Let $F:C^{0}[0,1]\rightarrow \mathbb{R}$, $F(f)=\displaystyle\int_{0}^{1} \frac{f(t)^2}{\sqrt{t}}dt$
How to prove that $F$ is continuous wrt the metric induced by $L^{p}$ norm for $2<p\leq \infty$
UPDATE: The statement doesn't hold true for $2<p\leq 4$. Thanks MaoWao and Sangchul Lee!
For $p \in (2, 4)$, pick $\alpha \in (\frac{1}{4}, \frac{1}{p})$ and define the sequence of continuous functions $(f_n)_{n\geq 2}$ by
$$ f_n(x) = \begin{cases} (2n)^{\alpha}, & \text{if $x \leq \frac{1}{2n}$}, \\ x^{-\alpha}, & \text{if $\frac{1}{2n} < x \leq \frac{1}{n}$}, \\ n^{\alpha}(2 - nx), & \text{if $\frac{1}{n} < x \leq \frac{2}{n}$}, \\ 0, & \text{if $x > \frac{2}{n}$}. \end{cases} $$
The graph of $f_n$ looks like:
We easily check that there exist absolute constants $c, c' > 0$ such that
$$ \int_{0}^{1} f_n(x)^p \, \mathrm{d}x \leq c n^{p\alpha-1} \qquad \text{and} \qquad \left| F(f_n) - F(0) \right| \geq c' n^{2\alpha - \frac{1}{2}} $$
holds as $n \to \infty$. Then by noting that $p\alpha -1 < 0$ and $2\alpha - \frac{1}{2} > 0$,
$$ \| f_n - 0 \|_{p} \xrightarrow[n\to\infty]{} 0 \qquad \text{but} \qquad |F(f_n)-F(0)| \xrightarrow[n\to\infty]{} \infty. $$
Therefore $F$ is not $L^p$-continuous when $p \in (2, 4)$. (I think $F$ is also discontinuous when $p = 4$, although it may need a more careful construction. I will skip this case anyway.)
On the other hand, if $p > 4$ then $F$ is indeed $L^p$-continuous. Let $f, g \in C([0,1])$. Then by the Hölder's inequality,
\begin{align*} |F(f) - F(g)| &\leq 2\|f\|_{\infty} \int_{0}^{1} \frac{|f(t) - g(t)|}{\sqrt{t}} \, \mathrm{d}t + \int_{0}^{1} \frac{|f(t) - g(t)|^2}{\sqrt{t}} \, \mathrm{d}t \\ &\leq 2\|f\|_{\infty} \| t^{-1/2} \|_{\frac{p}{p-1}}\|f - g\|_{p} + \|t^{-1/2}\|_{\frac{p}{p-2}} \| f - g\|_{p}^{2}. \end{align*}
Since $\frac{p}{p-1} < \frac{p}{p-2} < 2$, both $\| t^{-1/2} \|_{\frac{p}{p-1}}$ and $\| t^{-1/2} \|_{\frac{p}{p-2}}$ are finite. Now this bound shows that $F$ is $L^p$-continuous at each point $f$ in $C([0,1])$.
