Positive operator on a Hilbert Space 
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.    
 A: 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 $$\lim_{a \to \infty} (aL + I)^{-1} w = 0.$$
