Continuous operator on $L^\infty$

$1<p<\infty$ and $k\in L^\infty([0,1]^2)$

$(Tf)(s)=\int_{0}^{1}k(s,t)f(t)dt$

I want to show that it is a continuous operator $T:L^p([0,1]->L^p([0,1])$

Proof: What I need to show is that $\exists C>0$ with $||Tf||_{L^\infty([0,1]^2)}\le||C||||f||_{L^\infty([0,1])}$, correct?

I take squares: $||Tf||^2_{L^\infty([0,1]^2)}=\mathrm{ess } \sup\int_{0}^{1}(\int_{0}^{1}k(s,t)f(t)dt)^2ds$ right?

Now I would like to use Hölder inequality, but I do not know how, the essential supremum looks confusing.

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As $[0,1]$ has Lebesgue measure $1$, by Jensen's inequality, for $s\in [0,1]$, $$|T(f)(s)|^p\leqslant \int_{[0,1]}|k(s,t)|^p|f(t)|^pdt\leqslant \lVert k\rVert_{L^\infty([0,1]^2)}^p \int_{[0,1]}|f(t)|^pdt\leqslant \lVert k\rVert_{L^\infty([0,1]^2)}^p\lVert f\rVert_{L^p}^p.$$ Now integrate over $[0,1]$ with respect to $s$ in the last inequality. This gives $$\lVert T(f)\rVert_{L^p[0,1]}^p\leqslant \lVert k\rVert_{L^\infty([0,1]^2)}^p\lVert f\rVert_{L^p}^p.$$ As $T$ is linear, this gives continuity.
Almost, because we have to make the $p$-norm of $Tf$ appear. – Davide Giraudo Dec 1 '12 at 16:48