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 1


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.


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