Linear transformations

Theorem 3.1   Let $T: X\rightarrow Y$ linear transformation. TFAE (The following are equivalent)

1. $T$ is continuous at 0.
2. $T$ is continuous at any point $x\in X$.
3. If $V\subseteq Y$ is an open set containing 0 then $T^{-1}(V)$ contains an open set $U\subseteq X$ containing 0.
4. $T$ maps bounded sets to bounded sets. i.e., If $B\subseteq X$ is bounded then $TB\subseteq Y$ is bounded.
5. $T$ is uniformly continuous.
6. $A=\sup\{\Vert Tx\Vert: \Vert x\Vert\leq 1\}<\infty$.
7. $B=\sup\left\{\Vert Tx\Vert: \Vert x\Vert= 1\right\}<\infty$.
8. $C=\sup\left\{\frac{\Vert Tx\Vert}{\Vert x\Vert}:x\neq 0\right\}<\infty$.
9. $D=\inf\left\{M: \Vert Tx\Vert\leq M \Vert x\Vert,\, \forall x\in X\right\}<\infty$.

More over, $A=B=C=D$ and this number is called as norm of $T$, denoted by $\Vert T\Vert$.

$B(X,Y)=\{T:X\rightarrow Y: T$    is bounded and linear $\}$ $B(X)=B(X,X)$

Example 4 (Identity Operator)  

Example 5 (Unilateral, Bilateral shifts)   Right, Left shifts

Example 6 (Multiplication)  

Example 7 (Integral)  

Define convolution of two real valued Lebesgue measurable functions $f,g: \mathbb{R}^n\rightarrow \mathbb{R}$ as

$$(f*g)(x)=\int_{\mathbb{R}^n}(f(y)g(x-y)d\mu(y)$$

whenever the integral makes sense. The following inequality gives certain further information regarding the existence of the above defined integral.

Theorem 3.2 (Youngs inequality)   Let $1\leq p\leq r\leq \infty$ such that $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}+1$. Suppose $f\in L^p(\mathbb{R}^n)$ and $g\in L^q(\mathbb{R}^n)$ then

$$\Vert f*g\Vert _r\leq \Vert f\Vert _p\Vert g\Vert _q.$$

Let us look at the inequality at certain special values $p$, $q$ and $r$.

• Suppose $p=1$ and $q=1$. Then $r=1$ and we get $\Vert f*g\Vert _1\leq \Vert f\Vert _1\Vert g\Vert _1$. This proves that $L^1(\mathbb{R}^n)$ is closed under convolution.
• Suppose $p=r$. Then $q=1$ and we get $\Vert f*g\Vert _r\leq \Vert f\Vert _r\Vert g\Vert _1$. This proves the convolution with an $L^1(\mathbb{R}^n)$ function is closed in any $L^p(\mathbb{R}^n)$.

Example 8 (Convolution Operator)   From the above discussion by fixing a $g\in L^1(\mathbb{R}^n)$, we can view convolution as an operator $C:L^p(\mathbb{R}^n)\rightarrow L^p(\mathbb{R}^n)$ defined as

$$C(f)(x)=\int_{\mathbb{R}^n}(f(y)g(x-y)d\mu(y)$$

By Young's inequality $\Vert Cf\Vert\leq \Vert f\Vert _p\Vert g\Vert _1$ Hence $\Vert C\Vert\leq \Vert g\Vert _1$

The same convolution operator can be recognized as an integral operator as well with kernel $k(x,y)=g(x-y)$. With the same kernel, for $p=2$ this appears like a Hilbert-Schmidt operator. Let us first consider Schur's condition

$$C_1={\esssup}_{x\in \mathbb{R}^n}\int_{\mathbb{R}^n}\vert g(x-y)\vert d\mu(y)<\infty$$

$$C_1={\esssup}_{y\in\mathbb{R}^n}\int_{\mathbb{R}^n}\vert g(x-y)\vert d\mu(x)<\infty$$

But, even if $g\in L^2(\mathbb{R}^n)$,

$$\Vert K\Vert _2^2=\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}\vert g(x-y)\vert^2d\mu(y)d\mu(x)=\infty$$