Norm and compactness of the Operator $(Tu)(x)=\alpha(x)u'(x), u\in Y, x\in I$

Question

Let $I=[0,1]$ and call $X$ the Banach space $C(I)$, endowed with the uniform norm. Introduce

$Y=\{u\in X, u$ diffentiable on $I$ with $u'\in X\}$ and set $||u||_Y=||u||_\infty+||u'||_\infty, u\in Y, x\in I$.

Prove that $(Y, ||.||_Y)$ is a Banach space.

Let $\alpha$ be a nonzero element of $X$ and set

\begin{equation} (Tu)(x)=\alpha(x)u'(x), u\in Y, x\in I.\end{equation}

(i) Prove that $T\in L(Y,X)$ and find its norm. (ii) Establish if $T$ is compact and justify the answer.

proof

If $(u_n)$ is cauchy in $Y$. Then both $(u_n)$ and $(u_n')$ are cauchy in $X$. Thus, $Y$ is Banach.

(i) My idea is \begin{equation}|(Tu)x|=|\alpha(x)T_1u(x)|\leq ||\alpha(x)||_\infty ||T_1|| ||u(x)||_Y\end{equation} where $T_1u=u'$ is bounded by closed graph theory as $X,Y$ are Bananch spaces. Thus $||Tu||\leq M\|u\|$ where $M=||\alpha(x)||_\infty ||T_1||$ and $\|T\|\leq M$. Is this correct?

How do I show (ii) please?

Answer 1

I think $T$ is not compact. Let $u_n(x)=\frac {x^{n}} n$. Then $\{u_n\}$ is a bounded sequence in $Y$. If $\alpha (x)=1$ for all $x$ then $Tu_n(x)=u_n'(x)=x^{n-1}$ and no subsequence of thus converges uniformly on $[0,1]$.

As far as first part of i) is concerned you are making it too complicated. Isn't ti clear directly that $\|Tu\| \leq \|\alpha\|_{\infty} \|u\|_Y$?. There is no need to use closed graph theorem. BTW, you are also asked to find the norm of $T$. You have only found an upper bound, not the exact value of $\|T\|$.

Answer 2

I'm not sure how and why you use the closed graph theorem to prove that $T_1$ (an operator from $Y$ to $X$) is bounded: this should be almost trivial, actually: $$\Vert Tu(x)\Vert_X=\Vert \alpha(x)u'(x)\Vert\leq\Vert\alpha\Vert_X\Vert u'\Vert_X\leq\Vert\alpha\Vert_X\Vert u\Vert_Y.$$

A common way to verify that an operator is not compact is to try to prove that its image contains an infinite-dimensional Banach space. Your idea of "breaking up" the operator $T$ is useful here;

Let $T_1:Y\to X$, $T_1(u)=u'$. Then $T_1$ is a surjective bounded operator from $Y$ to $X$: $\Vert T_1 u\Vert_X=\Vert u'\Vert_X\leq\Vert u\Vert_Y$.

Let $M_\alpha:X\to X$, $M_\alpha(u)=\alpha u$, which is a bounded operator on $X$: $\Vert M_\alpha u\Vert_X=\Vert\alpha u\Vert_X\leq\Vert\alpha\Vert_X\Vert u\Vert_X$ (we use submultiplicativity of the supremum norm).

Then $T=M_\alpha\circ T_1$, which is another proof that $T$ is bounded.

If $\alpha$ were always nonzero, then $M_{\alpha^{-1}}$ would be an inverse of $M_\alpha$, so $M_{\alpha^{-1}}\circ T=T_1$ would be a surjective operator $Y\to X$, hence non-compact as $X$ is infinite-dimensional. Recall that composition by of bounded operator with a compact one gives a compact operator. In this case, as both $M_{\alpha^{-1}}$ and $T$ are bounded and their composition is non-compact, then $T$ is not compact.

However, $\alpha$ can be zero on parts of $I$ (it is only a nonzero function) (e.g. $\alpha(x)=\max(8|x-1/2|-1,0)$ is a nonzero function which is zero on $[3/8,5/8]$). So the argument needs to be modified. Instead of looking at the whole interval $I$, try to find a subinterval $J$ on which $\alpha$ is uniformly far from $0$: Then the restriction operator $R_J:X\to C(J)$, $u\mapsto u|_J$ is a bounded, surjective operator. If you prove that $R_j\circ T$ is not compact, basically with the same argument as above, then $T$ is non-compact.

Answer 3

$\|Tu\|=\sup|Tu(x)|=\sup |\alpha(x)u'(x)|\leq \|\alpha(x)\|_\infty \|u'(x)\|_\infty\leq \|\alpha(x)\|_\infty \|u(x)\|_\infty.$

This implies $\|T\|\leq \|\alpha\|_\infty$.

Let $u_0(x)=x$ so that $u_0'(x)=1$. Then $|Tu_0(x)|=|\alpha(x)|$. This implies $\|T\|\geq \|\alpha\|_\infty.$

Thus, $\|T\|=\|\alpha\|$.

Let $u_n(x)=\frac{(x+2)^{n}}{n}$. Then $u_n'(x)=(x+2)^{n-1}\rightarrow \infty$ on $I=[0,1]$. So $T$ is not compact.