Show that there exist a constant $C$ s.t. $\|F\| \leq C\|f\|$

Question

Let $X$ be a closed subspace of ${L^1 (0,2)}$. Suppose that for every $f \in L^1(0, 1)$ there exists an $F \in X $ whose restriction to $(0,1)$ is $f$. Show that there is a constant $C$ such that we can always choose an $F$ satisfying $\|F\| \leq C\|f\|$ .

Can anyone help me solve this problem?

Let $\rho \colon X \to L^1(0,1)$ be the restriction. Look at the induced map $\overline{\rho} \colon X/\ker \rho \to L^1(0,1)$. What do you know about $\overline{\rho}$? — Daniel Fischer, Sep 30 '16 at 12:53
Any thoughts of your own on the question? If this is from a textbook, is there any theorem/idea from the chapter that you expect applies to this question? Perhaps something about bounded inverses? — Ben Grossmann, Sep 30 '16 at 12:54
@Omnomnomnom This question was given after the open mapping theorem. So I think I need to apply that but don't know how to proceed with that? — user383517, Sep 30 '16 at 12:56
@DanielFischer are you making that into an answer, or should I? — Ben Grossmann, Sep 30 '16 at 12:58

Ben Grossmann · Accepted Answer · 2016-09-30T13:47:00.747

Following Daniel's hint:

Let $\rho:X \to L^{1}(0,1)$ be the map that takes a function and produces its restriction to $(0,1)$. Notably, this map is bounded with $\|\rho\|\leq 1$, and is surjective (by the properties of $X$ from the problem statement).

Consider the induced map $\bar \rho: X/\ker \rho \to L^1(0,1)$. This map is not only bounded, but is also injective, with the same range as $\rho$. In other words, $\bar \rho$ must be an invertible (bijective), bounded linear map between two Banach spaces.

By the open mapping theorem (more specifically, its corollary the bounded inverse theorem) we may conclude that $\bar \rho^{-1}$ is bounded. Now, if you consider what exactly it means for $\rho^{-1}$ to be bounded, you'll find that you have completely answered the question.

Show that there exist a constant $C$ s.t. $\|F\| \leq C\|f\|$

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