Modern Abstract Analysis

For functional analysis, measure theory, and related areas. M...
May 21, 2018 16:19
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Q: Using the Fredholm Alternative to Prove that a Transformation is a Bijection

Harry EvansLet $I$ be an interval in $\mathbb{R}.$ For $k \in \mathbb{N}$, let $g_k, h_k \in L^2 (I)$ where $$\sum_{k \in \mathbb{N}} \left\Vert g_k \right\Vert_2^2 < \infty \quad \text{and} \quad \sum_{k \in \mathbb{N}} \left\Vert h_k \right\Vert_2^2$$ and $$\left\Vert g \right \Vert_2 = \bigg [\int_{I...

 

 Calculus and analysis

For questions about calculus, real analysis, functional analys...
May 21, 2018 16:19
0
Q: Using the Fredholm Alternative to Prove that a Transformation is a Bijection

Harry EvansLet $I$ be an interval in $\mathbb{R}.$ For $k \in \mathbb{N}$, let $g_k, h_k \in L^2 (I)$ where $$\sum_{k \in \mathbb{N}} \left\Vert g_k \right\Vert_2^2 < \infty \quad \text{and} \quad \sum_{k \in \mathbb{N}} \left\Vert h_k \right\Vert_2^2$$ and $$\left\Vert g \right \Vert_2 = \bigg [\int_{I...

 

 Mathematics

Associated with Math.SE; for both general discussion & math qu...
May 21, 2018 16:17
Will @Daminark be here? Or @TedShifrin?
May 21, 2018 16:15
Munkres is a good author on Topology; I haven't seen his Algebraic Topology book, though
May 21, 2018 16:09
Thanks, @BalarkaSen...I wish Prof Ted were around
May 21, 2018 16:09
0
Q: Using the Fredholm Alternative to Prove that a Transformation is a Bijection

Harry EvansLet $I$ be an interval in $\mathbb{R}.$ For $k \in \mathbb{N}$, let $g_k, h_k \in L^2 (I)$ where $$\sum_{k \in \mathbb{N}} \left\Vert g_k \right\Vert_2^2 < \infty \quad \text{and} \quad \sum_{k \in \mathbb{N}} \left\Vert h_k \right\Vert_2^2$$ and $$\left\Vert g \right \Vert_2 = \bigg [\int_{I...

May 21, 2018 16:08
I posted it above, hoping you could answer, @BalarkaSen. Just in case you don't see it: math.stackexchange.com/questions/2790125/…
May 21, 2018 16:07
Have you tried Hatcher's book, @NicholasRoberts?
May 21, 2018 16:06
Anyone, @BalarkaSen?
May 21, 2018 15:37
I wish Prof Ted Shifrin were here
May 21, 2018 15:20
@Secret @LeakyNun do you have any ideas?
May 21, 2018 14:22
I guess the help I need is by showing that $\mathcal{N} (I - T) = \{0\}.$
May 21, 2018 14:21
I know that when the nullity is zero, then it is injective. By the Fredholm alternative, then it is injective if and only if it is surjective, thus, $I - T$ is bijective.
May 21, 2018 14:19
Let $I$ be an interval in $\mathbb{R}.$ For $k \in \mathbb{N}$, let $g_k, h_k \in L^2 (I)$ such that
$$\sum_{k \in \mathbb{N}} \left\Vert g_k \right\Vert_2^2 < \infty \quad \text{and} \quad \sum_{k \in \mathbb{N}} \left\Vert h_k \right\Vert_2^2$$
Define $K : I \times I \rightarrow \mathbb{R}$ by
$$K(x,y) = \sum_{n \in \mathbb{N}} g_n(x)h_n(y)$$
and $T : L^2(I) \rightarrow L^2(I)$ by
$$Tf(x) = \int_I K(x,y)f(y)dy.$$

Suppose that there exists $C \in (0, \infty)$ such that
$$\sum_{n \in \mathbb{N}} \left| \langle f,h_n \rangle \right| ^2 \ge C\left\Vert f \right\Vert ^2 \quad (\forall f \in L
May 21, 2018 14:18
Good morning!
May 20, 2018 15:55
I am studying for an exam in functional analysis and I need help on this item here:

Let $\{x_n \}_{n=1}^{\infty}$ be a sequence in a Hilbert space $\mathcal{H}$ such that $x_n \rightharpoonup 0.$ Recall that this means that $\langle x_n, u \rangle \rightarrow 0$ for all $u \in \mathcal{H}.$

1. By induction, prove that there exists a sub-sequence, $\{x_{n_k}\}_{k = 1}^{\infty}$ such that $\left| \langle x_{n_k}, x_{n_j} \rangle \right| \le \frac{1}{k}$ whenever $k > j.$

