I'm considering the operator $T$ given by (Tf)(x)=$\int_0^1k(x,y)f(y)dy$ with $dom(T)=\mathrm{L}^1([0,1])$, where $k\in\mathscr{C}([0,1]^2)$ and want to proof that it maps to $\mathscr{C}([0,1])$.
I know that for a given $x\in[0,1]$ and $\varepsilon>0$ the continuity of k gives me:$$\forall y\in[0,1]\exists\delta_y>0:\forall z\in[0,1]:\lvert x-z\rvert\lt\delta_y\implies\lvert k(x,y)-k(z,y)\rvert\lt\frac{\varepsilon}{\lvert\lvert f\rvert\rvert_1}$$ and also for any given $z$ $$\lvert(Tf)(x)-(Tf)(z)\rvert \le max_{y\in[0,1]}\lvert k(x,y)-k(z,y)\lvert\cdot\lvert\lvert f\rvert\rvert_1\le\lvert k(x,y_{max}-k(z,y_{max})\rvert\cdot\lvert\lvert f\rvert\rvert_{1}$$ but to prove continuity of $Tf$ I have to chose a $\delta$ before I can pick a $z$, I would expect that maybe $\sup_{[0,1]}\delta_y$ suffices, but then I would first have to show that it is finite?
Any hint is very welcome.