3
$\begingroup$

a) Let $C_{[a,b]}$ be the vector space of functions continuous on $a\leq x \leq b$, with the norm

$$\|f\| = \max_{a\leq x \leq b}|f(x)|$$.

Let $K(x,y) $ be a fixed function of two variables, continuous on $a\leq x \leq b , a\leq y \leq b $ and let $A$ be the operator

$$g(x)= Af(x)= \int_{a}^{b} K(x,y)f(y)dy $$

Prove that A is a continuous linear operator mapping $C_{[a,b]}$ into itself.

b) Let $C_{[a,b]}^2$ be the space of functions continuous on $[a,b]$ with norm

$$\|f\| = \sqrt{\int_{a}^{b}f^2(x)dx}$$,

and let $A$ be the same as in part a. Prove that $A$ is a continuous linear operator mapping $C_{[a,b]}^2$ into itself.

Thoughts: My first strategy is to try and show that the operator is in fact linear by showing that $A(\alpha x+\beta y)= \alpha Ax+ \beta Ay $. So far I have not been able to do this, but I think it would be possible for me to figure out. I can also easily show that the operator maps continuous functions to continuous functions. However, proving continuity of the operator, I am very lost.

$\endgroup$
3
$\begingroup$

The linearity is trivially since the integral is linear: $$\int^b_a K(x,y) [\alpha f(y) + g(y)] dy = \alpha \int^b_a K(x,y) f(y) dy + \int^b_a K(x,y) g(y) dy $$ for any $f,g \in C([a,b])$ and $\alpha \in \mathbb R$. This shows that $A(\alpha f + g) = \alpha Af + Ag.$

For continuity, note that $K$ is bounded (since it is continuous on the compact set $[a,b] \times [a,b]$). Say $\lvert K(x,y) \rvert \le M$ for all $a \le x,y \le b$. Then for any $f \in C([a,b])$ and $x \in [a,b]$, \begin{align*} \lvert Af(x) \rvert &= \left \lvert \int^b_a K(x,y) f(y) dy \right\rvert \\ &\le \int^b_a \lvert K(x,y) \rvert \lvert f(y) \rvert dy\\ &\le \int^b_a M \| f \|_\infty dy = M(b-a) \| f\|_\infty. \end{align*} Since this holds for all $x$, we can pass to the supremum showing that $$\| Af\|_\infty \le M(b-a) \| f\|_\infty.$$ This shows that $A$ is continuous. This concludes part (a).

For part (b), use the Cauchy-Schwarz inequality: for any $f \in C([a,b])$ and $x \in [a,b]$, \begin{align*} \lvert Af(x) \rvert &= \left \lvert \int^b_a K(x,y) f(y) dy \right\rvert \\ &\le \int^b_a \lvert K(x,y) \rvert \lvert f(y) \rvert dy\\ &\le M \int^b_a \lvert f(y) \rvert dy \\ &\le M \left(\int^b_a 1^2 dy \right)^{1/2}\left(\int^b_a f(y)^2 dy \right)^{1/2}\\ &= M (b-a)^{1/2} \| f\|_2. \end{align*} Thus squaring both sides, integrating and taking the square root gives $$\|A f\|_2 \le M(b-a) \| f\|_2$$ which completes part (b).

$\endgroup$
  • $\begingroup$ What exactly do you mean in part a when you use the infinity symbol under the norm? $\endgroup$ – Reginald Dick Apr 2 '17 at 20:48
  • 2
    $\begingroup$ That's the supremum norm: $$\|f\|_\infty := \sup_{a\le x \le b } \lvert f(x) \rvert.$$ It's the typical norm used on $C([a,b])$ since it is the norm which makes that space complete. EDIT: I see that in the question, you use the notation $\|f \|$ for this norm. $\endgroup$ – User8128 Apr 2 '17 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.