# Find $\|A\|$ for the operator $A$

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.

• 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 $\|A\|=1$. Nov 11 '21 at 19:09
• 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. Nov 11 '21 at 19:25
• Yes, you are correct @BrianMoehring. Part a was: Prove that A is a continuous linear operator mapping C[0, 1] into itself. 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 $$f\equiv 1$$.

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