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Let $\left{f_j\right}$ be a sequence of functions in $L^2(\mathbb{R})$. We say $f_j \rightarrow f$ strongly in $L^2$ if there exists $f \in L^2(\mathbb{R})$ such that
$$
\left|f_j-f\right|_{L^2(\mathbb{R})} \rightarrow 0
$$
Give an example of a sequence $\left{f_j\right}$ and a function $f$ such that $f_j \rightarrow f$ strongly in $L^2$ but $f_j$ does not converge to $f$ pointwise. Prove both of these facts about your example.
Consider the sequence of functions $\left{f_j\right}$ defined as follows:
$f_j(x)= \begin{cases}j, & \text { if } x \in\left[0, \frac{1}{j}\right] \ 0, & \text { otherwise }\end{cases}$
for $j\geq 1$.
Note that each $f_j$ is in $L^2(\mathbb{R})$ since
$\int_{-\infty}^{\infty}\left|f_j(x)\right|^2 d x=\int_0^{1 / j} j^2 d x=\frac{1}{j} \rightarrow 0 \quad$ as $j \rightarrow \infty$
Also, $f_j$ converges to $0$ pointwise everywhere on $\mathbb{R}$ since $\frac{1}{j}<x$ implies $f_j(x)=0$ for all $j\geq 1$.
To show that $f_j$ converges strongly to $0$ in $L^2(\mathbb{R})$, we calculate \begin{align*} \left|f_j-0\right|{L^2(\mathbb{R})}^2 &= \int{-\infty}^{\infty}|f_j(x)|^2dx\ &=\int_{0}^{1/j}j^2dx\ &=\frac{1}{j} \rightarrow 0 \quad \text{as }j\rightarrow \infty. \end{align*} Therefore, $f_j$ converges strongly to $0$ in $L^2(\mathbb{R})$.
However, $f_j$ does not converge pointwise to $0$ since, for example, $f_1(x)=1$ for $x\in[0,1]$ and $f_j(x)=0$ for $x\in[0,1/j]$ when $j\geq 2$.
Therefore, we have shown that $f_j$ converges strongly to $0$ in $L^2(\mathbb{R})$ but $f_j$ does not converge pointwise to $0$.
Let $X, Y$ be $n \times n$ matrices such that $X^3=Y^3$ and $X^2 Y=Y^2 X$. What are necessary and sufficient conditions on $X$ and $Y$ such that $X^2+Y^2$ is invertible?
$X^3=Y^3$ and $X^2 Y=Y^2 X$ taken together allow us to write
$$
\left(X^2+Y^2\right) X=X^3+Y^2 X=Y^3+X^2 Y=\left(X^2+Y^2\right) Y \text {. }
$$
Notice that if $X \neq Y$ then $X^2+Y^2$ cannot be invertible. Thus, a necessary condition is that $X=Y$.
Let $T: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ and $S: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ correspond to the transformations
$$
T(x, y, z)=(x, y) ; S(x, y, z)=(-x,-y,-z) .
$$
Notice that $T S$ is a well defined linear transformation. Find a matrix representation for $S$, $T$, and $T S$ using the basis ${(1,0,0),(1,1,0),(1,1,1)}$ for $\mathbb{R}^3$ and the basis ${(1,0),(1,-1)}$ for $\mathbb{R}^2$.
First note that $T(1,0,0)=(1,0)=1 \cdot(1,0)+0 \cdot(1,-1) ; \quad T(1,1,0)=$ $(1,1)=2 \cdot(1,0)-1 \cdot(1,-1) ; \quad T(1,1,1)=(1,1)=2 \cdot(1,0)-1 \cdot(1,-1)$ and thus the matrix for $T$ in these bases is
$$
\left(\begin{array}{ccc}
1 & 2 & 2 \
0 & -1 & -1
\end{array}\right)
$$
To find the matrix for $S$ we perform the same process:
$$
\begin{aligned}
& S(1,0,0)=(-1,0,0)=-1 \cdot(1,0,0)+0 \cdot(1,1,0)+0 \cdot(1,1,1) \
& S(1,1,0)=(-1,-1,0)=0 \cdot(1,0,0)-1 \cdot(1,1,0)+0 \cdot(1,1,1) \
& S(1,1,1)=(-1,-1,-1)=0 \cdot(1,0,0)+0 \cdot(1,1,0)-1 \cdot(1,1,1)
\end{aligned}
$$
Thus the matrix for this transformation is
$$
\left(\begin{array}{ccc}
-1 & 0 & 0 \
0 & -1 & 0 \
0 & 0 & -1
\end{array}\right)
$$
Thus we find the matrix for $T S$ by multiplication:
$$
\left(\begin{array}{ccc}
1 & 2 & 2 \
0 & -1 & -1
\end{array}\right)\left(\begin{array}{ccc}
-1 & 0 & 0 \
0 & -1 & 0 \
0 & 0 & -1
\end{array}\right)=\left(\begin{array}{ccc}
-1 & -2 & -2 \
0 & 1 & 1
\end{array}\right)
$$
