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The question posed is the following: Let $X$ be a Banach Space and let $T:X\to X$ be a Lipschitz-Continuous map. Show that, for $\mu$ sufficiently large, the equation \begin{equation} Tx+\mu x=y \end{equation} has, for any $y\in X$, a unique solution.

Note that $x,y$ are vectors, since our book (Mathematical Analysis by Mariano Giaquinta and Giuseppe Modica) generally ignores vector indicators, since it's all multivariable.

My proof is based on the Banach Fixed Point Theorem: Since $T$ is Lipschitz-continuous, we have $\|Tx\|\leq k\|x\|$ for $0<k\leq1$. So $\|Tx-\mu x\|\leq k\|x\| - \mu \|x\|$.

Then we can say

\begin{equation} \|Tx-\mu x\|\leq (k-\mu)\|x\| \end{equation}

So, if $\mu$ is large enough that $|k-\mu|<1$, we have a contractive map, and by the Banach Fixed Point theorem, there exists a unique fixed point $x_0$ for $(T-\mu)x$. Then, $Tx-\mu x=y$ has a unique solution, namely, $x_0$.

My question is whether this is a valid proof. I'm mostly foggy on if I applied the theorem correctly, and if I am allowed to say $Tx-\mu x=(T-\mu)x$, since $T$ is a map and $\mu$ is a constant (I think).

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  • $\begingroup$ What does $Tx \le kx$ mean? Both sides are vectors $\endgroup$ – mathworker21 Oct 7 '18 at 22:06
  • $\begingroup$ Hm, good point. I believe the definition for vectors is in relation to the norms. I'll correct the statement. $\endgroup$ – Khaled Allen Oct 7 '18 at 22:10
  • $\begingroup$ For the record, the notation $\overrightarrow x$ (or $\vec x$) has been abandoned, if it ever existed, in mathematics textbooks. To the best of my knowledge, at least. $\endgroup$ – Saucy O'Path Oct 7 '18 at 22:14
  • $\begingroup$ You cannot assume $k \le 1$. Further, Lipschitz continuity is a different property than $\|Tx\| \le k \|x\|$. Please look it up. Also, the inequality $\|Tx - \mu x \| \le k\|x\| - \mu \|x\|$ cannot be correct. Think about what this means for $\mu > k$. $\endgroup$ – Hans Engler Oct 7 '18 at 22:29
  • $\begingroup$ Ok. I misunderstood Lipschitz-continuity apparently. So, we can say $\|Tx_1-Tx_2\|\leq k\|x_1-x_2\|$, for $k>0$. Since this is matrix-vector multiplication, I'm pretty sure I can say $\|T(x_1-x_2)\|\leq k\|x_1-x_2\|$ If we let some $x=x_1-x_2$, can I then say $\|Tx\|\leq k\|x\|$? $\endgroup$ – Khaled Allen Oct 7 '18 at 22:44
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Rewrite the equation as $$ x=\frac1{\mu}\,(y-Tx)=Gx, $$ and show that $G$ is a contraction for large enough $\mu$.

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Here was my final answer for reference:

Let $Gx=\frac{1}{\mu}\left(y-Tx\right)$. Then \begin{align*} |Gx_1-Gx_2|&=\left|\frac{1}{\mu}\left(y-Tx_1\right)-\frac{1}{\mu}\left(y-Tx_2\right)\right|\\ &=\frac{1}{\mu}\left|Tx_1-Tx_2\right|\\ &\leq \frac{1}{\mu} k |x_1-x_2|\tag{For $0<k\leq 1$} \end{align*} Then for $\mu > k$, $G$ is a contractive map, and by the Banach Fixed Point theorem, has a fixed point $\bar{x}$ such that $G\bar{x}=\bar{x}$.

Then $G\bar{x}=\bar{x}=\frac{1}{\mu}(y-T\bar{x})$, and thus the equation $Tx-\mu x=y$ has the unique solution $\bar{x}$.

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