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

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

$$$$(Tu)(x)=\alpha(x)u'(x), u\in Y, x\in I.$$$$

(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 $$$$|(Tu)x|=|\alpha(x)T_1u(x)|\leq ||\alpha(x)||_\infty ||T_1|| ||u(x)||_Y$$$$ where $$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?

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.

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 $$\|T\|$$.

• How do I find the norm please. I tried $u_0(x)=1$ so that $\|u\|_y=1$. This make the right side zero. – Muhammad Mubarak Jan 22 at 13:07
• The sequence $u_n'(x)=x^{n-1}, x\in I$ is bounded above and below by 0 and 1 respectively in $\mathbb{R}$. So it has a convergent subsequence by Bolzano Weierstrass. – Muhammad Mubarak Jan 22 at 18:02
• @MuhammadMubarak It is not enough to have a pointwise convergent subsequence. You need a subsequence which converges in the norm of $X$ (i.e. uniform convergence). – Kavi Rama Murthy Jan 22 at 23:20

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

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.