For each $p \in (1,\infty]$ and $k \in \Bbb N$, there is $f_k : [0,1] \to \Bbb N$ continuous such that $\|f_k\|_1 = 1$ and $\lim_k \|f_k\|_p = \infty$ True or false:

For each $p \in (1,\infty) \cup \{\infty\}$ and $k \in \mathbb{N}$, there is a continuous function $f_k : [0,1] \to \mathbb{R}$ such that it fullfills that

*

*$\|f_k\|_1 = 1$ and

*$\displaystyle\lim_{k \to ∞} \|f_k(x)\|_p = \infty$.


Where $\|f_k\|_p$ is the $p$-norm of $f_k$.
P.D. I think that this argument is true though I can not think of a good example of $f_k$ that will fullfill it. If someone knows one can you tell me? And if this argument is false can you explain to me why?
 A: Let $C \in (0, 1)$ such that $\frac{2 - C}{2C} < 1$ (e.g. $C = \frac{3}{4}$). Define $\forall k \geq 2$, $$f_k(x) = \begin{cases} Ck, & 0 \leq \frac{1}{k} \\ Ck - \frac{Ck}{\theta_k}
\big(x - \frac{1}{k}\big),& \frac{1}{k} \leq x \leq \frac{1}{k} + \theta_k \\ 0, & \frac{1}{k} + \theta_k \leq x \leq 1 \end{cases}$$
where $\theta_k := \frac{2(1-C)}{kC}$. By our assumptions on $C$, we have that $\theta_k + \frac{1}{k} = \frac{2 - C}{kC} < 1$ (so each $f_k$ is constant on the intervals $[0, \frac{1}{k}]$ and $[\frac{1}{k} + \theta_k, 1]$ (on the latter of which it's identically 0) then linear in between so as to be continuous on all of $[0,1]$. Then $\forall k \geq 2$ we have:
$$||f_k||_1 = Ck\frac{1}{k} + \frac{1}{2}Ck\theta_k = C + (1 - C) = 1\\ \text{and } ||f_k||_p \geq Ck^{1 - \frac{1}{p}}$$ Since $p \in (1, \infty]$, $1 - \frac{1}{p}$ is strictly positive which gives $\lim_{k \rightarrow \infty}||f_k||_p \geq \lim_{k \rightarrow \infty}Ck^{1 - \frac{1}{p}} = \infty$. So the statement is true (just take your sequence to be $\{f_k\}_{k \geq 2}$). The key here is that the $L_p$ norms emphasize differently the peaks and decay of functions. When you have a bounded subset of $\mathbb{R}$ (as in your case), the peaks are what’s important. Though the intervals on which these functions maintain their highest values shrink and become a greater proportion of their overall area (as compared to the contribution on which they're linear), their $L_p$ norms grow since they they're taking higher values - an effect which becomes more pronounced when you take $p$ to $\infty$.
