Let $H$ be a Hilbert space and let $E\subset H$ be an orthonormal set in $H$. For $F\subset E$, let $P_{F}\colon H\to H$ denote the orthogonal projection on the closed linear span $\bigvee F$ of $F$. Now let $F_{1},F_{2},\ldots\subset E$ be pairwise disjoint subsets of $E$. I have shown that for every $h\in H$, we have $$\sum_{k\geq0}P_{F_{k}}h=P_{\bigcup_{k\geq0}}h.$$ Now I want to say something about the convergence (or non-convergence) of $\sum_{k\geq0}P_{F_{k}}$ in $\mathscr{B}(H)$ with respect to the operator norm. Sadly enough, I can't even think of an example that satisfies the conditions of this setting. Any suggestions to get me started would be greatly appreciated.
Does the sum of infinitely many orthogonal projections converge with respect to the operator norm
functional-analysishilbert-spacesorthogonalityprojectionsequences-and-series
Related Solutions
Let $x\in X$. We have $P_n(x)=\sum_{j=1}^n\langle x,e_j\rangle e_j$ so , by Bessel-Parseval equality $$\lVert P_nx-x\rVert^2=\sum_{j\geq n+1}|\langle x,e_j\rangle|^2.$$ As the latest series is convergent we have the result.
Now, let $K\subset H$ compact. Fix $\varepsilon >0$. Then we can find an integer $N$ and $x_1,\dots,x_N\in K$ such that for each $x\in K$, we can find $1\leq k \leq N$ such that $\lVert x-x_k\rVert\leq\varepsilon$. Fix $x\in K$. Then $$\lVert P_nx-x\rVert^2=\sum_{j\geq n+1}|\langle x-x_k+x_k,e_j\rangle|^2\leq \varepsilon^2+\max_{1\leq k\leq N}\sum_{j\geq n+1}|\langle x_k,e_j\rangle|^2.$$ As the RHS doesn't depend on $x$, we have $$\sup_{x\in K}\lVert P_nx-x\rVert^2\leq \varepsilon^2+\max_{1\leq k\leq N}\sum_{j\geq n+1}|\langle x_k,e_j\rangle|^2.$$ Now take the $\limsup_{n\to+\infty}$ to get the result.
- If $T$ is compact, then $K:=\overline{T(B(0,1))}$ is compact, so we apply the previous result to this $K$.
Note that the property of approximation of a compact operator by a finite rank operator is true in any Hilbert space, not only in separable ones. To see that, fix $\varepsilon>0$; then take $v_1,\dots,v_N$ such that $T(B(0,1))\subset \bigcup_{j=1}^NB(y_j,\varepsilon)$. Let $P$ the projection over the vector space generated by $\{y_1,\dots,y_N\}$ (it's a closed subspace). Consider $PT$: it's a finite ranked operator. Now take $x\in B(0,1)$. Then pick $j$ such that $\lVert Tx-y_j\rVert\leq\varepsilon$. We also have, as $\lVert P\rVert\leq 1$, that $\lVert PTx-Py_j\rVert\leq \varepsilon$. As $Py_j=y_j$, we get $\lVert PTx-Tx\rVert\leq 2\varepsilon$.
What you defined is not idempotent (i.e. $P^2\neq P$) in general.
Case N=1: when $v\neq 0$, the projection (= self-adjoint idempotent, i.e. range and nullspace orthgonal to each other) onto $\mbox{span} \,v$ is given by $$ P_vh=\frac{(h,v)}{\|v\|^2}v. $$ In your case, you used the formula $Ph=(h,v)v$ which yields $P^2=\|v\|^2P$. So that's an idempotent if and ony if $\|v\|=1$.
Note that more generally, an idempotent (not necessarily self-adjoint) with range $\mbox{span} \,v$ is given by $$ P_{u,v}h=(h,u)v\quad\mbox{with}\;(u,v)=1 $$ where $u$ is chosen so that $\ker P=u^\perp$. Note that in this case, we have $$ (P_{u,v})^*=(\cdot,v)u=P_{v,u}. $$ This describes all the rank one idempotents.
Case N=2: let us look at the simple case $H=\mathbb{C}^2$, $v_1=(1,0)$, $v_2=(1,1)$. Then $\mbox{span}(v_1,v_2)=\mathbb{C}^2$, so the projection onto the latter is just $P=Id$. Now if you compute the matrix of $P=(\cdot,v_1)v_1+(\cdot,v_2)v_2$ in, say, the canonical basis, you find $$ P=\pmatrix{2&3\\1&2} $$ which is far from $I_2$. It is not hard to see that such a formula can work if and only if $(v_1,v_2)=0$ and $\|v_1\|=\|v_2\|=1$, that is $(v_1,v_2)$ orthonormal. But what you can do is look for $(u_1,u_2)$ such that $$ Id=(\cdot,u_1)v_1+(\cdot,u_2)v_2\iff (v_i,u_j)=\delta_{ij}\quad 1\leq i,j\leq 2. $$ That is $(u_1,u_2)$ dual basis of $(v_1,v_2)$ with respect to the duality induced by the inner product.
General case: the formula $P=\sum_{j=1}^N(\cdot,v_j)v_j$ yields the projection onto $V$ if and only if $(v_j)$ is an orthonormal basis of $V$. Otherwise, $P$ is of the form
$$ P=\sum_{j=1}^nP_{u_j,v_j}=\sum_{j=1}^n(\cdot,u_j)v_j $$ where there is one and only one choice for the $u_j$'s, namely $$ u_1,\ldots,u_n\in V\qquad (u_i,v_j)=\delta_{ij}\quad 1\leq i,j\leq n. $$ That is $(u_j)$ is the dual basis of $(v_j)$ in $V^*$ identified with $V$ via the inner product and Riesz representation.
Best Answer
Not true. Let $(e_n)$ be an orthonormal basis for $H$. Let $F_k=\{e_k\}$. Then $P_{F_k}x= \langle x, e_k \rangle e_k$ and the series $\sum \langle x, e_k \rangle e_k$ does not converge in operator norm becasue the operator norm of the general term is $1$: $\sup \{\|\langle x, e_k \rangle e_k\|: \|x\| \leq 1\} =1$ for each $k$.