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Let $X$ be a Banach space and $U,V$ complementary subspaces. Let $A: U \to U, B: V \to V$ be continuous and let $T(x) = (A \oplus B)(u \oplus v)= A(u) \oplus B(v)$.

Does it hold that $\|T\| \le \|A \| + \|B\|$?

Student
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2 Answers2

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Not necessarily. $\lVert T\rVert$ can be bounded by $C(\lVert A\rVert + \lVert B\rVert)$, but the constant cannot always be chosen as $1$, it depends on how $U$ and $V$ lie relative to each other.

We can create a situation where the constant must be large even in $\mathbb{R}^2$ with the Euclidean norm. Let $U = \mathbb{R}\cdot (1,0)$ and $A = \operatorname{id}_U$, and $V = \mathbb{R}\cdot (\cos\varphi,\sin\varphi)$ with $B = -\operatorname{id}_V$ where $0 < \varphi < \pi$. Then

$$\begin{align} (0,1) &= -\cot\varphi\cdot (1,0) + \frac{1}{\sin\varphi}(\cos\varphi,\sin\varphi)\\ &\mapsto -\cot\varphi\cdot (1,0) - \frac{1}{\sin\varphi}(\cos\varphi,\sin\varphi)\\ &= \left(-2\cot\varphi,-1\right) \end{align}$$

and $\lVert T\rVert \geqslant \sqrt{1+4\cot^2\varphi}$ becomes arbitrarily large for $\varphi\to 0$ (or $\varphi\to \pi$).

Daniel Fischer
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  • Thank you. Is it possible to prove that such $C$ exists? I am in the process of proving that if $A,B$ are continuous then so is $A \oplus B$. My idea was to use that the projection onto complemented subspaces is continuous: $A$ is the map $p_U \circ T$ and $B=p_V \circ T$. Then I want to say that a map is continuous iff the projection onto each of its components. Do you think it is a valid argument? – Student Jun 08 '14 at 11:29
  • Does the $\mapsto$ in the displaystyle equation mean you apply $T$ to get from line 1 to line 2? – Student Jun 08 '14 at 11:31
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    Second comment first: Yes, it means applying $T = A\oplus B$. First comment: That $U$ and $V$ are complementary means that the natural map $U\oplus V \to X$ is an isomorphism, so there is a $K$ with $K^{-1}(\lVert u\rVert + \lVert v\rVert) \leqslant \lVert u+v\rVert \leqslant K(\lVert u\rVert + \lVert v\rVert)$. You can use that to obtain $C$. But to show that $T$ is continuous, the best way is to factor the map $$X \xrightarrow{\cong} U\oplus V \xrightarrow{A\oplus B} U\oplus V \xrightarrow{\cong} X$$ where every part is evidently continuous, hence so is the composition $T$. – Daniel Fischer Jun 08 '14 at 11:39
  • I seem to have found yet another proof. I want to post it here in this thread and I would be incredibly grateful if you could look at it later and tell me if it works. (I just discovered a new comment by you after posting this) – Student Jun 08 '14 at 11:40
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    Regarding your idea, there's a small mistake, $A$ is defined on $U$, not on $X$, so one doesn't have $A = p_U\circ T$, but $A\circ p_U = p_U\circ T$, similar for $B$. Apart from that small glitch, it is correct, since $U$ and $V$ are complementary, $T$ is continuous if and only if both $p_U\circ T$ and $p_V\circ T$ are continuous. – Daniel Fischer Jun 08 '14 at 11:43
  • Regarding those projections: this argument only works if the topology on $U \oplus V$ is the product topology. How do I see that this is indeed the case? – Student Jun 08 '14 at 12:07
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    That's by definition. The topology on $E\oplus F$ is, unless explicitly stated otherwise, the product topology. (It is the topology that makes it the categorical coproduct too, since there are only finitely many factors/summands; for infinitely many factors/summands, the product and the coproduct are not isomorphic, however.) – Daniel Fischer Jun 08 '14 at 12:15
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Yet another proof:

To show that $T$ is continuous if $A,B$ are one can use that a map $T$ is closed if given $x_n \to x \in X$ then $Tx_n \to Tx$.

Note that since $X = U \oplus V$ we have $x_n = u_n \oplus v_n$. Let $u = \lim u_n$ and $v= \lim v_n$. Since $A,B$ are continuous, $Au_n \to Au$ and $Bv_n \to Bv$ and since $U,V$ are closed it follows that $Au \in U$ and $Bv \in V$. Hence $$ Au_n \oplus Bv_n \to Au \oplus Bv$$ which is the same as $$ T(x_n) \to Tx$$ hence $T$ is continuous.

Student
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