I was trying to prove the following theorem: If $f(x, y)$ is continuous on some region $R \subseteq \mathbb R^2$ then any solution of IVP $$y'(x)=f(x,y(x)), y(x_0)=y_0 \tag{1}$$ is also a solution of integral equation $$y(x)=y_0+\int_{x_0}^x f(t,y(t)) \, dx \tag{2}$$ Converse also holds, meaning that, any continuous solution of $(2)$ is also a solution of $(1)$.
Proof: Let $\varphi$ be the solution of $(1)$ on some interval $\mathcal I$ which contains point $x_0$. Than we have $$\varphi'(t)=f(t,\varphi(t)),\forall t\in\mathcal I\tag{3}$$ and $$\varphi(x_0)=y_0\tag{5}$$ We would like for the right hand side of $(3)$ to be Riemman integrable, and it would be, provided that $f(t,\varphi(t))$ is continuous. Then we could simply integrate relation $(3)$ and obtain $$\int_{x_0}^x \varphi'(t) \,dt=\int_{x_0}^x f(t,\varphi(t)) \,dt \tag{6}$$ which would yield, after evaluating left integral $$\varphi(x)=y_0+\int_{x_0}^x f(t,\varphi(t)) \,dt,\forall x\in\mathcal I \tag{7}$$This would mean that $\varphi$ is indeed a solution of $(2)$ which would complete proof in one direction. Problem here lies in continuity of function $f(t,\varphi(t))$ on $\mathcal I$. $\varphi$ is continuous on $\mathcal I$ for sure because it's differentiable on $\mathcal I$ as given by $(3)$. Also by $(3)$ we know that for every $t\in\mathcal I$, point $(t,\varphi(t))$ belongs to domain of $f(x,y)$, but not necessarily the set $R$ for which we know that $f(x,y)$ is continuous on. How can we then guarantee continuity of $f(x,y)$ on $\mathcal I$ without additional assumptions like $$Dom(f(x,y))=R\tag{8}$$ or weaker one $\forall t\in\mathcal I:(t,f(t))\in R\tag{9}$ Do we have to introduce $(8)$ or $(9)$ into the statement of the theorem? Proving converse goes something like this, at least in all of the references that I could find:
Let $\varphi$ be the continuous solution of $(2)$ on some interval $\mathcal I$ which contains point $x_0$. Than we have $$\varphi(x)=y_0+\int_{x_0}^x f(t,\varphi(t)) \, dx,\forall t\in\mathcal I\tag{10}$$ Now we would like to differentiate relation $(10)$. By the FTC, if $f(t,\varphi(t))$ is continuous on $\mathcal I$, then $\int_{x_0}^x f(t,\varphi(t)) \, dx$ is differentiable on $\mathcal I$ and it holds that $$\frac{d}{dx}\int_{x_0}^x f(t,\varphi(t)) \, dx=f(x,\varphi(x))\tag{11}$$ So by differentiation of $(10)$ we obtained $$\varphi'(x)=f(x,\varphi(x)),\forall x\in\mathcal I\tag{12}$$Also, after plugging $x_0$ into $(10)$ we get $\varphi(x_0)=y_0$, and this would mean that $\varphi$ is a solution to an IVP $(1)$. Still, problem with continuity of $f(x,y)$ remains. $\varphi$ is continuous on $\mathcal I$ and $f(x,y)$ is continuous on some region $R$. As I see it, their composition can be continuous on $\mathcal I$ only if $(8)$ or $(9)$ holds. Maybe I am missing something and continuity of $f$ can be deduced by other means? I feel like learning this by heart without understanding its core would do me no good so in that case I would be better of studying poetry instead of math. Thanks to whoever read the whole thing and has any idea about this problem.
You're referencing Cauchy-Lipschitz or Picard-Lindelof theorem for the ODE, which usually assumes that that $ f \in \mathcal{C}([0,T] \times \mathbb{R}) $ in addition to being Lipschitz continuous in the second argument. Here I assume $x_0=0$ for convenience and $T>0$. Then you use your proof to equate the integral solution and the ODE and use Banach fixed point to guarantee a solution for $0 < t\leq T$.
There is a local version of this theorem which says that given any $r>0$ if $ f \in \mathcal{C} \left( [0,T] \times \overline{B(y_0;r)} \right)$ and $f$ is Lipschitz continuous in the second argument then the ODE admits a unique solution for $0 < t\leq \tau$ for some $\tau \in (0,T]$. Here $\overline{B(y_0;r)}$ is the closed ball centered at $y_0$ with radius $r$.
This works because the function $t \rightarrow f(t,\varphi(t)) $ is clearly continuous in a neighborhood of $x_0$ since $f$ is continuous in a neighborhood of $(x_0,y_0)$.