# Funcional analysis probelm

I have to solve the following problem.

Consider the spaces $X=C^0([0,1])$ and $Y=\{g\in C^1([0,1]): g(0)=0\}$ both provided with the supremum norm (so that $Y$ is not a Banach space). Prove that $$T:X\rightarrow Y,\qquad (Tf)(x)=\int_0^xf(t)dt,\quad x\in [0,1]$$ is a surjective linear bounded operator, but it is not an open map.

Have you some hints to solve it? Thanks

## 1 Answer

Boundeness is easy: It follows from

$$\|T(f)\| = \sup_{x\in [0,1]}\left| \int_0^x f(t)dt \right|\leq \sup_{x\in [0,1]} \int_0^x |f(t)|dt \leq \int_0^1 \|f\|dt = \|f\|.$$

For surjectivity, note that if $$f\in Y$$, then $$f'\in X$$ and $$T(f')=f$$.

Finally, suppose that $$T$$ is open. It is not hard to prove that $$T$$ is injective, thus $$T^{-1}$$ is continuous. But $$T^{-1}(f)=f'$$, and it is known that the restriction of $$T^{-1}$$ to the space of all polynomials with the supremum norm is not bounded (see for example Kreiszig, Introductory to Functional Analysis with applications, page 93), hence a contradiction.