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4

Dominated convergence and related theorems don't apply here anyway. The norm here is almost certainly the supremum norm since this is the norm that makes $C([0,1])$ a complete space. Also, you'll need to to know something about $K$ in order to do this. For example, if $K$ is continuous the following argument works: what you need to do is find a $C > 0$ ...


3

The question requires that $A$ has no zero row, because otherwise the result is not true. For instance, consider $$ A=\begin{bmatrix}1&1&0\\0&0&1\\0&0&0\end{bmatrix} $$ More than that, the question requires that the rows of $A$ are orthonormal (more properly, that all rows have the same norm). Otherwise, consider $$ ...


2

Note: It should be $A_{ij} = \langle A e_j, e_i \rangle$, not $\langle A e_i, e_j \rangle$. Observation 1: The $i$-th component of the transform $Av$ of any column vector $v = [v_1 \quad v_2 \quad \cdots \quad v_n]'$ is $(Av)_i = {\large\sum\limits_k} A_{ik}v_k$. Observation 2: The $i$-th component of the column vector representation of the $j$-th basis ...


2

First, let $$ a = Ae_i = \begin{bmatrix} A_{11} & \ldots & A_{1n} \\ A_{21} &\ldots &A_{2n} \\ \vdots &\ddots &\vdots \\ A_{n1} & \ldots & A_{nn} \end{bmatrix} \begin{bmatrix} 0 \\ \vdots \\ 1 \\ \vdots \\ 0\end{bmatrix} $$ $e_i$ has a $1$ at the $i$th row and $0$ everywhere else. Hence, on multiplying, it will "pick ...


2

I'm not totally sure what the hint is getting at (perhaps consider the integral as the $L^2$ inner product of $f(x) = \frac 1 {x^2} - a\frac 1 {x^3} - b \frac 1 {x^4}$ with itself), but regardless this seems like a simple calculus problem to me. Put \begin{align*} I(a,b) &= \int^\infty_1 \left( \frac 1 {x^2} - a\frac 1 {x^3} - b \frac 1 {x^4} \right)^2 ...


1

With the change of variable $x=\frac{1}{u}$ we are left with: $$ \min_{a,b} \int_{0}^{1}\left(x-ax^2-bx^3\right)^2\,dx $$ and now we may exploit the fact that shifted Legendre polynomials give a complete orthogonal base of $L^2(0,1)$ with respect to the usual inner product. Since: $$ P_0(2x-1) = 1,\quad P_1(2x-1)=2x-1,\quad P_2(2x-1)=6x^2-6x+1,\quad ...


1

For hermitian $T$ you have $\ker(T) \cap \rm{im}(T) = \{0\}$, as $\langle x,x \rangle = 0 \iff x=0$, so if $x = Tz$ and $x \in \ker(T)$ you have $\langle x, x\rangle=\langle x, Tz\rangle=\langle Tx,z\rangle = 0$. Now for a positive bounded operator $T$ on a Hilbert space you have a root $T^{1/2}$ that is also positive. Then $\langle x, Tx\rangle=\langle ...



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