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Exercises 5.9 Exercises

1.

Suppose that \(A\) is an \(m\times n\) matrix with rank \(n\text{,}\) and that \(m\ge n\text{.}\)
  1. Prove that there exists some \(\lambda \gt 0\) such that \(\Vert A\bx \Vert \ge \lambda \Vert\bx \Vert\) for all \(\bx\in \bbR^{n}\text{.}\)
  2. Prove that there exists some \(\delta >0\) such that any \(m\times n\) matrix \(B\) satisfying \(\Vert B-A\Vert \lt \delta\) also has rank \(n\text{.}\)

2.

Suppose that \(\bff\) is differentiable at \(\ba \in \bbR^{n}\) and takes values in \(\bbR^{m}\) with \(m\ge n\text{,}\) and that \(D\bff (\ba)\) has rank \(n\text{.}\)
  1. Prove that there exist some \(r >0\) and \(\lambda > 0\) such that \(\Vert \bff (\bx) -\bff (\ba)\Vert \ge \lambda \Vert \bx -\ba \Vert \) for all \(\bx\in \bbR^{n}\) such that \(\Vert \bx -\ba \Vert \le r\text{.}\)
  2. Prove that there exists some \(\delta >0\) such that for all \(\by \in B(\bff(\ba), \delta)\text{,}\) the open ball of radius \(\delta\) centered at \(\bff(\ba)\text{,}\)
    \begin{equation*} \min_{\bx\in \overline{B(\ba, r)}} \Vert \bff (\bx) - \by \Vert \end{equation*}
    is attained at some \(\bx^{*} \in B(\ba, r)\text{,}\) and that
    \begin{equation*} \left( \bff (\bx^{*}) - \by\right)\cdot \left( D\bff(\bx^{*})\bh \right)=0 \quad \text{ for all $\bh \in \bbR^{n}$.} \end{equation*}
    \(D\bff(\bx^{*})\bh\) is a tangent vector to the image of \(\bff\) at \(\bff(\bx^{*})\text{,}\) so the above statement shows that \(\bff(\bx^{*})\) is closest to \(\by\) among \(\bff(\bx), \bx\in \overline{B(\ba, r)}\) and \(\bff (\bx^{*}) - \by\) is orthogonal to all tangents to the image of \(\bff\) at \(\bff(\bx^{*})\text{.}\)
  3. In the set up above, assume furthermore that \(D\bff(\bx)\) is continuous in \(\bx\text{,}\) then prove that \(r>0\) can be adjusted so that \(D\bff(\bx)\) has rank \(n\) for all \(\bx \in B(\ba, r)\text{.}\) Assume in addition that \(m=n\text{.}\) Prove that \(\bff (\bx)=\by\) in the above and \(r>0\) above can be adjusted so that such an \(\bx\) is unique in \(B(\ba, r)\text{.}\)

3.

Verify that the function \(A\in \cM^{n\times n} \mapsto A^{2}\) is a differentiable function, where \(\cM^{n\times n}\) is the space of \(n\times n\) matrices. Then determine its derivative and check whether it is invertible in the case of \(A=\begin{bmatrix} 1 \amp 0 \\ 0 \amp 1\end{bmatrix}\) and \(A=\begin{bmatrix} 1 \amp 0 \\ 0 \amp -1\end{bmatrix}\) respectively.
Show that there exist neighborhoods \(U, V\) of \(\begin{bmatrix} 1 \amp 0 \\ 0 \amp 1\end{bmatrix}\) such that \(S:U\mapsto V\) has an inverse which is differentiable, namely, any matrix in \(V\) has a uniquely defined square root in \(U\text{.}\) Can one do the same around \(\begin{bmatrix} 1 \amp 0 \\ 0 \amp -1\end{bmatrix}\text{?}\)
Hint.
It may not be practical to work out the Jacobian matrix in the usual matrix form, but it suffices to obtain a linear function \(L(H)\in \cM^{n\times n}\) of \(H\in \cM^{n\times n}\) such that \(\Vert (A+H)^{2}-A^{2}-L(H)\Vert/\Vert H\Vert \to 0\) as \(\Vert H\Vert \to 0\text{.}\)

4.

Consider the \(N\)th Fourier series partial sum \(s_{N}(f; x)\) of \(f\in C[-\pi, \pi]\) as a linear map from \(C[-\pi, \pi]\) to itself. Using the integral expression for \(s_{N}(f; x)\) to show that its operator norm is \(\frac{1}{2\pi} \int_{-\pi}^{\pi} |D_{N}(t)|\,dt\text{.}\)

5.

Prove a lower bound of \(\frac{\int_{x_{1}}^{x_{2}}f(x)g(x)\, dx}{\sqrt{\left( \int_{x_{1}}^{x_{2}}f(x)^{2}\,dx\right) \left( \int_{x_{1}}^{x_{2}}g(x)^{2}\,dx\right) }}\) when \(f(x), g(x)\) are subject to positive upper and lower bounds. This problem is from #93 in Part II, Chapter 2 of Polya and Szeg’s classic β€œProblems and Theorems in Analysis I”
Let \(a, A, b, B\) be positive numbers such that \(a\lt A, b\lt B\text{.}\) If the two functions \(f(x) \) and \(g(x)\) are integrable over the interval \([x_{1}, x_{2}]\text{,}\) and \(a\le f(x)\le A, b\le g(x)\le B\) on the interval. Then
\begin{equation*} \frac{\int_{x_{1}}^{x_{2}}f(x)g(x)\, dx}{\sqrt{\left( \int_{x_{1}}^{x_{2}}f(x)^{2}\,dx\right) \left( \int_{x_{1}}^{x_{2}}g(x)^{2}\,dx\right) }}\ge \frac{2}{\sqrt{\frac{AB}{ab}+\sqrt{\frac{ab}{AB}}}}. \end{equation*}