Prove existence and uniqueness $f\in C([0,1])$ such that $f(t)+\frac{t}{2}\cos(f(t))=\int_0^1f(ts)s^2ds$

Question

Prove there is a unique $f\in C([0,1])$ such that $$ f(t)+\frac{t}{2}\cos(f(t))=\int_0^1f(ts)s^2ds. $$

Any idea how to approach the problem ?

I considered to use banach-theorem(fixed-point) but I am not entirly sure how to approach it.

Edit :

My attempt:

Let $f,g\in C([0,1])$ such that

$f(t)=(Tf)(t):=\int_0^1f(ts)s^2ds-\frac{t}{2}\cos(f(t)), g(t)=\int_0^1g(ts)s^2ds-\frac{t}{2}\cos(g(t))$.

$|(Tf)(t)-(Tg)(t)|\le |\int_0^1(f(ts)-g(ts))s^2ds|+\frac{t}{2}|\cos(f(t))-\cos(g(t))|\leq \int_0^1|f(ts)-g(ts)|s^2ds+\frac{t}{2}|f(t)-g(t)|$

How to continue ?

Appreciate any help.

Answer 1

Let $(Tf)(t)=\int\limits_0^1f(ts)s^2\,ds -{t\over 2}\cos f(t).$ Then $$|(Tf)(t)-(Tg)(t)|\le \int\limits_0^1|f(ts)-g(ts)|s^2\,ds+{1\over 2}|\cos f(t)-\cos g(t)|\\ \le \|f-g\|\int\limits_0^1s^2\,ds +{1\over 2}\|f-g\|={5\over 6}\|f-g\|$$ Hence $\|Tf-Tg\|\le {5\over 6}\|f-g\|,$ and the Banach fixed point theorem can be applied.