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.


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).

  • $\begingroup$ 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

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