Cfrgtkky

I've been trying to pin down the following intuitive statement:

Let $pi$ be a k-dimensional subspace in $mathbb R^N$, with ON basis $e_1,ldots, e_k$, and extend to an ON basis $e_1,ldots, e_k, f_1,ldots, f_{N-k}$ of $mathbb R^N$. Let $v_1,ldots, v_{N-k}$ be vectors in $mathbb pi.$

Now define the tilted plane $barpi$ by $$barpi={y+L(y) : yin pi} $$ where $$ L(y):= sum_{j=1}^{N-k}(v_jcdot y)f_j$$ Thus $barpi$ is $pi$ after being ``tilted'' in the directions of the $v_j$.

The estimate I want is the following: $$|P_pi-P_{barpi}|leqslant Csum_{j=1}^{N-k}|v_j|^2$$ (or possibly a $1$ for the last exponent) for a purely dimensional constant C not depending upon the $v_j$, where the $P_{-}$ are the orthogonal projections onto the respective planes.

As for progress I've made, a basis $zeta_1,ldots,zeta_k$ of $barpi$ defined by $$ zeta_i=e_i+L(e_i)=e_i+sum_{j=1}^{N-k}(v_jcdot e_i)f_j$$ should yield the metric $$g=I+L^tL$$ where $L$ is the matrix whose $i,j$ entry is $v_icdot e_j$.

This yields an estimate of the desired form $$|g-I|leqslant Csum_{j=1}^{N-k}|v_j|^2$$.

If the RHS was small enough, we could apply the Neumann series theorem and invert $g$ and get estimates, but I cant do this without choosing a new norm depending on the $v_j$, thus introducing non-dimensional dependencies into the constant C.

At any rate, just subtracting the projections on a vector $x$ of norm 1 yields something like $$P_{barpi}x-P_pi x=(delta^{ij}-g^{ij})(xcdot e_i)e_j-g^{ij}(xcdot L(e_i))e_j-g^{ij}(xcdot e_i)L(e_j)-g^{ij}(xcdot L(e_i))L(e_j).$$ Sooooo----any estimate has to come from the $g^{ij}$, and I am at a loss for how to do this. Of course, an estimate on the inverse metric would yield an estimate on $g^{ij}-delta^{ij}=g^{ik}(delta_{kj}-g_{kj})$ taking care of the first term as well. A potential issue I've come across in my scratch work is that it's very easy to introduce higher powers of the $|v_j|$ which I can't control, as far as I can tell.

Perhaps a related question is what may be said about estimating the norm of the matrix $(I+A)^{-1}$ where $A$ is symmetric. I've seen some posts around that give some rather convoluted expressions involving $A^{-1}$, which I would rather not deal with if possible since then I need estimates on the entries of the inverse.

Through and through all of this, I am quite shocked that this hasnt been more immediate--the statement is so intuitive after all!

Thanks so much!

A friend came up with the following simplified approach:

Let $T:=P_{barpi}-P_pi$, and let $C=sum_{j=1}^{N-k}|v_j|$. We aim to show that $$|Tx|leqslant C|x| $$ for all $xinmathbb R^N$.

It suffices to show this boundedness separately when $xin pi$ and when $xinpi^perp$, since in any $zinmathbb R^n$ has an orthogonal decomposition with components in $pi$ and $pi^perp$, allowing us to estimate $$|Tx|=|Tx^mathrm{T}+Tx^perp|leqslant |Tx^mathrm{T}|+|Tx^perp|leqslant C|x^mathrm{T}|+C|x^perp|leqslant sqrt{2}C|x|$$ where the last inequality follows from Jensen.

Thus, in case $xinpi$ and $|x|=1$, we have $$ |P_{pi}x-P_{barpi}x|=|x-P_{barpi}x|leqslant|x-y| $$ for all $yinbarpi$. In particular the inequality holds for $y=x+L(x)$, and the estimate follows.

In case $xinpi^perp$ and $|x|=1$, we have $$|P_{barpi} x-P_pi x|^2=|P_{barpi} x|^2=langle P_{barpi} x,P_{barpi} xrangle =frac{langle x,P_{barpi} xrangle^2}{|P_{barpi} x|^2}=frac{langle x,y+L(y)rangle^2}{|y+L(y)|^2}$$ for some $yin pi$ which yields

$$|P_{barpi} x-P_pi x|leqslantleft(frac{|L(y)|^2}{|y|^2+|L(y)|^2}right)^{1/2}leqslant =left(frac{|L(hat y)|^2}{1+|L(hat y)|^2}right)^{1/2}leqslant C$$ where we have written $hat y=y/|y|$.

And that's it!