This is my first question so I hope I'll do it well.
Let K be defined in a square $0\leq x,y\leq 1$ and $\mid K(x,y)\mid <1$. I want to prove that theres a continuous function f(x) in [0,1] such that $f(x)+\int_{0}^{1}K(x,y)f(y)dy=e^{x^{2}}$.
So, if I define a function $T(f)(t)=\int_{0}^{1}K(x,y)f(y)dy-e^{x^{2}}$, then $\mid T(f)(x)-T(g)(x)\mid=\mid \int_{0}^{1}K(x,y)f(y)dy-e^{x^{2}}-\int_{0}^{1}K(x,y)g(y)dy+e^{x^{2}}\mid = \mid\int_{0}^{1}K(x,y)f(y)-g(y)dy\mid \leq \int_{0}^{1}\mid K(x,y)\mid \mid f(y)-g(y)\mid dy$.
Now, since $\mid K(x,y)\mid<1$ theres a $\lambda$ s.t $\mid K(x,y)\mid \leq \lambda <1$. Then, $\mid T(f)(t)-T(g)(u)\mid \leq \lambda \mid \mid f-g \mid \mid_{\infty}$. [1]
Now, why exactly can I say that because of [1], $\mid \mid T(f)-T(g) \mid \mid_{\infty}\leq \lambda \mid \mid f-g \mid \mid_{\infty}$?
I had that question for a long time now and probably its a very basic fact, hope you guys can help me.
Equation (1) should be $$\lvert T(f)(x)-T(g)(x)\rvert\leq\lambda\lVert f-g\rVert_\infty\text{ for all }x\in [0,1].\tag{1}$$ Now take supremum over such $x$ to conclude $\lVert T(f)-T(g)\rVert_\infty\leq\lambda\lVert f-g\rVert_\infty$