Let $C[0, 1]$ be the linear space of all continuous functions on the interval $[0, 1]$ equipped with the norm $\|f\| = \max_{0≤x≤1} |f(x)|$. Let $K(x, y)$ be a fixed function of two variables, continuous on the square $[0, 1] \times [0, 1]$, and let $A$ be the operator defined by $$Af(x) = \int^{ 1}_0 K(x, y)f(y) dy.$$ For $K(x, y) = x$, find $\|A\|$.

Proof idea:

$A$ is a continuous linear operator mapping $C[0,1]$ on itself. Since $A$ is linear from linearity of integrals. Also $A$ is bounded since $K$ is continuous on a compact set $[0,1]\times[0,1]$ and thus bounded by some $M$. Therefore, $$|Af(x)|=\left|\int_0^1K(x,y)f(y)dy\right|\leq \int_0^1|K(x,y)||f(y)|dy\leq\int_0^1M\|f\|=M\|f\|.$$

Now, if I am correct about the above, $|Af(x)|\leq x\|f\|$ for $K(x,y)=x$.

$$\|A\|=\sup_{f(x)\neq0}\frac{\|Af(x)\|}{\|f(x)\|}=\sup_{f(x)\neq0}\frac{\|Af(x)\|}{\|f(x)\|}\leq\|f\|?$$ is $\|A\|=\sup|Af(x)|$? Is the final answer an inequality? I am confused about this.

  • 1
    $\begingroup$ The final answer is a specific positive real number, a supremum of a bounded set. Since $1$ bounds $K$, you also already know that this supremum is $\le 1$. If you find a function (or a sequence of functions) of norm $1$ that achieves (converges to) $1$, then you proved $\|A\|=1$. $\endgroup$
    – Berci
    Nov 11 '21 at 19:09
  • $\begingroup$ Is this part two of a two-part problem, with the first part being "Show $A$ is bounded"? You have a lot of extra work prior to "Now, if I am correct about the above", and then $|Af(x)| \leq x \|f\|$, while true, is currently unjustified. $\endgroup$ Nov 11 '21 at 19:25
  • $\begingroup$ Yes, you are correct @BrianMoehring. Part a was: Prove that A is a continuous linear operator mapping C[0, 1] into itself. $\endgroup$
    – Mihai.Mehe
    Nov 11 '21 at 19:26

Is $\|A\|=\sup|Af(x)|$?

This isn't even clear what you mean by the supremum, but I'm apt to say no*. The definition is $$\|A\| = \sup_{\|f\|=1} \|Af\| = \sup_{f\neq 0} \frac{\|Af\|}{\|f\|}$$ Since you also have the norm $$\|Af\| = \max_{0 \leq x \leq 1} |Af(x)| = \sup_{0 \leq x \leq 1} |Af(x)|$$ floating around, you need to be more clear about your intentions when you write the supremum.

[*If you meant $\|A\| = \sup_\limits{\|f\|=1 \\ 0 \leq x \leq 1} |Af(x)|$, then the answer would be "yes", but it would be an irregular use that doesn't match anything else you've written]

Is the final answer an inequality?

If you fix a couple notational issues (and one error) in your solution, you'll have shown an inequality $\|A\| \leq 1$. However, this is not the final answer, since you are tasked with actually computing $\|A\|$.

Let's clean up the notation. I'll run through a full solution (in particular, note where I've included $x$ and where I didn't):

First, setting $K(x,y)=x$, we have $$|Af(x)| = \left|\int_0^1 xf(y)\,dy\right| \leq x\int_0^1|f(y)|\,dy \leq x\int_0^1\|f\|\,dy = x\|f\|$$ Therefore $$\|Af\| = \max_{0\leq x \leq 1}|Af(x)| \leq \max_{0\leq x \leq 1} x\|f\| = \|f\|$$ and $$\|A\| = \sup_{f \neq 0}\frac{\|Af\|}{\|f\|} \leq 1,$$

To show the other direction, we need to go back to the definition. Noting that $f_0(x) = 1$ implies $\|f_0\| = 1$ and $Af_0(x) = x$, we have $$\|A\| \geq \frac{\|Af_0\|}{\|f_0\|} = \|Af_0\| = \max_{0 \leq x \leq 1} |Af_0(x)| = \max_{0 \leq x \leq 1} |x| = 1$$ so $\|A\| = 1$.


By rearranging what you have written (and remembering to divide by the norm of $f$ in your expression), you have shown the inequality $\|A\|\leq 1$, and in this instance, this is actually the norm.

To see this, note that the norm is achieved by $Af$, for $f\equiv 1$.

  • $\begingroup$ $$\|A\|=\sup_{f(x)\neq0}\frac{\|Af(x)\|}{\|f(x)\|}\leq||f||$$ since $||Af(x)||=sup_{f(x)\leq1}|Af(x)|=sup_{f(x)=1}|Af(x)|$ and $||f(x)||=sup_{x=1}|f(x)|$ then $||A||=1$. $\endgroup$
    – Mihai.Mehe
    Nov 11 '21 at 19:42

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