Let $H$ be a Hilbert space. Let $L: H \rightarrow H$ be a linear bounded positive operator (i.e. $\langle L(u), u \rangle \geq 0$ for all $u \in H$).

A) Prove that $I+aL$ is bijective for every $a > 0$ (I is the identity map)

B) Show that $\lim\limits_{a \rightarrow \infty}(I+aL)^{-1}u = Pu$ (where P is the orthogonal projection onto $Ker(L)$

For A, injectivity seems straightforward (if we use positivity). For surjectivity I defined:

$$S = S_a = \sum\limits_{n=0}^{\infty} (-aL)^n$$

Then, after observing that $(I+aL)(S(v)) = v$, I concluded that $S(v)$ is a pre-image of $(I+aL)$, which makes $(I+aL)$ surjective. But is S a valid operator?

For B, I decomposed $v = v_1 + v_2$ and $S(v) = u_1 + u_2$ (according to the orthogonal projection theorem) and then used uniqueness to show that:

$$L(u_2) = \frac{v_2-u_2}{a}$$

From there, I think that if $a \rightarrow \infty$ then $L(u_2) \rightarrow 0$ (and thus $u_2 \rightarrow 0$ and the proof follows). Is this correct? The fact that $u_2$ also shows up on the RHS makes me a bit uneasy.

  • $\begingroup$ you mean $\ker L$ instead of $T$? $\endgroup$ – el_tenedor Apr 6 '17 at 17:40
  • $\begingroup$ I find it hard to believe that the operator $S$ is well defined for all $a > 0$ how do we know it converges. I understand you probably tried 1/(1+aL) and used the series expansion. However, that is valid on if $\|aL\| < 1$. $\endgroup$ – Gregory Apr 6 '17 at 17:47
  • $\begingroup$ Why should the series $S_a$ converges? $\endgroup$ – user251257 Apr 6 '17 at 17:48
  • $\begingroup$ I do mean $Ker(L)$ thanks. @Gregory That was one of main concerns too. But how do we prove it's surjective then? It doesn't seem obvious to me. $\endgroup$ – Pellenthor Apr 6 '17 at 17:50

For A) you can use Lax-Milgram:

$$ \langle aLu + u, u \rangle = \langle aLu, u \rangle + \langle u, u \rangle \geq \|u\|^2. $$

This shows, that $I + aL$ is invertible with $\|(I + aL)^{-1}\| \leq 1$, for all $a > 0$.

For B): If $u \in \ker L$ we have $Lu = 0$. This implies $$ (aL + I) u = u, \text{ for all } a > 0 $$ and thus $$ (aL + I)^{-1}u = u, \text{ for all } a > 0. $$ Taking the limit $a \to \infty$ shows, that $(aL + I)^{-1}$ works as the identity on $\ker L$.

Note that since $L$ is positive it is in particular self adjoint. This gives you $$ (\ker L)^\perp = \overline{\operatorname{im} L^*} = \overline{\operatorname{im} L}. $$

Now if $u \in \operatorname{im} L$ there is some $v$ such that $u = Lv$. We can calculate $$ (aL + I)^{-1} Lv = (aL + I)^{-1} \frac{1}{a} (aL + I - I) v =\frac{1}{a} \|v\|- \frac{1}{a} (aL + I)^{-1} v, $$ which gives $$ \|(aL + I)^{-1} Lv\| \leq \frac{1}{a} v + \frac{1}{a} \|(aL + I)^{-1}\| \|v\| \leq \frac{2}{a} \|v\| \to 0, \text{ for } a \to \infty. $$

We just proved that $$\lim_{a \to \infty}(aL + I)^{-1}w$$ for all $w \in \operatorname{im}L$. But since the net $(aL + I)^{-1}$ is uniformly bounded in operator norm this lets us extend the convergence to the closure $\overline{\operatorname{im}L}$:

Let $w \in \overline{\operatorname{im} L}$. Let $\varepsilon > 0$ be given. Then, there exists $Lv \in \operatorname{im} L$ such that $$\|w - Lv\| \leq \frac{\varepsilon}{2}.$$ Furthermore, we have some $\tilde a > 0$ such that for all $a \geq \tilde a$ the inequality $$\|(aL + I)^{-1} Lv\| < \frac{\varepsilon}{2}$$ holds. Thus for all $a \geq \tilde a$ we have $$ \|(aL + I)^{-1} w\| = \|(aL + I)^{-1} (w - Lv + Lv) \| \leq \|w - Lv\| + \|(aL + I)^{-1} Lv\| \leq \varepsilon. $$ This proves $$\lim_{a \to \infty} (aL + I)^{-1} w = 0.$$

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    $\begingroup$ Thank you. So the full conclusion of Lax-Milgram (with all the details) would be something like this: ...and thus there is a unique $v \in H$ such that $\langle aLv + v, u\rangle = \langle u, u\rangle$ for all $u$, which means that $(I+aL)(v) = u$. This shows that $I +aL$ is invertible, On top of that $\Vert v \Vert = \Vert (I+aL)^{-1} \Vert \leq (1/c)*M = 1$ (where $c=1$ is the inequality constant and $M=1$ is the (trivial) operator norm. Is that correct? $\endgroup$ – Pellenthor Apr 6 '17 at 18:37
  • $\begingroup$ @Pellenthor: Sounds good! $\endgroup$ – el_tenedor Apr 6 '17 at 18:41
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    $\begingroup$ Isn't it the case that: $u \in Ker(L^*) \Leftrightarrow L^*(u) = 0 \Leftrightarrow \langle v, L^*(u) \rangle = 0 (\forall v \in H) \Leftrightarrow \langle L(v), u \rangle = 0 (\forall v \in H) \Leftrightarrow$ $u \in Im(L)^{\perp}$ and thus $Ker(L) = Ker(L^*) =Im(L)^{\perp}$? (so we don't have to worry about the closure) $\endgroup$ – Pellenthor Apr 6 '17 at 18:45
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    $\begingroup$ $\|(I+aL)^{-1}u \| \le \|(I+aL)^{-1}u_n \| + \|u-u_n\|$ where $u_n\in im(L)$. $\endgroup$ – user251257 Apr 6 '17 at 18:45
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    $\begingroup$ @Pellenthor: $((A^\perp)^\perp = \overline A$ $\endgroup$ – el_tenedor Apr 6 '17 at 19:00

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