Function continuous almost everywhere with $L^p(\mathbb{R})$ norm that exists for $p>1$.

Question

I have this little lemma to prove for a result concerning a linear operator I have.

If a function $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous everywhere except possibly at $x=0$. If also it is bounded on any interval $(-\infty,-a)$ or $(a,\infty)$ for $a>0$ and its $L^p(\mathbb{R})$ norm exists for all $p>1$. Then $f$ is bounded on $\mathbb{R}$ with possibly a finite jump-discontinuity at $x=0$.

My argument is a little heuristic and I am not sure how to prove this formally.

Answer 1

Counterexamples:

1. $f(x) = \dfrac{\ln |x|}{1+x^2}.$ This is unbounded in any neighborhood of $0.$

2. $f(x) = \dfrac{\sin(1/x)}{1+x^2}.$ This one is bounded, but the discontinuity at $0$ is not a jump discontinuity, no matter how $f(0)$ is defined.

Answer 2

In general it is false. Actually what you probably know is that a function $ f $ is in $ L^p $ for any $ p $ and $ \|f\|_p \leq C $ that does not depend on $ p $ then $ f \in L^{\infty} $. The issue in your Lemma is that $ \|f\|_p $ can explode as $ p $ goes to $ \infty $.

Let me show you a counterexample. Forgot for one second continuity which cannot help you. Consider the function

$$ f = \begin{cases} n \quad & \text{ in } \left[\frac{1}{2^{n}}, \frac{1}{2^{n+1}} \right] \\ 0 \quad & \text{ in } [-\infty,0]\cup [1, +\infty ] \end{cases}.$$

It holds

$$ \|f\|_{L^p}^p = \sum_{n} \frac{n^p}{2^n} < \infty. $$

I showed you an example of $ f $ satisfying the hypothesis of the lemma that goes to $ \infty $ as $ x $ tends to zero from the right.

To make an example of continuous function let notice that $ f/2 $ satisfy the same property. Moreover due to the fact that

$$ \frac{n+1}{2} < n $$

there exists a continuous function $ g $ such that $ f/2 < g < f $. The function $ g $ is then the counterexample for your lemma.