Is it possible that it exists a $C>0$ that verify for all continuous function on $[a;b]$; $\|f\|_{\infty} \leq C \|f\|_2 $?

Question

Question:

When $f \in C^2[a;b]$ we define $ \|f\|_2=\sqrt{\int_{[a;b]}|f|^{2}}$ and when $f$ is bounded on $[a;b]$ we define $\|f\|_{\infty} = \sup_{x \in [a;b]} |f(x)| $.
Is it possible that it exists a $C>0$ that verify for all continuous function on $[a;b]$; $||f||_{\infty} \leq C \|f\|_2 $ ?

Answer:

1- Let choose the set of functions $ f_k(x)=\frac{(x-a)^k}{(b-a)^k} , k \geq 1$.
Hence $ \|f_k(x)\|_2 ^2 = \int_a^b \frac{(x-a)^{2k}}{(b-a)^{2k}} dx = \frac{1}{(b-a)^{2k}} [\frac{(x-a)^{2k+1}}{2k+1}]_a^b = \frac{b-a}{2k+1} $

2- When $k \to \infty$ we have that for every $C$ we can choose: $ C \|f_k(x)\|_2 \to 0 $.

3- On an other side we have that $ \|f_k(x)\|_{\infty} = f(b)=1 $

4- Thus no matter which $C$ we can choose $ \exists K $ s.t. $ \forall k > K $ we will have $ \|f\|_{\infty} = 1 > C \|f\|_2 $
For exemple you can choose: $K = \frac{b-a-1}{2}$

Is this correct? Thank you.

Answer 1

The answer is no, there is no such $C$. For example, it is possible to construct a sequence of functions $(f_n)$ which are all continuous and satisfies both $$ \lim_{n \rightarrow \infty} \|f_n\|_2 \rightarrow 0 \hspace{1cm} \|f_n\|_\infty = 1 \ \forall \ n=1,2,... $$ The existence of such a sequence makes it impossible for a universal $C > 0$ to exists such that for all continuous $f$ it holds $\|f\|_\infty \leq C \|f\|_2$.

To see this fix an arbitrary $C>0$. By the fact that the limit goes to zero and each term is non-negative there must be some $n$ where $\|f_n\|_2 < 1/C$, but for such an $n$ we have $$\|f_n\|_\infty = 1 = C\frac{1}{C} > C\|f_n\|_2$$ which goes the opposite direction of what you are asking for. Since this holds for arbitrary $C > 0$ we can conclude that no universal $C$ exists.

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Here is one explicit construction of the series $f_n$.

$$f_n(x) = \begin{cases} 1 - n(x-a) & a \leq x \leq a+1/n \\ 0 & a + 1/n < x \leq b \end{cases}$$ Graphically, this a height 1 triangle where the base is of width $1/n$.

It is easy so show that this is continuous and $\|f\|_\infty = 1$ and one can explicitly compute (and I've skipped a few of the routine steps) $$\|f_n\|_2^2 = \int_0^{1/n} (1 - nx)^2 dx= \frac{(nx-1)^3}{3n} \bigg|_{0}^{1/n} = \frac{1}{3n}$$ which shows that $\lim_{n\rightarrow \infty} \|f_n\|_2$.

Answer 2

As you see based on Matt's example, a continuous function (even a smooth function) can't be controlled in $L^\infty$ by its $L^2$ norm. The example in that answer is a function which has a derivative $\approx n$ on an interval of length $\approx 1/n$, so the $L^2$ norm of the derivative is $\|f'\|_2 \approx n/\sqrt n = \sqrt n\gg \|f\|_{\infty}$.

Thus a new question you can ask yourself is whether it is possible that there is a universal constant $C>0$ such that $$ \sup_{x\in[0,1]} |f(x)|\le C\big(\|f\|_{L^2([0,1])} + \|f'\|_{L^2([0,1])}\big) $$ for all $f$ that are $C^1$ in $[0,1]$. This turns out to be true, and it's not too hard to prove. It's a version of the Sobolev embedding theorem.