Can You Prove This Advanced Mathematical Inequality?

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Here is this week's POTW:

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Let $K : [0,1]\times [0,1] \to \Bbb R$ be a continuous function such that $\sup\limits_{x\in [0,1]} \int_0^1 |K(x,y)|\, dy \le 1$ and $\sup\limits_{y\in [0,1]} \int_0^1|K(x,y)|\, dx \le 1$. Prove that
$$\int_0^1 \left(\int_0^1 K(x,y)f(y)\, dy\right)^2\, dx \le 1$$ for all continuous functions $f : [0,1]\to \Bbb R$ such that $\int_0^1 f(y)^2\, dy \le 1$.-----

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No one answered this week's problem. You can read my solution below.

Spoiler

Let $T : C[0,1] \to \Bbb R$ be defined by the equation

$$T(f)(x) = \int_0^1 K(x,y)f(y)\, dy$$

Then $T$ is a linear functional on $C[0,1]$. Let $f\in C[0,1]$ with $\int_0^1 f(y)^2\, dy \le 1$. We have

$$\int_0^1 \left(\int_0^1 K(x,y)f(y)\, dy\right)^2\, dx = \|Tf\|_2^2 = \sup_{\|g\|_2 = 1} \langle Tf,g\rangle^2$$

For fixed $g \in C[0,1]$ with $\|g\|_2 = 1$,

$$\langle Tf,g\rangle = \int_0^1 Tf(x)g(x)\, dx = \int_0^1 \int_0^1 K(x,y)f(y)g(x)\, dx\, dy$$ so that

$$\lvert \langle Tf,g\rangle\rvert \le \int_0^1 \int_0^1 \lvert K(x,y)\rvert \frac{f(y)^2 + g(x)^2}{2}\, dx\, dy$$
$$ = \frac{1}{2}\int_0^1 f(y)^2 \left(\int_0^1 \lvert K(x,y)\rvert\, dx\right)\, dy + \frac{1}{2}\int_0^1 g(x)^2 \left(\int_0^1 \lvert K(x,y)\rvert\, dy\right)\, dx $$
$$\le \frac{1}{2}\int_0^1 f(y)^2\, dy + \frac{1}{2}\int_0^1 g(x)^2\, dx \le 1 $$

Hence $$\sup_{\|g\|_2 = 1} \langle Tf,g\rangle^2 \le 1$$ as desired.