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I am stuck up with some of these questions.

Q1) Define $T:(C[0,1],\|\cdot\|_\infty)\to \mathbb{R}$ by $T(f)=\int_0^12x\:f(x)\:dx$. Then $\|T\|=$ ?
My attempt:

$$\|T\|=\sup_{f\in C[0,1]}\dfrac{\|T(f)\|}{\:\:\|f\|_\infty}$$ Therefore \begin{align} \|T(f)\|&=\Big|\int_0^12x\:f(x)\:dx\Big|\\&\le \int_0^12|x| |f(x)|dx\\&\le\sup_{x\in [0,1]}|f(x)|\int_0^12|x|dx=\|f\|_\infty \end{align} so that $\|T\|=1$. Is this right ?

Q2) Let $V=C^1[0,1]$ and $X=(C[0,1],\|\cdot\|_\infty)$ and $Y=(C[0,1],\|\cdot\|_2)$. How to check denseness of $V$ in $X$ and $Y$ ?

Q3) Let $H=L^2[0,\pi]$ under usual inner product and $u_n(t)=\sqrt{\dfrac{2}{\pi}}\sin nt$. Consider the set $E=\{u_n: n\in \mathbb{N}\}$. Then classify $E$.

I find that $\|u_n(t)\|=1$ and also $u_n$'s are orthogonal. So E is an orthonormal set. Hence E is linearly independent. To check whether E is a basis of $H$, we need to check $\text{span}(E)=H$. How to check this ?

Q4) Define $T:l^2\to l^2$ by $T(x_1,x_2,...,x_n,...)=(x_2-x_1,x_3-x_2,...,x_{n+1}-x_n,...)$. Then

(a)$\|T\|=1$ $\:\:$ (b) $\|T\|>2$ but bounded $\:\:$(c) $1<\|T\|\le 2\:\:$ (d) $\|T\|$ is bounded

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    Your reasoning for 1) shows $||T|| \le 1$. To prove eqaulity you need additional reasoning (e.g. by finding a function $g$ such that $||Tg|| = ||g||\neq 0$ or a sequence of function such that this hold in the limit. – 2017-01-08

1 Answers 1

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Q1) As Thomas indicated, you have shown $\|T\| \leq 1$. To show that $\|T\| = 1$, it suffices to compute $\|Tf\|$ with $f(x) = 1$.

Q2) Use the Weierstrass approximation theorem and the fact that you can pull the limit inside the integral in the case of uniform convergence.

Q3) The condition $\overline{\text{span}(E)} = H$ is equivalent to

$$ (x,u_{n}) = 0, \quad n \in \mathbb{N} \quad \Leftrightarrow \quad x = 0$$

What do you get for $x(t) = \cos(t)$?

Q4) Define $T_1(x_1,x_2,\dots) = (x_2,x_3,\dots)$. Then you have $$T = T_1 -I$$ where $I$ is the identity operator and therefore $$\|T\| \leq \|T_1\| + \|I\| = 2$$ To see that $\|T\| > 1$, let $(x_n)_{n\in\mathbb{N}} \in \ell^2$ with $x_n \geq 0$ and define $u_n = (-1)^n x_n$. Let $u = (u_n)_{n\in\mathbb{N}}$. Computing $\|Tu\|_{\ell^2}$ shows $\|T\| > 1$.

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    Wait. I will work out based on the points you have given and get back to you. Meanwhile +1 (up vote) for your efforts for the answer. – 2017-01-08