Problem 5

Exercise 11.5 (Scaling Mean Operator)   Let $X=C([0,1])$ be the Banach space of all continuous complex-valued functions on $[0,1]$ with the maximum norm. Consider the linear operator $A : X\to X$ defined by

$$(Af)(x) = x\int_0^1 f(y)dy, x\in[0,1].$$ (11.11)

1. Show that $A$ is bounded and determine its operator norm.
2. Determine the spectrum of the operator $A$.

Proof. To show $A$ is bounded, estimate

$$\Vert A\Vert = \sup_{\Vert f\Vert _\infty=1}\Vert Af\Vert _\infty$$ (11.12)

$$= \sup_{\Vert f\Vert _\infty=1}\sup_{x\in[0,1]}\left\vert x\int_0^1 f(y)dy\right\vert$$ (11.13)

$$\leq \sup_{\Vert f\Vert _\infty=1}\sup_{x\in[0,1]}\vert x\vert\int_0^1 \vert f(y)\vert dy$$ (11.14)

$$\leq \sup_{\Vert f\Vert _\infty=1}\int_0^1 \vert f(y)\vert dy$$ (11.15)

$$\leq 1.$$ (11.16)

To compute the operator norm, consider $A:L^2([0,1])\to L^2([0,1])$, extended by density. By applying Fubini's theorem

$$\langle Af, g\rangle = \int_0^1 \left(x\int_0^1f(y)dy\right) \overline{g(x)}dx$$ (11.17)

$$= \int_0^1 \int_0^1 xf(y)dy\overline{g(x)}dx$$ (11.18)

$$= \int_0^1 \int_0^1 xf(y)\overline{g(x)}dxdy$$ (11.19)

$$= \int_0^1 f(y) \int_0^1 x\overline{g(x)}dxdy$$ (11.20)

$$= \int_0^1 f(y) \overline{\int_0^1 \overline{x} g(x)}dxdy$$ (11.21)

$$= \int_0^1 f(y) \overline{\int_0^1 x g(x)}dxdy$$ (11.22)

$$= \langle f, A^*g\rangle$$ (11.23)

we can find the adjoint of $A$ is equal to

$$A^*f = \int_0^1 yf(y)dy.$$ (11.24)

which we can also verify is bounded. To apply the formula $\Vert A\Vert=\sqrt{\Vert A^*A\Vert}$, determine the composed operator

$$A^*Af = \int_0^1 y \left(y \int_0^1 f(x)dx\right) dy$$ (11.25)

$$= \int_0^1 y^2dy \int_0^1 f(x)dx$$ (11.26)

$$= \frac{1}{3} \int_0^1 f(x)dx$$ (11.27)

Taking norms shows

$$\Vert A^*A f\Vert \leq \frac{\Vert f\Vert _1}{3}$$ (11.28)

with equality on constant functions, so that $\Vert A^*A\Vert=1/3$ and $\Vert A\Vert=1/\sqrt{3}$.

To find the spectrum, we first prove that $A$ is a compact operator. Let $\{f_n\}$ be a sequence in the unit ball. Then $A(\{f_n\})$ is bounded because $A$ is bounded. To prove the sequence is equicontinuous, let $\epsilon>0$ be given. If $\vert x-y\vert<\epsilon$, then the estimate

$$\vert Af_n(x)-Af_n(y)\vert = \left\vert x\int_0^1f(t)dt - y\int_0^1 f(t)dt \right\vert$$ (11.29)

$$\leq \vert x-y\vert\int_0^1\vert f(t)\vert dt$$ (11.30)

$$\leq \vert x-y\vert$$ (11.31)

$$< \epsilon$$ (11.32)

is true for any $f_n$. Therefore the hypotheses of Arzelà-Ascoli apply, so that a converging subsequence exists, proving that $A$ is compact. Therefore it suffices to find the eigenvalues of $A^*$, which we do now. Let

$$A^*f = \lambda f$$ (11.33)

Because $A^*f$ is constant, this means $f(y)\equiv f$ is constant, so we in fact have

$$A^*f = \int_0^1 yfdy = f\int_0^1 ydy = \frac{f}{2}$$ (11.34)

so that $\lambda=1/2$. Therefore, $\sigma(A)=\{0,1/2\}$. $\qedsymbol$