Problem: Show that the identity operator from $(C([0,1]),\|\cdot\|_\infty)$ to $(C([0,1]),\|\cdot\|_1)$ is a bounded linear operator, but that the identity map from $(C([0,1]),\|\cdot\|_1)$ to $(C([0,1]),\|\cdot\|_\infty)$ is unbounded.

Attempt: The identity operator, $I$, is such that $I:C([0,1])\to C([0,1])$ where $If=f$ for $f\in C([0,1])$. Let's consider the first part of the problem. Now $I$ is bounded if $\exists M\in\mathbb R^+:\|If\|_1\le M\|f\|_\infty.$ Consider then,


Now, $|f(x)|\le \sup\{|f(x)|:x\in[0,1]\}=\|f\|_\infty$, so that,

$$\int_0^1|f(x)|dx\le\int_0^1\|f\|_\infty dx=\|f\|_\infty\int_0^1 dx=\|f\|_\infty$$

So that we have shown that $\|If\|_1\le1\cdot\|f\|_\infty$, highlighting that $\|I\|_{op}\le1$. Thus the identity operator from $(C([0,1]),\|\cdot\|_\infty)$ to $(C([0,1]),\|\cdot\|_1)$ is a bounded linear operator.

I'm a little unsure on the second part of the problem, where I have tried to proceed in an analogous manner to as before. Consider,


My first thought was to argue that this is greater than or equal to $|f(x)|$. I then thought to say that whilst $f\in C([0,1])$, it need not be bounded. Alternatively, saying that this is greater than or equal to $|f(x)|$, could we say that $|f(x)|=\frac d{dx}\int_0^1|f(x)|dx=\frac d{dx}\|f\|_1$. Then we would have that $\|If\|_\infty\ge\frac d{dx}\|f\|_1$, and argue that in this case $I$ is unbounded since differentiation is unbounded.

Would my differentiation being an unbounded operator argument be valid? I have a feeling there is an easier observation to make; what am I overlooking?


We have the identity map $I$ from $(C([0, 1], \Vert \cdot \Vert_1)$ to $(C[0, 1], \Vert \cdot \Vert_\infty)$; if it is bounded then there exists $0 < C \in \Bbb R$ with

$\Vert f \Vert_\infty \le C \Vert f \Vert_1, \; \forall f \in C([0, 1]); \tag 1$

consider the sequence $g_n \in C([0, 1])$, $n \in \Bbb N$, of continuous functions defined by

$g_k(x) = 1 - kx, \; x \in \left [0, \dfrac{1}{k} \right], \tag 2$

$g_k(x) = 0, \; x \in \left [ \dfrac{1}{k}, 0 \right ]; \tag 3$

then for every $k$,

$g_k(0) = 1, \; 0 \le g_k(x) < 1, \; x \in (0, 1]; \tag 4$


$\Vert g_k \Vert_\infty = 1; \tag 5$

on the other hand,

$\Vert g_k \Vert_1 = \displaystyle \int_0^1 \vert g_k(s) \vert \; ds = \int_0^1 g_k(s) \; ds$ $= \int_0^{1/k} (1 - ks)\; ds = \left (s - \dfrac{1}{2}ks^2 \right )_0^{1/k} = \dfrac{1}{2k} \to 0 \; \text{as} \; k \to \infty; \tag 6$

therefore for $k$ sufficiently large the inequality

$\Vert g_k \Vert_\infty \le C \Vert g_k \Vert_1 \tag 7$

fails, viz.:

$1 = \Vert g_k \Vert_\infty > \dfrac{C}{2k} = C \Vert g_k \Vert_1, \tag 8$

which is true for

$k > \dfrac{C}{2}; \tag 9$


$I: (C([0, 1]), \Vert \cdot \Vert_1) \to (C([0, 1]), \Vert \cdot \Vert_\infty) \tag{10}$

cannot be bounded.

  • $\begingroup$ This is a really nice counter example - I'll need to keep it in my back pocket - thanks! $\endgroup$ – Jeremy Jeffrey James Apr 29 '18 at 18:30
  • $\begingroup$ @JeremyJeffreyJames: Thank you sir, and thanks for the "acceptance"! Cheers! $\endgroup$ – Robert Lewis Apr 29 '18 at 18:32

The best way to handle the second part is to build functions $f_n\in C([0,1])$ such that $\|f_n\|_1=1$ while $\|f_n\|_\infty>n$ (or $\|f_n\|>M_n$ for some sequence $M_n\to\infty$). Taking bump functions with successively smaller supports does the trick.


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