Prove that A is a continuous linear operator mapping $C_2[0, 1]$

real-analysis

Let $C_2[0, 1]$ be the space of all continuous functions equipped with the norm
$$||f||_{L^2[0,1]} =(\int^1_0 f^2(x) dx)^{1/2}.$$
Let $K(x, y)$ be a fixed function of two variables, continuous on the square $[0, 1] × [0, 1]$, and let $A$ be the operator defined by $$Af(x) = \int^ 1_0 K(x, y)f(y) dy.$$
Prove that A is a continuous linear operator mapping $C_2[0, 1]$ into itself.

Proof idea:
first as a reminder:

  • $C_2[0, 1]$ is the space of continuous functions on C[0, 1] equipped with $L^2$ norm.
  • $L^2[0,1]= \{ f:[0,1]\rightarrow \mathbf {C}: \int^1_0 |f(x)|^2 dx<\infty \} $

Linearity $A(\alpha f +\beta g)(x)=\int^ 1_0 K(x, y)(\alpha f +\beta g)(y) dy = \alpha \int^ 1_0 K(x, y)f(y)dy +\beta \int^ 1_0 K(x, y)g(y)dy=\alpha Af(x)+\beta Ag(x)$. by linearity of the integrals.

Boundedness: $K\leq M$ is bounded by some $M$ finite since it is continuous on a closed set. Therefore, $$|Af(x)| = |\int^ 1_0 K(x, y)f(y) dy|\leq M |\int^ 1_0 f(y) dy \leq M ||f||_{L^2[0,1]} \int^ 1_0 dy=M||f||_{L^2[0,1]}<\infty$$

So $A$ is linear and continuous on $C_2[0,1]$.

I am not sure this is complete. Also, do I need also to find $||A||$? Also what is the implication of having $C_2[0,1]$, where do I use this?

Thanks and Regards,

Best Answer

Your proof for continuity is not quite right. To show $A: \mathcal{C}_{2}([0, 1]) \to \mathcal{C}_{2}([0, 1])$ is continuous, let $f \in \mathcal{C}_{2}([0, 1])$. Then:

$$||Af||_{L^{2}([0, 1])}^{2} = ||\int_{0}^{1}K(x, y)f(y)dy||_{L^{2}([0, 1])}^{2} = \int_{0}^{1}|\int_{0}^{1}K(x, y)f(y)dy|^{2}dx $$ $$\leq \int_{0}^{1}[(\int_{0}^{1}|K(x, y)|^{2}dy)(\int_{0}^{1}|f(y)|^{2}dy)]dx\text{, by Cauchy-Schwarz} $$ $$ = (\int_{0}^{1}\int_{0}^{1}|K(x, y)|^{2}dx dy)(\int_{0}^{1}|f(y)|^{2}dy) = ||K||_{L^{2}([0, 1]^{2})}^{2}||f||_{L^{2}([0, 1])}^{2} $$

$$\Rightarrow ||Af||_{L^{2}([0, 1]^{2})} \leq ||K||_{L^{2}([0, 1]^{2})}||f||_{L^{2}([0, 1])}$$

Note $||K||_{L^{2}([0, 1]^{2})}^{2} = \int_{0}^{1}\int_{0}^{1}|K(x, y)|^{2}dx dy \leq \text{sup}_{(x, y) \in [0, 1]^{2}}|K(x, y)|^{2} = ||K||_{\mathcal{C}_{0}([0, 1]^{2})}^{2} < \infty$, the last inequality from our assumption that $K$ is continuous. This shows that $A$ is a bounded operator, which implies it is continuous.

To find a bound for $||A||$, simply note that from the above, $||A|| \leq ||K||_{L^{2}([0, 1]^{2})}$.

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