1. **Problem Statement:** We are given two norms on the space of continuous functions on $[a,b]$: $$\|f\|_1 = \sup_{x \in [a,b]} |f(x)|, \quad \|f\|_2 = \int_a^b |f(x)| \, dx$$ We need to determine if these two norms are equivalent, i.e., if there exist constants $C_1, C_2 > 0$ such that for all $f \in C[a,b]$: $$C_1 \|f\|_1 \leq \|f\|_2 \leq C_2 \|f\|_1$$ 2. **Recall the definition of norm equivalence:** Two norms $\|\cdot\|_A$ and $\|\cdot\|_B$ on a vector space are equivalent if there exist positive constants $C_1, C_2$ such that for all vectors $v$: $$C_1 \|v\|_A \leq \|v\|_B \leq C_2 \|v\|_A$$ 3. **Check the inequality $\|f\|_2 \leq C_2 \|f\|_1$:** Since $|f(x)| \leq \|f\|_1$ for all $x \in [a,b]$, we have $$\|f\|_2 = \int_a^b |f(x)| \, dx \leq \int_a^b \|f\|_1 \, dx = (b - a) \|f\|_1$$ So we can take $C_2 = b - a$. 4. **Check the inequality $C_1 \|f\|_1 \leq \|f\|_2$:** We want to see if there exists $C_1 > 0$ such that $$C_1 \|f\|_1 \leq \int_a^b |f(x)| \, dx$$ for all $f$. 5. **Counterexample to disprove equivalence:** Consider a sequence of functions $f_n$ defined on $[a,b]$ such that each $f_n$ is zero everywhere except on a very small interval of length $\frac{1}{n}$ where $f_n(x) = 1$. - Then $\|f_n\|_1 = 1$ for all $n$. - But $\|f_n\|_2 = \int_a^b |f_n(x)| \, dx = \frac{1}{n}$. As $n \to \infty$, $\|f_n\|_2 \to 0$ but $\|f_n\|_1 = 1$ remains constant. 6. **Conclusion:** No constant $C_1 > 0$ can satisfy $C_1 \cdot 1 \leq \frac{1}{n}$ for all $n$, so the norms are not equivalent. **Final answer:** The norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are **not equivalent** on $C[a,b]$.