Let the set of all continuous mappings on the unit interval $[0,1]$ be denoted as $C[0,1]$ equipped with the $L^p$ norm $\|\cdot \|_p$. For $g\in C[0,1]$, define for all $x\in [0,1]$ the operator
$$T_gf(x)=g(x)f(x)$$
For what values of $1\leqslant p\leqslant \infty$ is $T_g$ a bounded linear operator from $L^2[0,1]$ to $L^p[0,1]$? Further for these values, what does $\|T_g\|_p$ equal?
I am not sure how to go about showing the above. I suspect the only values of $p$ that work are $1\leqslant p\leqslant2$.
That is, for $1\leqslant p<2$, by Holder's Inequality, we get
$$\|T_gf(x)\|_p=\|g(x)f(x)\|_p\leqslant \|f(x)\|_2 \|g(x)\|_{\frac {2p}{2-p}}$$
Since $g$ is continuous on the compact interval $[0,1]$, it follows that $g$ is bounded and hence $\|g(x)\|_{\frac {2p}{2-p}}\leqslant C$ for some non-negative constant $C$.
Hence
$$\|T_gf(x)\|\leqslant C\|f(x)\|_2$$
Implying $T_g$ is a bounded linear operator.
Similarily, for $p=2$,
$$\|T_gf(x)\|_p=\|g(x)f(x)\|_p\leqslant \|g\|_\infty \|f\|_2=C\|f\|_2$$ for the non-negative constant $C=\|g\|_\infty$
Hence again, $T_g$ is a bounded linear operator.
My question is, how can we show (assuming the claim is false) that this is not the case for $p>2$ and $p=\infty$? Further, is there an easy way to compute $\|T_g\|$ for $1\leqslant p \leqslant 2$?
Any help would be much appreciated. Thanks in advance!
For $p<2$ set $f = \frac{h}{\|h\|_2}$, where $h = g^{\frac p{2-p}}$. Then $\|Tf\|_p = \|g\|_{\frac{2p}{2-p}}$, so $\|T\|$ coincides with the latter value (and is even attained).
For $p=2$ choose some $x_0\in [0,1]$ such that $|g(x_0)| = \|g\|_\infty$. Let $\epsilon > 0$. Then there exists $\delta > 0$ such that for $x$ in the $\delta$-neighborhood $B_\delta$ of $x_0$ we have $|g(x)|^2\ge\|g\|_\infty^2-\epsilon$. Let $f\in L^2$ be such that $\|f\|_2=1$ and $\operatorname{supp}(f)\subset B_\delta$. Then $$ \|T\|^2\ge\|Tf\|_2^2 = \int_{B_\delta}|g(x)|^2|f(x)|^2\,dx \ge \|g\|_\infty^2-\epsilon. $$ Letting $\epsilon\to 0$ yields $\|T\| = \|g\|_\infty$.