2$ ( I won't prove $Gf$ is in $L^2$ since we did it in class )
Now let $$\begin{split} F \; : \; L^2([0,1]) &\to L^2([0,1])\\ L^2 \ni u &\mapsto F(u) := T(G(u)) \in L^2 \end{split}$$
it's easy to show that $u$ solves the problem if and only if $F(u) = u$.
I show an important inequality for $G$, let $N > 0$, let $$M = M(N) := \max_{\substack{% 0 \leq x \leq 1\\ |y| \leq N}}{|f(x,y)|} \hspace{0.5cm} \epsilon = \epsilon(N) := \sup_{\substack{% 0 \leq x \leq 1\\ |y| \geq N}}{\frac{|f(x,\xi)|}{|\xi|}}$$
clearly $M = M(N)$ is finite for all $N > 0$ and $\lim_{N \to \infty}{\epsilon(N)} = 0$
Then we find
$$||G(u)||^2_{L^2} = \int_{0}^{1}{|f(x,u(x))|^2 dx} = \int_{ |u(x)| \leq N }{ |f(x,u(x)|^2 dx} + \int_{|u(x)| > N}{ |f(x,u(x)|^2 dx} \leq \int_{ |u(x)| \leq N }{ M^2 dx} + \int_{ |u(x)| > N }{ \epsilon^2 \cdot |u(x)|^2 dx} \leq M^2 + \epsilon^2 ||u||^2_{L^2}$$
this shows that $G$ sends bounded sets in bounded sets, therefore $F = T \circ G$ is compact since $T$ is compact. Now let $N >> 0$ and $R >> 0$ ($R$ possibly depending on $N$) be such that $||T||^2(M^2 + \epsilon^2 R^2) \leq R^2$ then since $||F(u)||_{L^2} \leq ||T||\cdot||G(u)||_{L^2}$ I have that the set $$D := \{ u \in L^2([0,1]) \; : \; ||u||_{L^2} \leq R \}$$ is mapped into itself by $F$, meaning $||u||_{L^2} \leq R \implies ||F(u)||_{L^2} \leq R$, therefore if I restrict $F$ to $D$ by Schauder fixed point theorem there exists $u \in L^2$ such that $F(u) = u$, which proves the existence
I didn't managed to make any progress from here
I doubt that uniqueness holds under these assumptions. Consider any function $f\in C^1(\mathbb{R}, \mathbb{R})$ such that $f(\xi)=\pi^2\xi$ $(\xi \in [-1,1])$ and $f(\xi)=0$ $(|\xi| > 2)$. Clearly $$ \lim_{|\xi|\to \infty} \frac{f(\xi)}{\xi} = 0 $$ uniformly on $[0,1]$ (since $f$ is independent on $x$). Now the BVP $$ u''(x)+f(u(x))=0, \quad u(0)=u(1)=0 $$ has among others the solutions $u(x)=c\sin(\pi x)$ $(x \in [0,1])$ for each $c \in [-1,1]$.