Let $X$ be a normed space.

Prove that a linear functional $f:X \to \mathbb{R}$ is continuous if and only if there is a number $ c \in {0, \infty}$ such that $$|f(x)| \leq c\|x\|$$ for all $x \in X$

This is the proof:

Suppose there is a number $ c \in {0, \infty}$ such that $$|f(x)| \leq c\|x\|$$

Then if $x_n \to x_0$ in $X$, we have $$|f(x_n)-f(x_0)|=|f(x_n-x_0)| \leq c \|x_n -x_0\| \to 0$$ as $n \to \infty$. So $f$ is continuous

How does $c \|x_n -x_0\| \to 0$?

For the converse suppose that no such $c$ exists.

Then for all $n$ there exists $x_n \in X$ with $$|f(x_n)| > n\|x_n\|$$

Where does this come from?

We must have $x_n \neq 0$ so we can set $$y_n =\frac{1}{n} \frac{x_n}{\|x_n\|}$$

Where does this come from?

Also why does $\|y_n\| =\frac{1}{n}$?

The proof concludes by saying $$\left|f(y_n)\right|=\left|f\left(\frac{1}{n} \cdot\frac{x_n}{\|x_N\|}\right)\right|=\frac{1}{n} \frac{|f(x_n)|}{\|x_n\|} > 1$$ so f is not continuous.

Why greater than 1? And why is f not continuous?

  • 1
    $\begingroup$ you supposed $x_n \to x_0$ in $X$, which means precisely that $\|x_n-x_0\| \to 0$. and for the converse, $f$ unbounded precisely means that we can find a $x_n$ such that $|f(x_n)|$ is arbitrary large with respect to $\|x_n\|$, in particular we can find $|f(x_n)| > n \|x_n\|$ $\endgroup$
    – reuns
    Commented Apr 24, 2016 at 15:41

2 Answers 2


1) It is assumed that $x_n\to x_0$ in $X$ which, by definition, means $\|x_n-x_0\|_{X}\to 0$ as $n\to \infty$.

2) You are negating the condition $|f(x)|\leq c\|x\|$ for all $x\in X$. Negating this means that you can find a sequence of numbers $x_n\in X$ for which the condition does not hold, i.e. $|f(x_n)|>N\|x_n\|$ for these $x_n$, $n\geq 0$. We must have $x_n\neq 0$ because otherwise if $x=0$ then $0=|f(0)|>N\|x\|=0$, i.e. $0>0$ which is a contradiction so $x_n\neq 0$. (Recall that since $f$ is linear you must have $f(0)=0$).

3) $y_n = \frac{1}{n} \frac{x_n}{\|x_n\|}$. This you set by definition, since $x_n\neq 0$ then $\|x_n\|\neq 0$ and you can divide. Now, $\frac{1}{n}\frac{1}{\|x_n\|}$ is a real number, so the norm of $y_n$ is:

$$\|y_n\| = \|\frac{1}{n} \frac{x_n}{\|x_n\|}\| = \frac{1}{n}\frac{1}{\|x_n\|}\|x_n\|=\frac{1}{n},$$ where I pulled out the factor $\frac{1}{n}\frac{1}{\|x_n\|}$ from inside the norm since it is a scalar.

4) Remember that you assumed that $|f(x_n)|>N\|x_n\|$, hence $\frac{1}{N\|x_n\|}|f(x_n)|>1$.

$f$ is not continuous because you constructed a sequence of numbers $y_n$ which converge to 0 (since $\|y_n\|=1/n\to 0$) and such that $f(y_n)$ is always bounded from below. If $f$ were continuous then one should immediately have $\lim_n f(y_n)= f(\lim_n y_n)=f(0)=0$ (again $f(0)=0$ in this case because $f$ is linear) but in your case you showed that $|f(y_n)|>1$ hence a contradiction which implies $f$ is not continuous.

Comment: This property holds actually for more general topological vector spaces (not only linear functionals from a Banach space into the reals) but the proof is different.

  • $\begingroup$ In your comment, you said that this holds for more general topological vector space. I know that this holds for Frechet spaces for example. Do you if it does hold for F-spaces? $\endgroup$
    – Student
    Commented Apr 17, 2020 at 15:43

First question: this is the direct translation of $x_n\rightarrow x$ into the relation of $x_n$ to $x$ in norm

Second question: this is the negation of the statement that there exists a number $c\in (0,\infty)$ such that $|f(x)| \le c|x|$ for every $|x|$.

Third question: this is just a definition of $y_n$. The fact that $x\neq 0$ implies $|x|\neq 0$ so we may divide by this number. Also, it should be $\frac{1}{N}$ here. $|y|= 1/N$ (not: $1/n$) because it's $1/n$ times a vector of length $1$

fourth question a): this is $> 1$ because this is what was conlcuded earlier. Here you also have $1/N$ instead of $1/n$ throughout the entire line. It's not continuous because $y_n\rightarrow 0$ from the previous step but $f(y_n)$ has norm $>1$ hence does not converge to $f(0)=0$.


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