about properties of the operator $T_f(g) := f\cdot g$.

Question

let $f \in C([0,1])$ and $T_f: L^2([0,1]) \to L^2([0,1])$ and $T_f(g) := f\cdot g$ prove :

1)$T_f$ is well define , linear and bounded and find $|| T_f ||$ .

2 )if $T_f$ be compact operator then $f=0$

i can prove $T_f$ is linear.

Answer 1

Answer of the second part: Let $a \in [0,1]$ Define $g_n(x)=\sqrt n$ for $|x-a|<\frac 1n$ and $0$ for all other $x$. Then $(g_n)$ is bounded in $L^{2}$. If $T$ is compact then $(fg_n)$ has convergent subsequence. Note that $(fg_n) \to 0$ almost eveywhere. Hence the subsequence can only converge to $0$. If $|f(a)| >0$ then $\int |fg_n|^{2} \geq c \int |g_n|^{2}$ for some $c>0$ and $n$ sufficiently large. We got a contradiction since $\int |g_n|^{2}$ does not tend to $0$. Thus $f$ vanishes at every point $a$.

Hints for first part: The norm of $T_f$ is $M$, the supremum of $|f|$. Clearly $\|Tf\| \leq M\|g\|$ so $\|T_f\| \leq M$. There exist $x$ such that $M=|f(x)|$. Now there exists $r>0$ such that $|f| >M-\epsilon$ on $(x-r,x+r)$. Take $g(y)=\frac 1 {\sqrt {2r}} $ for $|y-x| <r$ and $0$ for all other $y$. Then $\|g\|=1$. Can you show that $\|T_f\| \geq \|T_f(g)| \geq M-\epsilon$?.

Answer 2

Hints:

For 1) Show $T_f g \in L^2$ and use the Cauchy Schwartz inequality.

For 2) Show that $T_f$ is self adjoint. Then $T_f$ has a non zero eigenvalue. This says something about $f$. And the compactness allows you to deduce $f=0$.

Answer 3

With $f$ continuous you can get away with an "easy" argument: assume that $f$ is not identically zero, so that it's, say, positive on an open interval; call it $I$. On $I$ you can build an orthonormal sequence by normalizing appropriate bump functions with disjoint supports. Elements of this sequence all have unit norm, but any limit point of the images would have to be zero on $I$ (here we use the definition of $I$).