# Showing that $T+S$ is firmly nonexpansive

Show that $T+S$ is firmly nonexpansive considering that $T$ and $S$ are firmly nonexpansive mappings from $\mathbb R^n$ to $\mathbb R^n$.

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

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It is false. Let $T = S = I$. Then $T+S = 2I$ which once you plug in you see that is not firmly nonexpansive. Do you perhaps mean to consider $\frac12(T+S)$? In which case the desired conclusion follows from triangle inequality applied many times. – Willie Wong Jan 9 '13 at 10:31
So how am I suppose to show that , Z=T(2S−I)+I−S is firmly nonexpansive? I know that , 2S-1 and I-S are firmly non expansive . So firmly nonexpansive of a firmly nonexpansive is firmly nonexpansive?? – Ilkay Jan 9 '13 at 13:00
Huh? What does your last comment have to do with the question you asked above? – Willie Wong Jan 9 '13 at 13:06
I know that 2S-I and I-S are firmly nonexpansive . I thought maybe 'nonexpansive expansion of a nonexpansive function is nonexpansive' and 'sum of two nonexpansive functions are nonexpansive' , so I can say that Z i nonexpansive too. – Ilkay Jan 9 '13 at 13:56
Please edit your question to reflect the question you actually want to ask, not the intermediate step (which is not true by my first comment). – Willie Wong Jan 9 '13 at 14:04

It is false. Let $T = S = I$. Then $T+S = 2I$ which once you plug in you see that is not firmly nonexpansive. Do you perhaps mean to consider $\frac12(T+S)$? In which case the desired conclusion follows from triangle inequality applied many times.