2. For $N \in \mathbb{N}$, define
May 19, 2018 13:49
Exactly
May 19, 2018 13:38
since $\alpha_n$ is bounded, there exists $M \in \mathbb{R}$ such that $\alpha_n \le M$.
May 19, 2018 13:38
Does that make sense?
May 19, 2018 13:38
I think my flow is
$\left |Tx \right| \le \sum_{n \in \mathbb{N}} \left| \alpha_n \langle x, e_n \rangle f_n\right| \le \sum_{n \in \mathbb{N}} \alpha_n \left\Vert x \right\Vert ^2$
May 19, 2018 13:31
Hey good morning, @Secret!
May 19, 2018 13:19
I'm preparing for a test and I encountered this problem. I think in order for me to show that $T$ is bounded, I need to show that there exists $C \in \mathbb{R}$ such that $\left\Vert Tx \right\Vert \le C \left\Vert x\right\Vert$.
May 19, 2018 13:16
Let $\{\alpha_n\}_{n \in \mathbb{N}}$ be a bounded sequence of real numbers. Suppose $\{e_n\}_{n \in \mathbb{N}}$ and $\{f_n\}_{n \in \mathbb{N}}$ are orthonormal sequences in a Hilbert space, $\mathcal{H}$. Define $T : \mathcal{H} \rightarrow \mathcal{H}$ by
$$Tx = \sum_{n \in \mathbb{N}} \alpha_n \langle x, e_n \rangle f_n.$$
1. Show that $T$ is bounded.
2. Show that $T$ is compact if and only if $\alpha_n \rightarrow 0$.
May 19, 2018 08:32
I see, so it is possible that $K$ is not in $L^2 (I \times I)$. I was just thinking that is $K$ is, then by a theorem we used in class, I could show that $T : L^2(I) \rightarrow L^2(I)$ defined by
$$Tf(x) = \int_I K(x,y)f(y)dy$$
is compact.
May 19, 2018 08:16
If I check whether
$$\int_{I \times I} \left| g_n(x) h_n (y) \right|^2 dx dy < \infty$$

Does that work?
May 19, 2018 08:09
If say $g_k, h_k \in L^2 (I)$, I'm interested to know if $K : I \times I \rightarrow \mathbb{R}$ defined by
$$K(x,y) = \sum_{n \in \mathbb{N}} g_n(x)h_n(y)$$
is in $L^2 (I \times I).$
May 12, 2018 05:30
Thanks much, Prof!!!
May 12, 2018 05:30
Yes, I only live near my school
May 12, 2018 05:29
I will see if I can drop by and have you sign an autograph
May 12, 2018 05:29
Oh nice!
May 12, 2018 05:28
Thanks, Prof @TedShifrin, I think I'm good! :)
May 12, 2018 05:26
Actually Cauchy Schwarz?
May 12, 2018 05:26
Can we use the Holder inequality there?
May 12, 2018 05:24
I have always thought of the interval as a subset of the set $C$
May 12, 2018 05:24
OK, prof @TedShifrin, help me understand so I can come out of my confusion :)
May 12, 2018 05:23
If $u$ is $a$ and $v$ is $b$, right?
May 12, 2018 05:22
So from that definition, one of the functions must be less than the other
May 12, 2018 05:22
The definition of convexity I know is this: if $a,b \in C$, then $[a,b] := \{ta + (1-t)b : t \in [0,1] \} \subseteq C.$
May 12, 2018 05:20
for all $t\in [0,1]$, can we assume that $u(t) \le v(t)$
May 12, 2018 05:19
$u < v$, so $uv \le v^2$, correct?
May 12, 2018 05:17
$s^2u^2 + 2s(1-s)uv + (1-s)^2v^2$
May 12, 2018 05:15
But can't I just factor our $(1-s)^2$?
May 12, 2018 05:14
I will show the sum is less than 1
May 12, 2018 05:14
Yup, hear me out, see if it works
May 12, 2018 05:13
Consider
$$\int_0^1 \left| su(t) \right|^2 dt \quad \text{and} \int_0^1 \left| (1-s) v(t) \right|^2 dt$$
May 12, 2018 05:12
OK, let me see if this is correct
May 12, 2018 05:08
So my idea is I let $a, b \in [0,1]$ and then define $u(at + (1-b)t)$, integrate, etc
May 12, 2018 03:54
Is it correct if I use the definition of convexity here?
May 12, 2018 03:53
Let $C([0,1])$ denote the vector space of continuous functions $u : [0,1] \rightarrow \mathbb{R}$. On $C([0,1])$, consider the norm defined by
$$\left\Vert \varphi \right\Vert = \sup _{t \in [0,1]} \left| \varphi(t) \right| \quad (\forall \varphi \in C([0,1]).$$
Set
$$C := \bigg \{ u \in C([0,1]) : \int_0^1 \left| u(t) \right| ^2 dt < 1 \bigg \}$$

Show that $C$ is an open and convex subset of $C([0,1]).$
May 12, 2018 03:53
I have one last question