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Suppose $T$ and $S$ are firmly nonexpansive mappings from $\mathbb R^n$ to $\mathbb R^n$. Let $I$ be identity operator. I want to show that $Z=T(2S−I)+I−S$ is firmly nonexpansive.

Definition. We say that $F$ is firmly nonexpansive if: $$\|F(x)-F(y)\|^2+\|(I-F)(x)-(I-F)(y)\|^2\le \|x-y\|^2$$


In the question it is said that $T$ and $S$ are firmly nonexpansive. So we have: $$\|T(x)-T(y)\|^2+\|(I-T)(x)-(I-T)(y)\|^2\le \|x-y\|^2$$ $$\|S(x)-S(y)\|^2+\|(I-S)(x)-(I-S)(y)\|^2\le \|x-y\|^2$$

Now I have to show that: $$\|Z(x)-Z(y)\|^2+\|(I-Z)(x)+(I-Z)(y)\|^2\le \|x-y\|^2$$ where $Z=T(2S−I)+I−S$.

I cant find this expression from those two expression given. It gets very complicated when I want to solve it. Here I have to show that if S is firmly non-expansive , so 2S−I and I-S is firmly nonexpansive too . then I have to show that S(T)(x) is non expansive too. but how?!

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Please do not post the same question twice. –  user53153 Jan 7 '13 at 16:50
    
I spent some time formatting this copy before realizing it's a duplicate. Adding to the frustration, I'm 3 points short of voting to close... –  user53153 Jan 7 '13 at 16:55
    
possible duplicate of Sow that $Z = T(2S-I)+I-S$ is firmly nonexpansive –  Clive Newstead Jan 7 '13 at 17:26
    
I've deleted the previous post . –  Ilkay Jan 8 '13 at 9:28
    
sorry for that Pavel . –  Ilkay Jan 8 '13 at 11:02
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1 Answer

Unfortunately, your idea of

Here I have to show that if S is firmly non-expansive , so 2S−I and I-S is firmly nonexpansive too . then I have to show that S(T)(x) is non expansive too.

is incorrect. As you can see from the proof below, the conclusion requires delicate cancellations between different parts of $Z$, so naively splitting up $Z$ into parts and doing each part separately will not work.


Let's try by brute force. For convenience we write $x' = (2S - I)(x)$ and $y' = (2S - I)(y)$. Note that $Z = T(2S - I) + I - S = (T-I)(2S - I) + S$.

$$ \begin{align} \| Z(x) - Z(y) \|^2 &+ \|(Z-I)(x) + (Z-I)(y)\|^2 \\ & = \|(T-I)(x') - (T-I)(y') + S(x) - S(y)\|^2 + \|T(x') - T(y') - S(x) + S(y)\|^2 \\ & = \|(T-I)(x') - (T-I)(y')\|^2 + \|T(x')- T(y')\|^2 + 2 \|S(x) - S(y)\|^2 \\ & \qquad - 2 \langle T(x') - T(y'), S(x) - S(y)\rangle + 2\langle (T-I)(x') - (T-I)(y'), S(x) - S(y) \rangle\\ &= \|(T-I)(x') - (T-I)(y')\|^2 + \|T(x')- T(y')\|^2 + 2 \|S(x) - S(y)\|^2 \\ & \qquad - 2\langle (2S - I)(x) - (2S-I)(y), S(x)- S(y)\rangle \\ &= \|(T-I)(x') - (T-I)(y')\|^2 + \|T(x')- T(y')\|^2 \\ & \qquad - 2\langle (S-I)(x) - (S-I)(y), S(x) - S(y) \rangle \end{align}$$

So far we haven't used any estimates, and just used algebraic manipulations of the squared norm. The first two terms after the last equality can be controlled by the nonexpansivity of $T$, so we have

$$ \begin{align} & \leq \|(2S - I)(x) - (2S - I)(y)\|^2 - 2\langle (S-I)(x) - (S-I)(y), S(x) - S(y) \rangle \\ & = \|S(x) - S(y)\|^2 + \|(S - I)(x) - (S-I)(y)\|^2 \\ & \qquad + 2 \langle S(x) - S(y), (S-I)(x) - (S-I)(y) \rangle - 2 \langle S(x) - S(y), (S-I)(x) - (S-I)(y) \rangle \end{align}$$

Again the equality is by expanding using $$ \|A + B\|^2 = \|A\|^2 + \|B\|^2 + 2 \langle A,B\rangle $$ for the Euclidean inner product. Observe now that the last two terms in the final expression cancel. So we can finally use the nonexpansivity of $S$

$$ \begin{align} &= \|S(x) - S(y)\|^2 + \|(S - I)(x) - (S-I)(y)\|^2 \\ & \leq \|x - y\|^2 \end{align}$$

Q.E.D.

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