# Direct sum for linear closed and bounded below operators.

Let $U$ and $V$ be Hilbert linear spaces. A is a linear closed and bounded below operator $A:U\to V$

How to show that: \begin{equation*} V=R(A)\oplus {R(A)}^\perp. \end{equation*} I have read a theorem about the direct sum of the range and kernel of linear operators but I cannot apply it for orthogonal subspace of $A$. Any help will be appreciated.

• Well for any closed subspace $V_0$ of $V$, we have $V=V_0\oplus V_0^\perp$. This just amounts to showing that $R(A)$ is closed. – Aweygan May 10 '17 at 0:28
• Could you please explain to me why is that ? – user429382 May 10 '17 at 23:37

Hint: Let $y_n=A(x_n)$, suppose that $y_n$ converges towards $y$, this implies that $y_n$ is a Cauchy sequence, since $A$ is bounded below, $x_n$ is also a Cauchy sequence, so $x_n$ converges towards $x$ since $U$ is complete, we have $y=A(x)$ so $R(A)$ is closed.
Since $A$ is bounded below, there exists $d$ such that $\|A(u)\|\geq d\|u\|$,
Since $y_n$ is a Cauchy sequcence, for every $c>0$, there exists an integer $N_c$ such that $n,m>N_c$ implies that $\|y_n-y_m\|<cd$. , we deduce that $\|y_n-y_m\|=\|A(x_n-x_m)\|\geq d\|x_n-x_m\|$. This implies that for every $n,m>N_c, \|x_n-x_m\|<c$. We deduce that $(x_n)$ is a Cauchy sequence.
• Could you please show me how the convergence of $y_n$ implies that $x_n$ is Cauchy? Many thanks, @tsemo – user429382 May 10 '17 at 0:52
• By definition of bounded below $||Tx_n-Tx_m||\geq c||x_n-x_m||$ hence $y_n$ Cauchy sequence implies $x_n$ Cauchy sequence – JJR May 10 '17 at 0:59