First, recall that if $X$ is a normed space, $X_1$, $X_2$ are linear subspaces of $X$, and $X = X_1 \oplus X_2$, then the natural projector $P$ onto $X_1$ parallel to $X_2$ is defined as follows:

given $x\in X$, decompose it as $x = x_1 + x_2$, where $x_1 \in X_1$, $x_2 \in X_2$ (by assumption, this decomposition exists and is unique), and then put $P(x) = x_1$.

This linear projector $P$ is continuous if and only if there is a constant $C > 0$ such that for every $x_1 \in X_1$, $x_2 \in X_2$ the inequality $\|x_1\| \le C \|x_1 + x_2\|$ holds true.

The most natural way of constructing examples of closed subspaces with non-closed sum is based on the following theorem (which is a standard application of the closed graph theorem):

Let $X$ be a Banach space, $X_1$ and $X_2$ be closed subspaces of $X$, and $X = X_1 \oplus X_2$. Then the natural projector $P$ onto $X_1$ parallel to $X_2$ is a continuous operator.

Consequently, if in a Banach space $E$ you construct two closed subspaces $X_1$ and $X_2$ with zero intersection, denote $X = X_1 \oplus X_2$ and find that the projector $P$ onto $X_1$ parallel to $X_2$ is discontinuous, then $X$ must be non-closed (otherwise $X$ would be a Banach space and by the previous theorem $P$ should be continuous).

There are many such examples of $X_1$ and $X_2$. For example, if in the Hilbert space $\ell_2$ you take an orthonormal basis $e_n$, $n = 1, 2, ...$, and take as $X_1$ the closed linear hull of all $e_{2k}$, $k = 1, 2, ..$ and as $X_2$ the closed linear hull of vectors $e_{2k} + 2^{- k} e_{2k-1}$, $k = 1, 2, ..$, then these $X_1$ and $X_2$ will be the desired example.

Dear Vladimir

I thank you a lot for clarifications. I tried by chance to treat the case of odd degrees polynoms and even ones. The way that you show me is simple and instructive. Thank you again

Hi everybody,

In addition to the negative of Vladimir, I think that there is a positive answer in a Hilbert space (to be verified). If $X_1$ and $X_2$ are closed subspaces of a Hilbert space $H$, and if there exists some positive constant $c

Dear Lahbib,

• The inequality you suggest is satisfied in every hilbet space (Schwartz inequality) for c=1.
• If C[0,1] is equipped with the L^1norm, then the function 1 is in the sum because it is equal almost everywhere to 1_{[0;1/2]}+1_{[1/2;1]}. Don't forget that in this case the impact of 1/2 image is without effect since Lebesgue measure of any point set {a} is zero and has not a contribution on L^1 norm.
• Thank you for interaction. It's a pleasure to read your answer after these clarifications

Dear Ahmed,

Concerning the inequality, the constant c must be strictly less than 1 but not 1.

As to the second point, the functin you have given is not continuous. Then it does even not belong to the whole space C[0,1 ].

It's clear that the function S that I proposed (the sum above) does not belong to C[0,1] (it is clearly discontinuous at 1/2, BUT:

with L^1 norm, one usually can find a continuous represantant for S. When we equip C[0,1] with L^1 norm, its elements become classes or orbits then the sum S above, although its dicontinuity, may be considered an element of (C[0,1],\|f\|_1).

Thanks mister Lhabib

Dear Ahmed

The space C[ 0,1] is clearly defined. The continuity belongs to its definition and your function is not continous. Moreover, the only continuous representation 1 of S cannot be writen as a sum of an element of $I$ and another of $J$. Good luck

Dear Lhabib,

with L^1 topology, C[0,1] is not a Banach space. So the desired example loses sense if we dont embedd C[0,1] in a banach one (here L^1[0,1]). In this case, the two ideals you proposed will be equipped with L^1 topology, so the continuity of 1 function becomes as the same "continuity" of the sum 1_{[0,1/2]}+1_{[1/2,1]} .

I think that the difficulty is now clear.

Faithfully yours and thanks again

For people interested with the question, a partial answer is at attached file.