# why is this a contraction map?

Let $B$ be a Banach space and $V$ be a normed linear space and $L_0$, $L_1$ be bounded lienar operators from $B$ to $V$. For each $t\in[0,1]$, set

$$L_t=(1-t)L_0+tL_1$$

and suppose that there is a constant $C$such that $$||x||\leq C||L_tx||\quad (*)$$ for $t\in[0,1]$. Then $L_1$ maps $B$ onto $V$ if and only if $L_0$ maps $B$ onto $V$.

Proof: Suppose $L_s$ is onto for some $s\in [0,1,]$. By (*), $L_s$ is one-to-one and hence the inverse mapping $L^{-1}_s:V\rightarrow B$ exists. For $t\in [0,1]$ and $y\in V$, the equation $L_tx=y$ is equivalent to the equation

$$L_s(x) = y +(L_s-L_t)x=y+(t-s)L_0x -(t-s)L_1x$$

which in turn, is equivalent to

$$x = L_s^{-1}y+(t-s)L_s^{-1}(L_0-L_1)x$$

The mapping $T$ from $B$ into itself given by $Tx = L_s^{-1}y+(t-s)L^{-1}_s(L_0-L_1)x$ is clearly a contraction mapping if $$|s-t|<\delta=[C(||L_0||+||L_1||)]^{-1}$$ and hence the mapping $L_t$ is onto for all $t\in [0,1]$, satisfying $|s-t|<\delta$

I do not really understand the part in bold. Firstly I cannot see why it is a contraction map and why implies $L_t$ is onto.

Take an arbitrary $y\in V$. $$Tx_1-Tx_2=(t-s)L_s^{-1}(L_0-L_1)(x_1-x_2)$$ and hence $$||Tx_1-Tx_2||\le |t-s|||L_s^{-1}||||L_0-L_1||||x_1-x_2||\le |t-s|C(||L_0||+||L_1||)||x_1-x_2||$$ Thus, $T$ is a contraction for $|s-t|<\delta$ and has a fixed point $x_*$, $Tx_*=x_*$, by the contraction mapping theorem. But then $L_tx_*=y$.
• this is what I do not understand, does $y$ not depend on $x_1$ or $x_2$? Has the author just took an arbitrarily fixed $x$ and $y$? – Lost1 Jul 1 '14 at 18:52
• You take and fix $y$ arbitrarily and define $T$ by your formula. Thus, the operator $T$ depends on $y,s,$ and $t$. (It is not a linear operator, by the way, because $y$ is fixed). $T$ maps $B$ into $B$. If, for some $x$, $Tx=x$, then, for the same $x$, $L_tx=y$. We have proved that $T$ is a contraction whatever $y$ we take. (Provided that $|t-s|$ is small enough.) Thus, $L_tx=y$ is solvable for every $y$. – Vladimir Jul 1 '14 at 18:55