# Show that there exists C>0C>0 such that ||f||1≤||f||2||f||_1 \leq ||f||_2 for ff in Pn[x]P_n[x]

Let Pn[x]P_n[x] be a vector space consisting of all polynomials on [0,1][0,1] with real coefficients and degree ≤n\leq n.

I am trying to show that there exists C>0C>0 such that Csupt∈[0,1]|f(t)|≤∫10|f(t)|dtC\sup_{t\in[0,1]} |f(t)| \leq \int_0^1|f(t)|dt for all ff in Pn[x]P_n[x].

My attempt at a solution: Note that supt∈[0,1]|f(t)|\sup_{t\in[0,1]} |f(t)| and ∫10|f(t)|dt\int_0^1|f(t)|dt are both norms on C([0,1],R)C([0,1],\mathbb{R}), the vector space on continuous real-valued function on [0,1][0,1]. Moreover, Pn[x]⊂C([0,1],R)P_n[x]\subset C([0,1],\mathbb{R}) and so the above two norms on C([0,1])C([0,1]) are also norms on Pn[x]P_n[x]. But Pn[x]P_n[x] has finite dimension and so all norms on Pn[x]P_n{[x]} are equivalent which gives use (more than) the desired result.

I would just like to check if this argument is correct and weather there is perhaps a more elegant proof that I am missing.

Thanks.

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Do you mean C=supt∈[0,1]|f(t)| C = \sup_{t\in[0,1]} |f(t)|\ or supt∈[0,1]|f(t)| ⩽1C∫10|f(t) dt ?\sup_{t\in[0,1]} |f(t)|\ \leqslant \frac1C \int_0^1 |f(t)\ \mathsf dt\ \mathrm ?
– Math1000
2 days ago

1

The second one is what I mean
– Dman
2 days ago

In that case, I suppose ‖⋅‖1\|\cdot\|_1 would be the \sup\sup norm and \|\cdot\|_2\|\cdot\|_2 the L^1L^1 norm?
– Math1000
2 days ago

Yes sorry didnt, I didnt use the same notarion
– Dman
2 days ago

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