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Let $(X,||.||_X)$ be a normed space, M,N two subspaces with norms $||.||_M,||.||_N$

The identity maps are cont. Now I can define the norm $||x||_{M+N}=inf\{||m||_M+||n||_N:m\in M, n\in N, x=m+n\}$

1)I already showed that $||x||_{M+N}=0$ iff $x=0$ and homogenity. But what about the traingle inequality ? I am not sure how to write $||x+y||_{M+N}$.Maybe $||x+y||_{M+N}=inf\{||m||_M+||n||_N:m\in M, n\in N, x+y=m+n\}$ ?

2) If $M,||.||_M$ and $N,||.||_N$ are complete (i.e $||x_n-x_m||<\epsilon$ and $||y_n-y_m||<\epsilon$) how can I follow that $(M+N,||.||_{M+N})$ is complete ?

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  1. Let $x,y\in M+N$, $k>1$ fixed; there are $m_k,m'_k\in M$, $n_k,n'_k\in N$ such that $x=m_k+n_k$ and $y=m'_k+n'_k$ and $\lVert m_k\rVert_M+\lVert n_k\rVert_N<\lVert x\rVert_{M+N}+k^{-1},\quad \lVert m'_k\rVert_M+\lVert n'_k\rVert_N<\lVert y\rVert_{M+N}+k^{-1}.$ We have $x+y=\underbrace{m_k+m'_k}_{\in M}+\underbrace{n_k+n'_k}_{\in N}$, so $\lVert x+y\rVert_{M+N}\leqslant \lVert m_k+m'_k\rVert_M+\lVert n_k+n'_k\rVert_N< \lVert x\rVert_{M+N}+\lVert y\rVert_{M+N}+2k^{—1}.$

  2. This will give ideas.

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    Since it's true for all $k$, take the limit $k\to +\infty$.2012-12-12
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1) Temporarily forget about the inf, and pretend that $\|x\|_{M+N}=\|m_x\|_M+\|n_x\|_N$ for some $m_x\in M$ and $n_x\in N$ with $x=m_x+n_x$, and similarly for $Y$. Then figure out the proper answer with the inf by following this pattern.

2) I got in a tangle trying to do this directly from the definitions. Instead, I suggest using the fact that a normed vector space is complete if and only if every absolutely convergent series is convergent: if $\sum_k \|x_k\|_{M+N}<\infty$, then it's not hard to see that you can write $x_k=m_k+n_k$ where $\sum_k \|m_k\|+\|n_k\|<\infty$, which forces $\sum_k m_k$ and $\sum_k n_k$ to converge in the complete spaces $M$ and $N$, respectively. Then show that this implies that $\sum_k x_k$ converges to $\sum_k m_k + \sum_k n_k$ in $M+N$.

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    That's right, yes.2012-12-12